line equation problem solving

Word Problems on Linear Equations

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers. 

Solution: Then the other number = x + 9 Let the number be x.  Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9)  ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides)  ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers?  Solution:   Let the common ratio be x.  Let the common ratio be x.  Their difference = 48 According to the question,  7x - 3x = 48  ⇒ 4x = 48  ⇒ x = 48/4  ⇒ x = 12 Therefore, 7x = 7 × 12 = 84           3x = 3 × 12 = 36  Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle.  Solution: Let the breadth of the rectangle be x,  Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72  ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x                       = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages. 

Solution: Let Ron’s present age be x.  Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4.  According to the question;  Ron will be twice as old as Aaron.  Therefore, x + 4 = 2(x - 5 + 4)  ⇒ x + 4 = 2(x - 1)  ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts.  Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x  ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x  ⇒ x = 30/2  ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40  The two parts are 15 and 25. 

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages.  Solution: Let Robert’s age be x years.  Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question,  4x + 5 = 3(x + 5)  ⇒ 4x + 5 = 3x + 15  ⇒ 4x - 3x = 15 - 5  ⇒ x = 10 ⇒ 4x = 4 × 10 = 40  Robert’s present age is 10 years and that of his father’s age = 40 years.  

7. The sum of two consecutive multiples of 5 is 55. Find these multiples.  Solution: Let the first multiple of 5 be x.  Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2  ⇒ x = 25  Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.  

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles.  Solution: Let the angle be x.  Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51  Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair.  Solution: The table cost $ 40 more than the chair.  Let us assume the cost of the chair to be x.  Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x)  Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165. 

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?  Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2  According to the question,  3/5 ᵗʰ of the number is 4 more than 1/2 of the number.  ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.  

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

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Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

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Practice Test on Word Problems on Linear Equations

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Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

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Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

• Linear Equations
• Application of Linear Equations
• Two-Variable Linear Equations
• Linear Equations and Half Planes
• One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

• Substitution method
• Elimination method
• Cross multiplication method
• Graphical method

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Linear Equations

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A linear equation is an algebraic equation that forms a straight line when graphed. Each term is either a constant, or the product of a constant and a single variable.

A linear equation can have one or more dependent variables . For example, the following equation expresses the total cost of buying \(a\) apples at $0.50 each and \(b\) bananas for $0.30 each. \( \text{Cost } = a \times $0.50 + b \times $0.30. \) The total cost is dependent, and will vary if \(a\) or \(b\) is changed.

The following concepts must be understood to manipulate and use linear equations: linear rate of change, slope, points, and intercepts. Once these are mastered, they can be used in various ways to define a line when information is missing.

Linear Rate of Change

Finding the slope of a line, finding points on a line, finding intercepts of a line, linear equations word problems, linear equations - problem solving.

A linear rate of change means that the value being discussed changes by a fixed amount in a given unit of time. This means that the slope of the equation will be a constant value and that a graph of the relationship will be a straight line.

When relationships involve a linear rate of change, it is usually simplest to represent them as a linear equation in slope-intercept form :

\[ y = mx+ b. \]

In this formula, \( y \) represents our final amount, \( m \) is the rate of change or the slope of the line, \( x \) is the time elapsed (in the same units as \(m\)), and \( b \) is the amount we started with at time \( x=0\).

Tom makes \($9\) per hour. How much does Tom earn if he works \(20\) hours? Answer Solution 1: The amount of money Tom earns can be represented by the equation \[ y = 9x + 0.\] The slope is \( 9\) and the \(y\)-intercept is \(0 \) because Tom earns \($9\) per hour and starts out with \($0.\) Evaluating \( y \) when \( x = 20 \) yields \[ y = 20 \times 9 = 180. \] Thus, Tom earns \($180\) after working \(20\) hours. \(_\square\) Solution 2: Since Tom earns \($9\) per hour and worked \(20\) hours, we can see that his wages are \[ 20 \times $9 = $180.\ _\square\]

The library loses 100 books every week. If they currently have 10,000 books, how many weeks will it be before they have only 8400 books? Answer The number of books the library possesses can be expressed as the equation \[ y = -100x + 10000.\] To solve the problem, the value of \( x \) that corresponds to \( y = 8400 \) must be found: \[\begin{align} 8400 &= -100x + 10000 \\ -1600 &= -100x \\ x &= 16 . \end{align}\] Thus, the answer is \(16\) weeks. \(_\square\)

The slope-intercept formula of a line is given by \( y = mx + b, \) where \(m\) and \(b\) are fixed numbers. The set of points satisfying this equation describes a straight line in the \(xy\)-plane.

The slope of a line is the ratio between the change of \(y\) and the change of \(x\). For example, for two different points on a line, \( p_1= (x_1, y_1) \) and \( p_2 = (x_2, y_2) \), the slope is equal to the ratio \( \frac{ y_2 - y_1 } { x_2 - x_1} \). This is also known as the rate of change of \(y\) with respect to \(x\).

Slope is sometimes expressed as rise over run . Students who forget which component is the numerator might want to visualize determining slope as walking up a flight of stairs. The vertical change comes first (is the numerator), and is followed by the horizontal change (the denominator).

For the line \(y=mx+b\), the rate of change of \(y\) with respect to \(x\) is always equal to the slope \(m\) for any pair of distinct points on the line, which can be calculated as follows:

\[ \frac{ y_2 - y_1 } { x_2 - x_1 } = \frac{ (mx_2 + b) - (mx_1 + b) } { x_2 - x_1} = \frac{ m (x_2 - x_1) } { x_2 - x_1} = m. \]

Note: An exception to this rule occurs for lines of the form:

• \( x = k \) for some constant \(k\). In this case, the slope is undefined.
• \( y = k \) for some constant \(k\). In this case, the slope is \(0.\)

By solving the equation of the line when \(x=0\), we obtain \(y = m \cdot 0 + b,\) so \((0,b)\) is a point on the line. Since this point lies on the \(y\)-axis, it is the \(y\)-intercept of the line. This explains the name slope-intercept form for the line \( y = mx + b, \) where \(m\) is the slope and \(b\) is the \(y\)-intercept.

Find the slope and the \(y\)-intercept of the line \( x = 2 y + 6 .\) Answer At a quick glance, you might be tempted to say the slope of the line is \(2.\) However, the given equation is not in slope-intercept form. In order to convert the equation into slope-intercept form, make \(y\) the subject, which gives \( y = \frac{x}{2} - 3\). Hence, the slope is \( \frac{1}{2} \) and the \(y\)-intercept is \(-3\). \(_\square\)

Let \((a-3)x-y+b-5=0\) be the equation of a line that forms a \(45^\circ\) angle with the positive side of the \(x\)-axis. If the \(y\)-intercept is \(-2,\) what is \(a-b?\) Answer Rewriting the equation gives: \[\begin{align} (a-3)x-y+b-5&=0\\ y&=(a-3)x+(b-5)\\ a-3&=\tan 45^\circ, \quad b-5=-2\\ a&=4, \quad b=3\\ \Rightarrow a-b&=4-3\\ &=1. \ _\square \end{align}\]

If we want to find a point on the line, we simply substitute the coordinates into the equation, and solve the resulting linear equation. Typically, we are given either the \(x\)-coordinate, or the \(y\)-coordinate, and asked to find the other. We could also be given the coordinates of the actual point, and asked to determine the slope or \(y\)-intercept.

If the point \(P = (a, 5) \) lies on the line \( y = 3x + 2 \), what is the value of \(a\)? Answer Substituting the coordinates of the point into the line, \( 5 = 3 \times a + 2 \), which implies \( a = 1 \). \(_\square\)

If the point \( P = (1, 3) \) lies on the line \( y = mx - 4 \), what is the value of \(m\)? Answer Substituting the coordinates of the point into the line, \( 3 = m \times 1 - 4 \), which implies \( m = 7 \). \(_\square\)

The \(y\)-intercept is the point where the line crosses the \(y\)-axis. Since the \(y\)-axis corresponds to points which satisfy \( x = 0 \), the \(y\)-intercept is obtained by setting \( x = 0 \), which gives the standard form \( y = m \times 0 + b = b\). The \(y\)-intercept is \( (0, b ) \).

The \(x\)-intercept is the point where the line crosses the \(x\)-axis. Since the \(x\)-axis corresponds to the point that satisfies \( y= 0 \), the \(x\)-intercept is obtained by setting \( y = 0 \), which gives us \( 0 = m \times x + b \), or \( x = - \frac{ b}{m}\).

Note: A common mistake is to switch the definitions of the \(x\)- and \(y\)-intercepts. It can be helpful to draw some examples to ensure the definitions are clear.

What are the \(x\) and \(y\) intercepts of the line \( 2x + 3y = 12 \)? Answer To find the \(y\)-intercept, we set \( x = 0 \), which give us \( 2 \times 0 + 3 \times y = 12\), or \( y = 4\). Hence, the \(y\)-intercept is \( (0, 4) \). To find the \(x\)-intercept, set \( y = 0\), which gives \( 2 \times x + 3 \times 0 = 12 \), or \(x = 6 \). Hence, the \(x\)-intercept is \( (6,0) \). \(_\square\)

Sunni has \(300\) bricks on his wall on Monday and \(425\) bricks on his wall on Thursday. If he works at a constant rate every day, how many bricks would we expect him to have on Sunday? Answer If Sunni is working at a constant rate, the number of bricks will increase linearly. The difference between Thursday and Monday is 3 days, and the difference between \(425\) and \(300\) is \(125,\) making the rate of increase \( \dfrac{125 \text{ bricks}}{3 \text{ days}} \). After another three days, Sunni will have added an additional \(125\) bricks, which makes \( 425 + 125 = 550 \) bricks. Thus, the answer is \(550\) bricks. \(_\square\)

Henry has $125 in his pocket, but he is spending $20 per hour at the arcade. If he leaves in 2.5 hours, how much money will he have left? Answer Henry's money can be represented by \( y = -20x + 125 \), where \(y\) is the amount of money left and \(x\) is the time (in hours) spent in the arcade. The problem requires evaluating this expression when \( x = 2.5 \), which gives \[ y = -20 (2.5) + 125 = 125-50 = 75. \] Henry will have \($75\) left when he leaves. \(_\square\)

If a baseball and a bat together cost \($110\) and the bat costs \($100\) more than the ball, then how much does the ball cost?

What is the equation of the line that passes through \( (1,5) \) and has a \(y\)-intercept of \(6?\) Answer Since the slope through \( (1,5) \) and \( (0,6) \) is \( m = \dfrac{5-6}{1-0}=-1 \), and the \(y\)-intercept is 6, the equation of the line is \[ y = -x+6.\ _\square \]

Let \(\sqrt{3}x+ay+b=0\) be the equation of a line passing through the point \((1, 3)\) and forming a \(60^\circ\) angle with the positive side of the \(x\)-axis. What is \(a+b?\) Answer Rewriting the given equation, we have \[\begin{align} \sqrt{3}x+ay+b&=0\\ y&=-\frac{\sqrt{3}}{a}x-\frac{b}{a} \qquad (1) \\ -\frac{\sqrt{3}}{a}&=\tan 60^\circ\\ &=\sqrt{3}\\ \Rightarrow a&=-1. \end{align}\] Then \((1)\) becomes \(y=\sqrt{3}x+b.\) Now, substituting in the coordinates \((1, 3)\) gives \(3=\sqrt{3}+b \Rightarrow b=3-\sqrt{3}.\) Therefore, \[a+b=-1+(3-\sqrt{3})=2-\sqrt{3}.\ _\square \]

What is the length of the line segment that lies in the second quadrant of the \(xy\)-plane and is part of the line passing through the two points \((-3, -1)\) and \((4, 6)?\) Answer The equation of the line passing through the two points \((-3, -1)\) and \((4, 6)\) is \[\begin{align} y-(-1)&=\frac{6-(-1)}{4-(-3)}(x-(-3))\\ \Rightarrow y&=x+2. \end{align}\] Now, the \(x\)-intercept of the line is \((-2, 0).\) Similarly, the \(y\)-intercept of the line is \((0, 2).\) Therefore, the length of the line segment that lies in the second quadrant is the length between these two points, which is \[\sqrt{(0+2)^2+(2-0)^2}=2\sqrt{2}.\ _\square \]

Find all the values of \(k\) such that the three points below lie on a straight line: \[(2, 0), \quad (6, k-1), \quad (-3+k, 3).\] Answer Observe that the slope of the line passing through \((2, 0)\) and \((6, k-1)\) is the same as the slope of the line passing through \((6, k-1)\) and \((-3+k, 3).\) \[\begin{align} \frac{(k-1)-0}{6-2}&=\frac{3-(k-1)}{(-3+k)-6}\\ (k-1)(k-9)&=4(-k+4)\\ k^2-6k-7&=0\\ (k+1)(k-7)&=0. \end{align}\] Then the values of \(k\) such that \((2, 0), (6, k-1)\) and \((-3+k, 3)\) lie on a straight line are \(k=-1\) and \(k=7.\) \(_\square\)

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Linear Equations

A linear equation is an equation for a straight line

These are all linear equations:

Let us look more closely at one example:

Example: y = 2x + 1 is a linear equation:

The graph of y = 2x+1 is a straight line

• When x increases, y increases twice as fast , so we need 2x
• When x is 0, y is already 1. So +1 is also needed
• And so: y = 2x + 1

Here are some example values:

Check for yourself that those points are part of the line above!

Different Forms

There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:

• Exponents (like the 2 in x 2 )
• Square roots , cube roots , etc

Examples: These are NOT linear equations:

Slope-Intercept Form

The most common form is the slope-intercept equation of a straight line :

Example: y = 2x + 1

• Slope: m = 2
• Intercept: b = 1

Point-Slope Form

Another common one is the Point-Slope Form of the equation of a straight line:

Example: y − 3 = (¼)(x − 2)

It is in the form y − y 1 = m(x − x 1 ) where:

General Form

And there is also the General Form of the equation of a straight line:

Example: 3x + 2y − 4 = 0

It is in the form Ax + By + C = 0 where:

There are other, less common forms as well.

As a Function

Sometimes a linear equation is written as a function , with f(x) instead of y :

And functions are not always written using f(x):

The Identity Function

There is a special linear function called the "Identity Function":

And here is its graph:

It is called "Identity" because what comes out is identical to what goes in:

Constant Functions

Another special type of linear function is the Constant Function ... it is a horizontal line:

No matter what value of "x", f(x) is always equal to some constant value.

Using Linear Equations

You may like to read some of the things you can do with lines:

• Finding the Midpoint of a Line Segment
• Finding Parallel and Perpendicular Lines
• Finding the Equation of a Line from 2 Points

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• \frac{3}{4}x+\frac{5}{6}=5x-\frac{125}{3}
• \sqrt{2}x-\sqrt{3}=\sqrt{5}
• 7y+5-3y+1=2y+2
• \frac{x}{3}+\frac{x}{2}=10
• What is a linear equation?
• A linear equation represents a straight line on a coordinate plane. It can be written in the form: y = mx + b where m is the slope of the line and b is the y-intercept.
• How do you find the linear equation?
• To find the linear equation you need to know the slope and the y-intercept of the line. To find the slope use the formula m = (y2 - y1) / (x2 - x1) where (x1, y1) and (x2, y2) are two points on the line. The y-intercept is the point at which x=0.
• What are the 4 methods of solving linear equations?
• There are four common methods to solve a system of linear equations: Graphing, Substitution, Elimination and Matrix.
• How do you identify a linear equation?
• Here are a few ways to identify a linear equation: Look at the degree of the equation, a linear equation is a first-degree equation. Check if the equation has two variables. Graph the equation.
• What is the most basic linear equation?
• The most basic linear equation is a first-degree equation with one variable, usually written in the form of y = mx + b, where m is the slope of the line and b is the y-intercept.

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Linear Equations Questions

The linear equations questions and answers will assist students to understand the concepts better. Linear Equation is a topic that is covered in basically every class. The NCERT guidelines will be followed for preparing the questions. Linear Equations are used in mathematics as well as in everyday life. So, the basics of this concept must be grasped. For students of all levels, the problems here will include both the basics and more challenging problems. As a result, students will be able to use it to solve problems involving linear equations. Learn more about Linear Equations by clicking here .

Here, we’ll go through a variety of linear equations problems with solutions, based on various concepts.

Linear Equations Questions with Solutions

1. Write the statement as an equation: A number increased by 8 equals 15.

Given statement: A number increased by 8 equals 15.

Let the number be “x”.

So, x is increased by 8 means x + 8.

Hence, x increased by 8 equals 15 means x + 8 = 15, which is the equation for the given statement.

2. Write the statement for the given equation: 2x = 18.

Given equation: 2x = 18.

The statement for the given equation is “Twice the number x equals 18”.

3. Solve the equation: x + 3 = -2

Given equation: x + 3 = -2.

Now, keep the variables on one side and constants on the other side. Hence, the equation becomes,

Hence, the value of x is -5.

4. Verify that x = 4 is the root of the equation 3x/2 = 6.

To verify whether the given root is the solution of the given equation, substitute x = 4 in the equation 3x/2 = 6.

⇒ (3(4))/2 = 6

⇒ (12/2) = 6

Hence, x = 4 is the root of the equation 3x/2 = 6.

5. If 5 is added to twice a number, the result is 29. Determine the number.

The equation for the given statement is 5+2x = 29.

To find the number “x”, we have to solve the equation.

⇒ 2x = 29 – 5

Hence, the required number is 12.

6. If x = 2, then 2x – 5 = 7. Check whether the statement is true or false.

Given equation: 2x – 5 = 7

= 2(2) – 5

= 4 – 5 = -1

Hence, the given statement is false.

7. The sum of two consecutive numbers is 11. Find the numbers.

Let the number be x.

Hence, the two consecutive numbers are x and x+1.

According to the given statement, the equation becomes

⇒ x + x + 1 = 11

⇒ 2x + 1 = 11

⇒ x = 10/2 = 5

If x = 5, then x + 1 = 5 + 1 = 6

Hence, the two numbers are 5 and 6.

8. Express the equation x = 3y in the form of ax+by+c = 0 and find the values of a, b and c.

Given equation: x = 3y

We know that the standard form of linear equation in two variables is ax+by+c = 0 …(1)

Now, rearranging the given equation, we get

⇒ x – 3y = 0

This can be written as

⇒ 1(x) + (-3)y + (0)c = 0 …(2)

On comparing equation (1) and (2), we get

⇒ a = 1, b = -3 and c = 0.

9. Find three solutions for the equation 2x + y = 7.

To find the solutions for the equation 2x + y = 7, substitute different values for x.

When x = 0,

⇒ 2(0) + y = 7

Therefore, the solution is (0, 7).

When x = 1,

⇒ 2(1) + y = 7

⇒ y = 7 – 2

Hence, the solution is (1, 5).

When x = 2,

⇒ 2(2) + y = 7

⇒ 4 + y = 7

Hence, the solution is (2, 3).

Therefore, the three solutions are (0, 7), (1, 5) and (2, 3).

10. Solve the following equations using the substitution method:

3x + 4y = 10 and 2x – 2y = 2

3x + 4y = 10 …(1)

2x – 2y = 2 …(2)

Equation (2) can be written as:

2(x – y) = 2

x – y = 1

x = 1+y …(3)

Now, substitute (3) in (1), we get

3 (1+y) + 4y = 10

3 + 3y + 4y = 10

7y = 10 – 3

Hence, y = 1.

Now, substitute y = 1 in (3), we get

Hence, x = 2 and y = 1 are the solutions of the given equations.

Practice Questions

• Write the statement as an equation: Twice a number subtracted from 19 is 11.
• The sum of the two numbers is 30 and their ratio is 2: 3. Find the numbers.
• If the point (3, 4) lies on the graph of equation 3y = ax + 7, determine the value of a.
• Solve the equations using the elimination method: (x/2)+(2y/3) = -1 and x – (y/3) = 3.

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2.2 Linear Equations in One Variable

Learning objectives.

In this section, you will:

• Solve equations in one variable algebraically.
• Solve a rational equation.
• Find a linear equation.
• Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
• Write the equation of a line parallel or perpendicular to a given line.

Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $15.00/hr, and she opened a savings account with an initial deposit of $400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure 1 .

Solving Linear Equations in One Variable

A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form a x + b = 0 a x + b = 0 and are solved using basic algebraic operations.

We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation.

The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for x x will make the equation true.

A conditional equation is true for only some values of the variable. For example, if we are to solve the equation 5 x + 2 = 3 x − 6 , 5 x + 2 = 3 x − 6 , we have the following:

The solution set consists of one number: { − 4 } . { − 4 } . It is the only solution and, therefore, we have solved a conditional equation.

An inconsistent equation results in a false statement. For example, if we are to solve 5 x − 15 = 5 ( x − 4 ) , 5 x − 15 = 5 ( x − 4 ) , we have the following:

Indeed, −15 ≠ −20. −15 ≠ −20. There is no solution because this is an inconsistent equation.

Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.

Linear Equation in One Variable

A linear equation in one variable can be written in the form

where a and b are real numbers, a ≠ 0. a ≠ 0.

Given a linear equation in one variable, use algebra to solve it.

The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x = _________, x = _________, if x is the unknown. There is no set order, as the steps used depend on what is given:

• We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero.
• Apply the distributive property as needed: a ( b + c ) = a b + a c . a ( b + c ) = a b + a c .
• Isolate the variable on one side of the equation.
• When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.

Solving an Equation in One Variable

Solve the following equation: 2 x + 7 = 19. 2 x + 7 = 19.

This equation can be written in the form a x + b = 0 a x + b = 0 by subtracting 19 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.

The solution is 6.

Solve the linear equation in one variable: 2 x + 1 = −9. 2 x + 1 = −9.

Solving an Equation Algebraically When the Variable Appears on Both Sides

Solve the following equation: 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) . 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) .

Apply standard algebraic properties.

This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, x = − 5 3 . x = − 5 3 .

Solve the equation in one variable: −2 ( 3 x − 1 ) + x = 14 − x . −2 ( 3 x − 1 ) + x = 14 − x .

Solving a Rational Equation

In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation .

Recall that a rational number is the ratio of two numbers, such as 2 3 2 3 or 7 2 . 7 2 . A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.

Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).

Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.

Solve the rational equation: 7 2 x − 5 3 x = 22 3 . 7 2 x − 5 3 x = 22 3 .

We have three denominators; 2 x , 3 x , 2 x , 3 x , and 3. The LCD must contain 2 x , 3 x , 2 x , 3 x , and 3. An LCD of 6 x 6 x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6 x . 6 x .

A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as ( x + 1 ) . ( x + 1 ) . Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x , x , x − 1 , x − 1 , and 3 x − 3. 3 x − 3. First, factor all denominators. We then have x , x , ( x − 1 ) , ( x − 1 ) , and 3 ( x − 1 ) 3 ( x − 1 ) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of ( x − 1 ) . ( x − 1 ) . The x x in the first denominator is separate from the x x in the ( x − 1 ) ( x − 1 ) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x , x , one factor of ( x − 1 ) , ( x − 1 ) , and the 3. Thus, the LCD is the following:

So, both sides of the equation would be multiplied by 3 x ( x − 1 ) . 3 x ( x − 1 ) . Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.

Another example is a problem with two denominators, such as x x and x 2 + 2 x . x 2 + 2 x . Once the second denominator is factored as x 2 + 2 x = x ( x + 2 ) , x 2 + 2 x = x ( x + 2 ) , there is a common factor of x in both denominators and the LCD is x ( x + 2 ) . x ( x + 2 ) .

Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.

We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.

Multiply a ( d ) a ( d ) and b ( c ) , b ( c ) , which results in a d = b c . a d = b c .

Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.

Rational Equations

A rational equation contains at least one rational expression where the variable appears in at least one of the denominators.

Given a rational equation, solve it.

• Factor all denominators in the equation.
• Find and exclude values that set each denominator equal to zero.
• Find the LCD.
• Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.
• Solve the remaining equation.
• Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator.

Solving a Rational Equation without Factoring

Solve the following rational equation:

We have three denominators: x , x , 2 , 2 , and 2 x . 2 x . No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is 2 x . 2 x . Only one value is excluded from a solution set, 0. Next, multiply the whole equation (both sides of the equal sign) by 2 x . 2 x .

The proposed solution is −1, which is not an excluded value, so the solution set contains one number, −1 , −1 , or { −1 } { −1 } written in set notation.

Solve the rational equation: 2 3 x = 1 4 − 1 6 x . 2 3 x = 1 4 − 1 6 x .

Solving a Rational Equation by Factoring the Denominator

Solve the following rational equation: 1 x = 1 10 − 3 4 x . 1 x = 1 10 − 3 4 x .

First find the common denominator. The three denominators in factored form are x , 10 = 2 ⋅ 5 , x , 10 = 2 ⋅ 5 , and 4 x = 2 ⋅ 2 ⋅ x . 4 x = 2 ⋅ 2 ⋅ x . The smallest expression that is divisible by each one of the denominators is 20 x . 20 x . Only x = 0 x = 0 is an excluded value. Multiply the whole equation by 20 x . 20 x .

The solution is 35 2 . 35 2 .

Solve the rational equation: − 5 2 x + 3 4 x = − 7 4 . − 5 2 x + 3 4 x = − 7 4 .

Solving Rational Equations with a Binomial in the Denominator

Solve the following rational equations and state the excluded values:

• ⓐ 3 x − 6 = 5 x 3 x − 6 = 5 x
• ⓑ x x − 3 = 5 x − 3 − 1 2 x x − 3 = 5 x − 3 − 1 2
• ⓒ x x − 2 = 5 x − 2 − 1 2 x x − 2 = 5 x − 2 − 1 2

The denominators x x and x − 6 x − 6 have nothing in common. Therefore, the LCD is the product x ( x − 6 ) . x ( x − 6 ) . However, for this problem, we can cross-multiply.

The solution is 15. The excluded values are 6 6 and 0. 0.

The LCD is 2 ( x − 3 ) . 2 ( x − 3 ) . Multiply both sides of the equation by 2 ( x − 3 ) . 2 ( x − 3 ) .

The solution is 13 3 . 13 3 . The excluded value is 3. 3.

The least common denominator is 2 ( x − 2 ) . 2 ( x − 2 ) . Multiply both sides of the equation by x ( x − 2 ) . x ( x − 2 ) .

The solution is 4. The excluded value is 2. 2.

Solve − 3 2 x + 1 = 4 3 x + 1 . − 3 2 x + 1 = 4 3 x + 1 . State the excluded values.

Solving a Rational Equation with Factored Denominators and Stating Excluded Values

Solve the rational equation after factoring the denominators: 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . State the excluded values.

We must factor the denominator x 2 −1. x 2 −1. We recognize this as the difference of squares, and factor it as ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Thus, the LCD that contains each denominator is ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.

The solution is −3. −3. The excluded values are 1 1 and −1. −1.

Solve the rational equation: 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 . 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 .

Finding a Linear Equation

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y = m x + b , y = m x + b , where m = slope m = slope and b = y -intercept . b = y -intercept . Let us begin with the slope.

The Slope of a Line

The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.

If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure 2 . The lines indicate the following slopes: m = −3 , m = −3 , m = 2 , m = 2 , and m = 1 3 . m = 1 3 .

The slope of a line, m , represents the change in y over the change in x. Given two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) , ( x 2 , y 2 ) , the following formula determines the slope of a line containing these points:

Finding the Slope of a Line Given Two Points

Find the slope of a line that passes through the points ( 2 , −1 ) ( 2 , −1 ) and ( −5 , 3 ) . ( −5 , 3 ) .

We substitute the y- values and the x- values into the formula.

The slope is − 4 7 . − 4 7 .

It does not matter which point is called ( x 1 , y 1 ) ( x 1 , y 1 ) or ( x 2 , y 2 ) . ( x 2 , y 2 ) . As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result.

Find the slope of the line that passes through the points ( −2 , 6 ) ( −2 , 6 ) and ( 1 , 4 ) . ( 1 , 4 ) .

Identifying the Slope and y- intercept of a Line Given an Equation

Identify the slope and y- intercept, given the equation y = − 3 4 x − 4. y = − 3 4 x − 4.

As the line is in y = m x + b y = m x + b form, the given line has a slope of m = − 3 4 . m = − 3 4 . The y- intercept is b = −4. b = −4.

The y -intercept is the point at which the line crosses the y- axis. On the y- axis, x = 0. x = 0. We can always identify the y- intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute x = 0 x = 0 and solve for y.

The Point-Slope Formula

Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.

This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.

Given one point and the slope, the point-slope formula will lead to the equation of a line:

Finding the Equation of a Line Given the Slope and One Point

Write the equation of the line with slope m = −3 m = −3 and passing through the point ( 4 , 8 ) . ( 4 , 8 ) . Write the final equation in slope-intercept form.

Using the point-slope formula, substitute −3 −3 for m and the point ( 4 , 8 ) ( 4 , 8 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .

Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.

Given m = 4 , m = 4 , find the equation of the line in slope-intercept form passing through the point ( 2 , 5 ) . ( 2 , 5 ) .

Finding the Equation of a Line Passing Through Two Given Points

Find the equation of the line passing through the points ( 3 , 4 ) ( 3 , 4 ) and ( 0 , −3 ) . ( 0 , −3 ) . Write the final equation in slope-intercept form.

First, we calculate the slope using the slope formula and two points.

Next, we use the point-slope formula with the slope of 7 3 , 7 3 , and either point. Let’s pick the point ( 3 , 4 ) ( 3 , 4 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .

In slope-intercept form, the equation is written as y = 7 3 x − 3. y = 7 3 x − 3.

To prove that either point can be used, let us use the second point ( 0 , −3 ) ( 0 , −3 ) and see if we get the same equation.

We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.

Standard Form of a Line

Another way that we can represent the equation of a line is in standard form . Standard form is given as

where A , A , B , B , and C C are integers. The x- and y- terms are on one side of the equal sign and the constant term is on the other side.

Finding the Equation of a Line and Writing It in Standard Form

Find the equation of the line with m = −6 m = −6 and passing through the point ( 1 4 , −2 ) . ( 1 4 , −2 ) . Write the equation in standard form.

We begin using the point-slope formula.

From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.

This equation is now written in standard form.

Find the equation of the line in standard form with slope m = − 1 3 m = − 1 3 and passing through the point ( 1 , 1 3 ) . ( 1 , 1 3 ) .

Vertical and Horizontal Lines

The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as

where c is a constant. The slope of a vertical line is undefined, and regardless of the y- value of any point on the line, the x- coordinate of the point will be c .

Suppose that we want to find the equation of a line containing the following points: ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , and ( −3 , 5 ) . ( −3 , 5 ) . First, we will find the slope.

Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x- coordinates are the same and we find a vertical line through x = −3. x = −3. See Figure 3 .

The equation of a horizontal line is given as

where c is a constant. The slope of a horizontal line is zero, and for any x- value of a point on the line, the y- coordinate will be c .

Suppose we want to find the equation of a line that contains the following set of points: ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , and ( 5 , −2 ) . ( 5 , −2 ) . We can use the point-slope formula. First, we find the slope using any two points on the line.

Use any point for ( x 1 , y 1 ) ( x 1 , y 1 ) in the formula, or use the y -intercept.

The graph is a horizontal line through y = −2. y = −2. Notice that all of the y- coordinates are the same. See Figure 3 .

Finding the Equation of a Line Passing Through the Given Points

Find the equation of the line passing through the given points: ( 1 , −3 ) ( 1 , −3 ) and ( 1 , 4 ) . ( 1 , 4 ) .

The x- coordinate of both points is 1. Therefore, we have a vertical line, x = 1. x = 1.

Find the equation of the line passing through ( −5 , 2 ) ( −5 , 2 ) and ( 2 , 2 ) . ( 2 , 2 ) .

Determining Whether Graphs of Lines are Parallel or Perpendicular

Parallel lines have the same slope and different y- intercepts. Lines that are parallel to each other will never intersect. For example, Figure 4 shows the graphs of various lines with the same slope, m = 2. m = 2.

All of the lines shown in the graph are parallel because they have the same slope and different y- intercepts.

Lines that are perpendicular intersect to form a 90° 90° -angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is −1 : m 1 ⋅ m 2 = −1. −1 : m 1 ⋅ m 2 = −1. For example, Figure 5 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of − 1 3 . − 1 3 .

Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither

Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: 3 y = − 4 x + 3 3 y = − 4 x + 3 and 3 x − 4 y = 8. 3 x − 4 y = 8.

The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.

First equation:

Second equation:

See the graph of both lines in Figure 6

From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.

The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.

Graph the two lines and determine whether they are parallel, perpendicular, or neither: 2 y − x = 10 2 y − x = 10 and 2 y = x + 4. 2 y = x + 4.

Writing the Equations of Lines Parallel or Perpendicular to a Given Line

As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.

Given an equation for a line, write the equation of a line parallel or perpendicular to it.

• Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.
• Use the slope and the given point with the point-slope formula.
• Simplify the line to slope-intercept form and compare the equation to the given line.

Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point

Write the equation of line parallel to a 5 x + 3 y = 1 5 x + 3 y = 1 and passing through the point ( 3 , 5 ) . ( 3 , 5 ) .

First, we will write the equation in slope-intercept form to find the slope.

The slope is m = − 5 3 . m = − 5 3 . The y- intercept is 1 3 , 1 3 , but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y- intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.

The equation of the line is y = − 5 3 x + 10. y = − 5 3 x + 10. See Figure 7 .

Find the equation of the line parallel to 5 x = 7 + y 5 x = 7 + y and passing through the point ( −1 , −2 ) . ( −1 , −2 ) .

Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Find the equation of the line perpendicular to 5 x − 3 y + 4 = 0 5 x − 3 y + 4 = 0 and passing through the point ( − 4 , 1 ) . ( − 4 , 1 ) .

The first step is to write the equation in slope-intercept form.

We see that the slope is m = 5 3 . m = 5 3 . This means that the slope of the line perpendicular to the given line is the negative reciprocal, or − 3 5 . − 3 5 . Next, we use the point-slope formula with this new slope and the given point.

Access these online resources for additional instruction and practice with linear equations.

• Solving rational equations
• Equation of a line given two points
• Finding the equation of a line perpendicular to another line through a given point
• Finding the equation of a line parallel to another line through a given point

2.2 Section Exercises

What does it mean when we say that two lines are parallel?

What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?

How do we recognize when an equation, for example y = 4 x + 3 , y = 4 x + 3 , will be a straight line (linear) when graphed?

What does it mean when we say that a linear equation is inconsistent?

When solving the following equation:

2 x − 5 = 4 x + 1 2 x − 5 = 4 x + 1

explain why we must exclude x = 5 x = 5 and x = −1 x = −1 as possible solutions from the solution set.

For the following exercises, solve the equation for x . x .

7 x + 2 = 3 x − 9 7 x + 2 = 3 x − 9

4 x − 3 = 5 4 x − 3 = 5

3 ( x + 2 ) − 12 = 5 ( x + 1 ) 3 ( x + 2 ) − 12 = 5 ( x + 1 )

12 − 5 ( x + 3 ) = 2 x − 5 12 − 5 ( x + 3 ) = 2 x − 5

1 2 − 1 3 x = 4 3 1 2 − 1 3 x = 4 3

x 3 − 3 4 = 2 x + 3 12 x 3 − 3 4 = 2 x + 3 12

2 3 x + 1 2 = 31 6 2 3 x + 1 2 = 31 6

3 ( 2 x − 1 ) + x = 5 x + 3 3 ( 2 x − 1 ) + x = 5 x + 3

2 x 3 − 3 4 = x 6 + 21 4 2 x 3 − 3 4 = x 6 + 21 4

x + 2 4 − x − 1 3 = 2 x + 2 4 − x − 1 3 = 2

For the following exercises, solve each rational equation for x . x . State all x -values that are excluded from the solution set.

3 x − 1 3 = 1 6 3 x − 1 3 = 1 6

2 − 3 x + 4 = x + 2 x + 4 2 − 3 x + 4 = x + 2 x + 4

3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 ) 3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 )

3 x x − 1 + 2 = 3 x − 1 3 x x − 1 + 2 = 3 x − 1

5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3 5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3

1 x = 1 5 + 3 2 x 1 x = 1 5 + 3 2 x

For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form.

( 0 , 3 ) ( 0 , 3 ) with a slope of 2 3 2 3

( 1 , 2 ) ( 1 , 2 ) with a slope of − 4 5 − 4 5

x -intercept is 1, and ( −2 , 6 ) ( −2 , 6 )

y -intercept is 2, and ( 4 , −1 ) ( 4 , −1 )

( −3 , 10 ) ( −3 , 10 ) and ( 5 , −6 ) ( 5 , −6 )

( 1 , 3 )  and   ( 5 , 5 ) ( 1 , 3 )  and   ( 5 , 5 )

parallel to y = 2 x + 5 y = 2 x + 5 and passes through the point ( 4 , 3 ) ( 4 , 3 )

perpendicular to 3 y = x − 4 3 y = x − 4 and passes through the point ( −2 , 1 ) ( −2 , 1 ) .

For the following exercises, find the equation of the line using the given information.

( − 2 , 0 ) ( − 2 , 0 ) and ( −2 , 5 ) ( −2 , 5 )

( 1 , 7 ) ( 1 , 7 ) and ( 3 , 7 ) ( 3 , 7 )

The slope is undefined and it passes through the point ( 2 , 3 ) . ( 2 , 3 ) .

The slope equals zero and it passes through the point ( 1 , −4 ) . ( 1 , −4 ) .

The slope is 3 4 3 4 and it passes through the point ( 1 , 4 ) ( 1 , 4 ) .

( –1 , 3 ) ( –1 , 3 ) and ( 4 , –5 ) ( 4 , –5 )

For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.

y = 2 x + 7 y = − 1 2 x − 4 y = 2 x + 7 y = − 1 2 x − 4

3 x − 2 y = 5 6 y − 9 x = 6 3 x − 2 y = 5 6 y − 9 x = 6

y = 3 x + 1 4 y = 3 x + 2 y = 3 x + 1 4 y = 3 x + 2

x = 4 y = −3 x = 4 y = −3

For the following exercises, find the slope of the line that passes through the given points.

( 5 , 4 ) ( 5 , 4 ) and ( 7 , 9 ) ( 7 , 9 )

( −3 , 2 ) ( −3 , 2 ) and ( 4 , −7 ) ( 4 , −7 )

( −5 , 4 ) ( −5 , 4 ) and ( 2 , 4 ) ( 2 , 4 )

( −1 , −2 ) ( −1 , −2 ) and ( 3 , 4 ) ( 3 , 4 )

( 3 , −2 ) ( 3 , −2 ) and ( 3 , −2 ) ( 3 , −2 )

For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.

( −1 , 3 )  and   ( 5 , 1 ) ( −2 , 3 )  and   ( 0 , 9 ) ( −1 , 3 )  and   ( 5 , 1 ) ( −2 , 3 )  and   ( 0 , 9 )

( 2 , 5 )  and   ( 5 , 9 ) ( −1 , −1 )  and   ( 2 , 3 ) ( 2 , 5 )  and   ( 5 , 9 ) ( −1 , −1 )  and   ( 2 , 3 )

For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y -intercept occurs. State your ymin and ymax values.

0.537 x − 2.19 y = 100 0.537 x − 2.19 y = 100

4,500 x − 200 y = 9,528 4,500 x − 200 y = 9,528

200 − 30 y x = 70 200 − 30 y x = 70

Starting with the point-slope formula y − y 1 = m ( x − x 1 ) , y − y 1 = m ( x − x 1 ) , solve this expression for x x in terms of x 1 , y , y 1 , x 1 , y , y 1 , and m m .

Starting with the standard form of an equation A x + B y = C A x + B y = C solve this expression for y y in terms of A , B , C A , B , C and x x . Then put the expression in slope-intercept form.

Use the above derived formula to put the following standard equation in slope intercept form: 7 x − 5 y = 25. 7 x − 5 y = 25.

Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular.

( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) ( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) and ( 0 , 4 ) ( 0 , 4 )

Find the slopes of the diagonals in the previous exercise. Are they perpendicular?

Real-World Applications

The slope for a wheelchair ramp for a home has to be 1 12 . 1 12 . If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope.

If the profit equation for a small business selling x x number of item one and y y number of item two is p = 3 x + 4 y , p = 3 x + 4 y , find the y y value when p = $ 453 and   x = 75. p = $ 453 and   x = 75.

For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be y = 45 + .25 x , y = 45 + .25 x , where x x is the number of miles traveled.

What is your cost if you travel 50 mi?

If your cost were $ 63.75 , $ 63.75 , how many miles were you charged for traveling?

Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel?

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Linear Equations in the real world

A page on how to find the equation and how to graph real world applications of linear equations.

• Interpreting Graphs of Real World Linear Equations
• Worksheet on real world linear equations (worksheet version of this web page)
• slope of a line
• y-intercept
• interactive linear equation
• equation given slope and a point
• equation from graph of a line
• real world applications

A cab company charges a $3 boarding rate in addition to its meter which is $2 for every mile. What is the equation of the line that represents this cab company's rate?

A cab company charges a $5 boarding rate in addition to its meter which is $3 for every mile. What is the equation of the line that represents this cab company's rate?

A cab company charges a $3 boarding rate in addition to its meter which is $½ for every mile. What is the equation of the line that represents this cab company's rate?

A cab company charges a $4 boarding rate in addition to its meter which is $ ¾ for every mile. What is the equation of the line that represents this cab company's rate?

A cab company does not charge a boarding fee but then has a meter of $4 an hour. What equation represents this cab company's rate?

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Solve Systems of Equations Using Matrices

A system of linear equations is a set of equations where each equation is linear in nature, meaning it forms a straight line when graphed. Solving such systems is crucial in various fields, including engineering, physics, economics, and computer science. Matrices provide an efficient way to handle and solve these systems, especially when dealing with large sets of equations.

What Are Systems of Linear Equations?

A system of linear equations consists of two or more linear equations involving the same set of variables.

The solution to a system of linear equations is the set of values for the variables that satisfy all equations simultaneously.

Example of Systems of Linear Equations

Two Equations with Two Variables

Three Equations with Three Variables

x + y + z = 6

• 2x − y + 3z = 14
• 4x + 3y + 2z = 24

What Are Matrices?

Matrices are rectangular arrays of numbers, symbols, or expressions arranged in rows and columns. In the context of solving linear equations, matrices are used to represent the coefficients of the equations and manipulate them to find the solutions.

Example of Systems of Linear Equations as Matrices

Matrix Representation of a System of Equations for the system

The matrix form is [Tex]\begin{pmatrix} 1 & 3 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}[/Tex]

Augmented Matrix, The augmented matrix combines the coefficient matrix and the constants i.e.,

[Tex]\left[ \begin{array}{cc|c} 1 & 3 & 5 \\ 2 & -1 & 4 \end{array} \right] [/Tex]

How to Solve Problems Step-by-Step Solution

To solve a system of linear equations using matrices, follow these steps

Step 1. Form the Augmented Matrix: Write the system of equations as an augmented matrix. Step 2. Perform Row Operations: Use row operations to simplify the matrix to row echelon form or reduced row echelon form. Step 3. Back-Substitution: If using row echelon form, solve for the variables starting from the last row and moving upwards. Step 4. Check Your Solution: Substitute the values of the variables back into the original equations to verify the solution.

Example: Solve the system using Gaussian elimination

Step 1: Form the augmented matrix.

[Tex]\left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & -1 & 3 & 14 \\ 4 & 3 & 2 & 24 \end{array} \right] [/Tex]

Step 2: Perform row operations to simplify.

• R 2 = [2, -1, 3, 14] – 2 × [1, 1, 1, 6] = [2 – 2, -1 – 2, 3 – 2, 14 – 12] = [0, -3, 1, 2]
• R 3 = [4, 3, 2, 24] – 4 × [1, 1, 1, 6] = [4 – 4, 3 – 4, 2 – 4, 24 – 24] = [0, -1, -2, 0]
• Further simplify to row echelon form.

Step 3: Back-substitute to solve for x, y, and z.

From row 3:

-1y – 2z = 0

y + 2z = 0 (multiply by -1 to simplify)

Now, Solve for y in terms of z

From row 2:

-3y + z = 2

Substitute y = -2z into this equation:

-3(-2z) + z = 2

Now, Solve for y

Now that we have z = 2/7, substitute this into y = -2z:

y = -2 × 2/7 = -4/7

Now, Solve for x

From row 1:

Substitute y = – 4/7and z = 2/7 into this equation:

x – 4/7 + 2/7 = 6

x – 2/7 = 6

7x – 2 = 42

Step 4: Verify the solution by substituting back into the original equations.

Thus, the solution to the system is:

x = 44/7, y = -4/7, z = 2/7

Solved Problems on Systems of Equations Using Matrices

Problem 1: Solve the system of equations using matrices.

• 2x − 3y = -4

Form the Augmented Matrix: [Tex]\left[ \begin{array}{cc|c} 1 & 1 & 5 \\ 2 & -3 & 4 \end{array} \right] [/Tex] Row Operations: [Tex]Subtract 2×R 1 ​  from R 2 ​ : \left[ \begin{array}{cc|c} 1 & 1 & 5 \\ 0 & -5 & -14 \end{array} \right][/Tex] Back-Substitute: From R 2 : [Tex]? = 14/ 5[/Tex] Substitute into R 1 : [Tex]x+5/14=5[/Tex] , solve for x. Solution: x = 5/11 ,y = 5/14

Problem 2: Solve using the inverse matrix method.

• 3x + 2y = 6

Matrix Form: [Tex]A = \begin{bmatrix} 3 & 2 \\ 1 & -1 \end{bmatrix} B = \begin{bmatrix} x \\ y \end{bmatrix} C = \begin{bmatrix} 6 \\ 4 \end{bmatrix}[/Tex] AB = C Find Inverse of A: [Tex]A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} -1 & -2 \\ -1 & 3 \end{bmatrix} =\begin{bmatrix} \frac{1}{-5} \cdot (-1) & \frac{1}{-5} \cdot (-2) \\ \frac{1}{-5} \cdot (-1) & \frac{1}{-5} \cdot 3 \end{bmatrix} = \begin{bmatrix} 0.2 & 0.4 \\ 0.2 & -0.6 \end{bmatrix}[/Tex] Compute B = A − 1 C: [Tex]B = \begin{bmatrix} 0.2 & 0.4 \\ 0.2 & -0.6 \end{bmatrix} \begin{bmatrix} 6 \\ 4 \end{bmatrix} \begin{bmatrix} 0.2 \cdot 6 + 0.4 \cdot 4 \\ 0.2 \cdot 6 – 0.6 \cdot 4 \end{bmatrix} = \begin{bmatrix} 2.8 \\ -1.2 \end{bmatrix}[/Tex] Answer: x = 2.8, y = -1.2

Problem 3: Solve the system using Gaussian elimination.

• 2x + 5y = 14

Form the Augmented Matrix: [Tex]\left[ \begin{array}{cc|c} 1 & 3 & 7 \\ 2 & 5 & 14 \end{array} \right] [/Tex] Row Operations: Subtract 2×R 1 from R 2 : [Tex]\left[ \begin{array}{cc|c} 1 & 3 & 7 \\ 0 & -1 & 0 \end{array} \right] [/Tex] Solve for y: From R 2 : y = 0 ​Substitute y = 0 in R 1 to find x: x + 3(0) = 7⟹x = 7 Solution: x = 7, y = 0

Problem 4: Solve the system using the inverse matrix method.

• x − 2y = -3

​Matrix Form: [Tex]A = \begin{bmatrix} 4 & 1 \\ 1 & -2 \end{bmatrix} B = \begin{bmatrix} x \\ y \end{bmatrix} C = \begin{bmatrix} 9 \\ -3 \end{bmatrix} [/Tex] Find Inverse of A: [Tex]A^{-1} = \frac{1}{(-4-1)} \begin{bmatrix} -2 & -1 \\ -1 & 4 \end{bmatrix} =\begin{bmatrix} \frac{1}{-5} \cdot (-2) & \frac{1}{-5} \cdot (-1) \\ \frac{1}{-5} \cdot (-1) & \frac{1}{-5} \cdot 4 \end{bmatrix} = \begin{bmatrix} 0.4 & 0.2 \\ 0.2 & -0.8 \end{bmatrix}[/Tex] Compute B = A − 1 C: [Tex]B = \begin{bmatrix} 0.4 & 0.2 \\ 0.2 & -0.8 \end{bmatrix} \begin{bmatrix} 9 \\ -3 \end{bmatrix} \begin{bmatrix} 0.4 \cdot 9 + 0.2 \cdot (-3) \\ 0.2 \cdot 9 – 0.6 \cdot (-3) \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \end{bmatrix}[/Tex] Solution: x = 3, y = -0

Problem 5: Solve the system using Cramer’s Rule.

• 2x + 3y = 8

Form the Coefficient Matrix A and its Determinant:​ [Tex]A = \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} ,\text{det}(A) = (2 \cdot (-1)) – (3 \cdot 1) = -2 – 3 = -5 [/Tex] Calculate Determinants for x and y: ​ [Tex]A_x = \begin{bmatrix} 8 & 3 \\ 1 & -1 \end{bmatrix} ,\text{det}(A_x) = (8 \cdot (-1)) – (3 \cdot 1) = -8 – 3 = -11[/Tex] [Tex]A_y = \begin{bmatrix} 2 & 8 \\ 1 & 1 \end{bmatrix} ,\text{det}(A_y) = (2 \cdot 1) – (8 \cdot 1) = 2 – 8 = -6[/Tex] Apply Cramer’s Rule: [Tex]x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-11}{-5} = 2.2, y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-6}{-5} = 1.2[/Tex] ​ Solution: x = 2.2, y = 1.2

Problem 6: Solve the system using Gaussian elimination.

• x + y + z = 10
• 2x − y + z = 8
• − x + 3y − 2z = 4

Form the Augmented Matrix: [Tex]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 10 \\ 2 & -1 & 1 & 8 \\ -1 & 3 & -2 & 4 \end{array}\right][/Tex] Row Operations: [Tex]R_2 = R_2 – 2R_1, R_3 = R_3 + R_1[/Tex] ​ [Tex]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 10 \\ 0 & -3 & -1 & -12 \\ 0 & 4 & -1 & 14 \end{array}\right][/Tex] [Tex]R_3 = R_3 + \frac{4}{3}R_2[/Tex] [Tex]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 10 \\ 0 & -3 & -1 & -12 \\ 0 & 0 & -2 & -4 \end{array}\right][/Tex] Solve for z, y, and x: [Tex]From ?_3: z=2, From R_2:-3y – 1(2) = -12 \implies y = 4, From R_1:x + 1(4) + 1(2) = 10 \implies x = 4[/Tex] Solution: x = 4, y = 4, z = 2

Practice Problems on Systems of Equations Using Matrices

Problem 1: Solve the system using Gaussian elimination

• x + 2y − z = 4
• 3x − y + 2z = 5
• 2x + y − z = 1

Problem 2: Use the inverse matrix method to solve

• 4x + 3y = 18
• − x + 2y = 1

Problem 3: Use Gaussian elimination to solve.

• 5x + 2y = 13

Problem 4: Apply Cramer’s Rule to solve.

Problem 5: Solve the system using Gaussian elimination:

Problem 6: Solve the system using Gauss-Jordan elimination:

• 3x + 2y − z = 4
• x − y + 4z = 3
• 2x + y + 3z = 5

​ Answer Key

• x = 1, y = 1, z = -1
• x = 3, y = 2
• x = 2, y = 1
• x = 1, y = 2
• x = 1, y = 4, z = 5
• x = 2, y = − 1, z = 1
• Types of Matrices
• Elements of Matrix
• Algebra of Matrices
• Matrix Operations
• Solve Systems of Linear Equations by Graphing

FAQs: Systems of Equations Using Matrices

What are the benefits of using matrices to solve linear equations.

Matrices allow for systematic and compact representation and manipulation of systems of linear equations, making them ideal for computer algorithms and handling large-scale problems.

What should I do if a system of equations has no solution or infinitely many solutions?

If a system has no solution, it is inconsistent. If it has infinitely many solutions, it is dependent. In both cases, special techniques or additional constraints may be needed to analyze the system further.

Can matrices be used for non-linear systems?

No, matrices are specifically designed for linear systems. Non-linear systems require other methods such as numerical algorithms or graphical analysis.

How does the determinant of a matrix affect its solvability?

The determinant indicates whether a matrix is invertible. A non-zero determinant means the matrix is invertible, and the system has a unique solution. A zero determinant means the matrix is not invertible, and the system may have no solution or infinitely many solutions.

What are the differences between Gaussian elimination and Gauss-Jordan elimination?

Gaussian elimination transforms the matrix into row echelon form, allowing for back-substitution to find solutions. Gauss-Jordan elimination goes further by transforming the matrix into reduced row echelon form, providing solutions directly without back-substitution.

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Rangers’ Vesey Could Solve Right Wing Problem on Kreider/Zibanejad Line

Maybe in the New York Rangers ‘ endless search for a running mate for Chris Kreider and Mika Zibanejad, the answer has been under their nose all along.

Perhaps the solution isn’t another go-round with former No. 2 overall draft pick Kaapo Kakko, new acquisition Reilly Smith or a promising youngster such as Brennan Othmann or Brett Berard. Perhaps the right course of action is being overlooked as the club heads into training camp for the 2024-25 season.

Could Jimmy Vesey , the Rangers’ highly-successful reclamation project who’s heading into the third season of his second stint with the team, be ready for a return to the top six – the role for which he seemed destined as a highly sought-after free-agent prospect eight years ago? Unlikely as it may seem, Vesey’s track record with Zibanejad and Kreider – thanks to a style that has meshed well with the BFFs – make it an appealing option that should at least be experimented with during camp.

This space was used recently to examine the options, none of which were wildly appealing, to fix the Rangers’ right-wing depth issues that have plagued them since the trade of Pavel Buchnevich. Kakko, who despite being just 23 has a murky future with the organization , has never been able to find chemistry with Kreider and Zibanejad despite multiples opportunities to do so. Smith profiles more as a useful middle-six option, and Othmann and Berard are unproven prospects. There’s at least a good chance that the club might be looking for outside help at the trade deadline for the fourth straight season next spring.

Vesey hasn’t seemed to be in the conversation as a candidate for the assignment, but past results indicate that he should be. While the 6-foot-3, 202-pound forward has played limited minutes with the long-time duo over the past two seasons, the numbers indicate that it would behoove the Blueshirts to change that.

Metrics Show Vesey to Be a Strong Fit With Kreider and Zibanejad

In nearly 200 minutes together at 5-on-5 over 157 games, Kreider-Zibanejad-Vesey has posted a 59.4 expected goal share. The unit has outchanced opponents 118-82, including 50-37 in high-danger chances, while outscoring the opposition 11-9. Vesey’s effect on the BFFs is clear, as Kreider and Zibanejad’s expected goal share drops to 51.3 without him during that span, while the pair was outchanced 1,076-1,063 with other linemates.

Such encouraging metrics make one wonder why this alignment hasn’t been tried more consistently. Vesey’s strong all-around game seems to complement those of Kreider and Zibanejad well, as his defensive abilities and responsible play in his own zone appear to shore up that element of the trio. Yet Vesey’s strongest attribute – his relentless forechecking and hunting of pucks – might be the key for why the line works.

Fans and media have for years speculated on the best type of linemate for Kreider and Zibanejad – a speed guy, a shooter, a pass-first player, a big grinder. The duo has played with all kinds since Buchnevich’s departure threw the right-wing spot on their unit into a perpetual state of flux. Maybe Vesey is a fit because he incorporates a little bit of all of those elements.

Kreider and Zibanejad are elite offensive players, but they have a tendency to become passive at times. Vesey plays with high intensity, possibly giving them a jolt of energy – his all-out style no doubt the product of his path that nearly led him out of the NHL. He works the walls and is especially effective pressuring pucks in open ice, which leads to turnovers and recovery that results in puck possession for his linemates. He’s a quick skater with an underrated shot who’s not afraid to pull the trigger, as well as being a heady player who moves the puck well.

In short, Kreider and Zibanejad seem to need a variety of skills from their third wheel, and Vesey might be the only candidate on the roster who can provide all of them – unless the Rangers believe the solution is to move rising young winger Alexis Lafreniere from the Artemi Panarin-Vincent Trocheck line, which doesn’t appear to be a sound overall solution.

Would putting Vesey in the top six on a permanent basis result in him finally becoming the offensive player that most NHL teams believed he could become after a high-scoring career at Harvard? It’s been a long and winding road for Vesey, who was in high demand coming out of college following his refusal to sign with the Nashville Predators after they drafted him 66th overall in 2012, and inked a contract with the Rangers in 2016 intead.

Related: Rangers No Closer to Solving Right Wing Conundrum

That began a mostly discouraging march through five teams that resulted in Vesey eventually hitting rock bottom during the 2020-21 season, then reinventing himself as a solid bottom-six penalty-killing option after making the New Jersey Devils roster off of a PTO in 2021. He did the same with the Rangers the next season, then signed a two-year, $1.6 million extension that takes him through this season and has proved to be a huge bargain.

Can Vesey Still Become a Scorer if He Plays Consistently in Top-6?

What if Vesey’s metamorphosis from supposed scorer to versatile penalty killer, however, was the first step back toward fulfilling his offensive promise, rather than his destined NHL identity? While he might not say it, Vesey may very well have felt the pressure of trying to live up to heightened expectations brought on by the competition for his services in his first three years in a Blueshirt. The timing wasn’t right then. Vesey played often in the top six in his first stint on Broadway and was out of place, a 23-year-old rookie on a mostly rebuilding team, but an Original Six one in a major media market with a fan base that always carried with it big expectations.

Though he scored 50 goals with 40 assists in those three seasons from 2016-19, he never looked comfortable. Clearly at ease now – he cited a sense of familiarity as his reason for signing his extension in January 2023 – and more mature, it’s possible that a promotion to the Kreider-Zibanejad line might unlock some of the offensive potential that made him so sought-after eight years ago. Vesey scored 13 goals in 80 games despite playing often on the fourth line and in his familiar penalty-killing role last season.

“I always envisioned myself as a 20-goal guy and when I left here after three years, I thought I pretty much had figured it out,” Vesey said in September 2022 after making the roster. (From ‘Jimmy Vesey Reinvented Himself For Second Rangers Act After Miserable Career Turn’, New York Post , 9/26/22)

Vesey has never reached 20 goals in the NHL, and it’s probably wildly optimistic at this point in his career to expect him to become the sniper that he was at Harvard, where he totaled 56 goals in 70 games his final two seasons. A full-time assignment with gifted offensive players such as Kreider and Zibanejad, though, could mean he becomes a dependable scorer who also lifts the play of linemates with whom he clicked before – and in the process finally stabilizes a top-two forward unit that has been largely a two-man operation for the past three seasons.

If that were to happen, it would certify Vesey’s journey as truly a full-circle one – and maybe the result he thought he’d get all along when he reached the NHL, even if it happens about eight years later than expected.