linear inequality problem solving examples

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Last modified on April 18th, 2024

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Linear inequalities.

Linear inequalities are algebraic expressions where the power of the unknown variable is no more than one, and the variable is connected with an inequality sign (>, <, ≤, or  ≥).   

7x – 12 > 16 and 5x + 11 < 2 are examples of linear inequalities

Rules to Solve

For adding or subtracting .

Adding or subtracting a number on or from both sides of the inequality does not change its direction. 

If a < b, then a + c < b + c

If a > b, then a + c > b + c

If a < b, then a – c < b – c

If a > b, then a – c > b – c

For Multiplying or Dividing

Like adding and subtracting, multiplying, or dividing an inequality by the same positive number also does not change the direction of the inequality. 

If a < b and c > 0, then ac < bc

If a > b and c > 0, then ac > bc

If a < b and c < 0, then ac > bc

If a > b and c < 0, then ac < bc

If a < b and c > 0, then ${\dfrac{a}{c} <\dfrac{b}{c}}$

If a > b and c > 0, then ${\dfrac{a}{c} >\dfrac{b}{c}}$

If a < b and c < 0, then ${\dfrac{a}{c} >\dfrac{b}{c}}$

If a > b and c < 0, then ${\dfrac{a}{c} <\dfrac{b}{c}}$

Solving and Graphing

When solving a linear inequality, the solution is typically represented as an ordered pair (x, y) that satisfies the inequality, which is then graphed on a number line.

Using the above rules, we solve the inequality x + 3 > 10

Step 1: Using the Subtraction Property

x + 3 – 3 > 10 -3

Step 3: Graphing the Solution

For One Variable

Let us try solving the inequality with one variable as 4x + 3 < 23

4x + 3 – 3 < 23 – 3

⇒ 4x < 20

Step 2: Using the Division Property

${\dfrac{4x}{4} <\dfrac{20}{4}}$

Thus, the solution is x < 5

For Two Variables:

Now, we solve the inequality 7y – 5x ≤ 6y – 3x + 3

Step 1: Solve for ‘x’ Using the Addition Property

7y – 5x + 5x ≤ 6y – 3x + 5x + 3

⇒ 7y ≤ 6y + 2x + 3

Step 2: Solve for ‘y’ Using the Subtraction Property

7y – 6y ≤ 6y + 2x + 3 – 6y

⇒ y ≤ 2x + 3

To plot a graph of the inequality y ≤ 2x + 3, we consider ‘≤’ as ‘=’ sign. Now, plotting the graph y = 2x + 3, we get

Since the inequality is ‘≥,’ the graph of the equation is formed as a solid line. 

Now, considering a point (x, y) as (-2, 0), we get 

2x + 3 = 2(-2) + 3 = -4 + 3 = -1, which is not greater than 0

Thus, y ≰ 2x + 3 at (-2, 0)

Now, shading the region that does not contain (-2, 0) shows the following graph.

Solved Examples

Solve and Graph the inequality with one variable: 2x + 5 < 3x

Given, 2x + 5 < 3x Step 1: Using the Subtraction Property 2x + 5 – 2x < 3x – 2x ⇒ 5 < x ⇒ x > 5, which is graphed on the number line as shown.

Solve and Graph the inequality with the variables on both sides: 7x – 5 > 3x + 13

Given, 7x – 5 > 3x + 13 Step 1: Using the Addition and Subtraction Property 7x – 5 + 5 – 3x > 3x + 13 + 5 – 3x  ⇒ 4x > 18 Step 2: Using the Division Property ⇒ x > ${\dfrac{18}{4}}$ (by division property) ⇒ x > ${\dfrac{9}{2}}$ However, linear inequalities with the same solution are called equivalent inequalities . Thus, 7x – 5 > 3x + 13 is equivalent to x > ${\dfrac{9}{2}}$ or x > ${4\dfrac{1}{2}}$ By plotting the solution on the number line, we get the shown graph.

Special Cases

Sometimes, plotting the solution of linear inequalities can form a horizontal line parallel to the x-axis or a vertical line parallel to the y-axis.

Case 1: When we plot the graph of the linear inequality y > 1 (‘y’ is greater than 1, excluding 1), the graph includes the entire region above the line y = 1, without the line y = 1 (shown in dotted form). 

Case 2: When we plot the graph of the linear inequality x < 4 (‘x’ is less than 4, excluding 4), the graph includes the entire region left of line x = 4, without the line x = 4 (shown in dotted form).

System of Linear Inequalities

A system of linear inequalities consists of two or more linear inequalities with the same variables. Its solution includes all ordered pairs that simultaneously satisfy each inequality in the system.

Graphically, the solution set is depicted as the intersection of the regions represented by each individual inequality on the coordinate plane. 

Let us consider the system of inequalities:

5y – 3x < 15 and 5x + 3y > 15

First, we graph each inequality separately, as shown.

Now, let us pick a point that is not on the two given lines and verify whether this point satisfies the inequalities. 

Considering the point (x, y) as (2, 1), we get 

5y – 3x = 5(1) – 3(2) = 5 – 6 = -1, which is less than 15

5x + 3y = 5(2) + 3(1) = 10 + 3 = 13, which is not greater than 15

Thus, the individual shaded regions and their overlapped region are shaded:

Here, the graphs are formed as dotted lines for the strict inequalities (‘<’ and ‘>’).

Solve and graph the linear inequality 6(x – 3) > 30

Here, 6(x – 3) > 30 ⇒ x – 3 > 5 (by division property) ⇒ x > 5 + 3 (by addition property) ⇒ x > 8 Thus, the solution is x > 8, and its graph is as follows.

Solve the inequality 6(y – 5) < 5(4 + y)

Here, 6(y – 5) < 5(4 + y) ⇒ 6y – 30 < 20 + 5y ⇒ 6y – 5y – 30 + 30 < 20 + 5y – 5y + 30 (by addition and subtraction property) ⇒ y < 50 Thus, the solution is y < 50

Which linear inequality is represented by the following graph?

Here, the given graph represents the linear inequality x ≥ -2 for any variable x.

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Examples of Solving Harder Linear Inequalities

Intro & Formatting Worked Examples Harder Examples & Word Prob's

Once you'd learned how to solve one-variable linear equations, you were then given word problems. To solve these problems, you'd have to figure out a linear equation that modelled the situation, and then you'd have to solve that equation to find the answer to the word problem.

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Solving Inequalities

So, now that you know how to solve linear inequalities — you guessed it! — they give you word problems.

• The velocity of an object fired directly upward is given by V = 80 − 32 t , where the time t is measured in seconds. When will the velocity be between 32 and 64 feet per second (inclusive)?

This question is asking when the velocity, V , will be between two given values. So I'll take the expression for the velocity,, and put it between the two values they've given me. They've specified that the interval of velocities is inclusive, which means that the interval endpoints are included. Mathematically, this means that the inequality for this model will be an "or equal to" inequality. Because the solution is a bracket (that is, the solution is within an interval), I'll need to set up a three-part (that is, a compound) inequality.

I will set up the compound inequality, and then solve for the range of times t :

32 ≤ 80 − 32 t ≤ 64

32 − 80 ≤ 80 − 80 − 32 t ≤ 64 − 80

−48 ≤ −32 t ≤ −16

−48 / −32 ≥ −32 t / −32 ≥ −16 / −32

1.5 ≥ t ≥ 0.5

Note that, since I had to divide through by a negative, I had to flip the inequality signs.

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Note also that you might (as I do) find the above answer to be more easily understood if it's written the other way around, with "less than" inequalities.

And, because this is a (sort of) real world problem, my working should show the fractions, but my answer should probably be converted to decimal form, because it's more natural to say "one and a half seconds" than it is to say "three-halves seconds". So I convert the last line above to the following:

0.5 ≤ t ≤ 1.5

Looking back at the original question, it did not ask for the value of the variable " t ", but asked for the times when the velocity was between certain values. So the actual answer is:

The velocity will be between 32 and 64 feet per second between 0.5 seconds after launch and 1.5 seconds after launch.

Okay; my answer above was *extremely* verbose and "complete"; you don't likely need to be so extreme. You can probably safely get away with saying something simpler like, "between 0.5 seconds and 1.5 seconds". Just make sure that you do indeed include the approprioate units (in this case, "seconds").

Always remember when doing word problems, that, once you've found the value for the variable, you need to go back and re-read the problem to make sure that you're answering the actual question. The inequality 0.5 ≤  t  ≤ 1.5 did not answer the actual question regarding time. I had to interpret the inequality and express the values in terms of the original question.

• Solve 5 x + 7 < 3( x  + 1) .

First I'll multiply through on the right-hand side, and then solve as usual:

5 x + 7 < 3( x + 1)

5 x + 7 < 3 x + 3

2 x + 7 < 3

2 x < −4

x < −2

In solving this inequality, I divided through by a positive 2 to get the final answer; as a result (that is, because I did *not* divide through by a minus), I didn't have to flip the inequality sign.

• You want to invest $30,000 . Part of this will be invested in a stable 5% -simple-interest account. The remainder will be "invested" in your father's business, and he says that he'll pay you back with 7% interest. Your father knows that you're making these investments in order to pay your child's college tuition with the interest income. What is the least you can "invest" with your father, and still (assuming he really pays you back) get at least $1900 in interest?

First, I have to set up equations for this. The interest formula for simple interest is I = Prt , where I is the interest, P is the beginning principal, r is the interest rate expressed as a decimal, and t is the time in years.

Since no time-frame is specified for this problem, I'll assume that t  = 1 ; that is, I'll assume (hope) that he's promising to pay me at the end of one year. I'll let x be the amount that I'm going to "invest" with my father. Then the rest of my money, being however much is left after whatever I give to him, will be represented by "the total, less what I've already given him", so 30000 −  x will be left to invest in the safe account.

Then the interest on the business investment, assuming that I get paid back, will be:

I = ( x )(0.07)(1) = 0.07 x

The interest on the safe investment will be:

(30 000 − x )(0.05)(1) = 1500 − 0.05 x

The total interest is the sum of what is earned on each of the two separate investments, so my expression for the total interest is:

0.07 x + (1500 − 0.05 x ) = 0.02 x + 1500

I need to get at least $1900 ; that is, the sum of the two investments' interest must be greater than, or at least equal to, $1,900 . This allows me to create my inequality:

0.02 x + 1500 ≥ 1900

0.02 x ≥ 400

x ≥ 20 000

That is, I will need to "invest" at least $20,000 with my father in order to get $1,900 in interest income. Since I want to give him as little money as possible, I will give him the minimum amount:

I will invest $20,000 at 7% .

• An alloy needs to contain between 46% copper and 50% copper. Find the least and greatest amounts of a 60% copper alloy that should be mixed with a 40% copper alloy in order to end up with thirty pounds of an alloy containing an allowable percentage of copper.

This is similar to a mixture word problem , except that this will involve inequality symbols rather than "equals" signs. I'll set it up the same way, though, starting with picking a variable for the unknown that I'm seeking. I will use x to stand for the pounds of 60% copper alloy that I need to use. Then 30 −  x will be the number of pounds, out of total of thirty pounds needed, that will come from the 40% alloy.

Of course, I'll remember to convert the percentages to decimal form for doing the algebra.

How did I get those values in the bottom right-hand box? I multiplied the total number of pounds in the mixture ( 30 ) by the minimum and maximum percentages ( 46% and 50% , respectively). That is, I multiplied across the bottom row, just as I did in the " 60% alloy" row and the " 40% alloy" row, to get the right-hand column's value.

The total amount of copper in the mixture will be the sum of the copper contributed by each of the two alloys that are being put into the mixture. So I'll add the expressions for the amount of copper from each of the alloys, and place the expression for the total amount of copper in the mixture as being between the minimum and the maximum allowable amounts of copper:

13.8 ≤ 0.6 x + (12 − 0.4 x ) ≤ 15

13.8 ≤ 0.2 x + 12 ≤ 15

1.8 ≤ 0.2 x ≤ 3

9 ≤ x ≤ 15

Checking back to my set-up, I see that I chose my variable to stand for the number of pounds that I need to use of the 60% copper alloy. And they'd only asked me for this amount, so I can ignore the other alloy in my answer.

I will need to use between 9 and 15 pounds of the 60% alloy.

Per yoozh, I'm verbose in my answer. You can answer simply as " between 9 and 15 pounds ".

• Solve 3( x − 2) + 4 ≥ 2(2 x − 3)

First I'll multiply through and simplify; then I'll solve:

3( x − 2) + 4 ≥ 2(2 x − 3)

3 x − 6 + 4 ≥ 4 x − 6

3 x − 2 ≥ 4 x − 6

−2 ≥ x − 6            (*)

Why did I move the 3 x over to the right-hand side (to get to the line marked with a star), instead of moving the 4 x to the left-hand side? Because by moving the smaller term, I was able to avoid having a negative coefficient on the variable, and therefore I was able to avoid having to remember to flip the inequality when I divided through by that negative coefficient. I find it simpler to work this way; I make fewer errors. But it's just a matter of taste; you do what works for you.

Why did I switch the inequality in the last line and put the variable on the left? Because I'm more comfortable with inequalities when the answers are formatted this way. Again, it's only a matter of taste. The form of the answer in the previous line, 4 ≥ x , is perfectly acceptable.

As long as you remember to flip the inequality sign when you multiply or divide through by a negative, you shouldn't have any trouble with solving linear inequalities.

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Linear Inequalities

In mathematics, inequality occurs when a non-equal comparison is made between two mathematical expressions or two numbers. In general, inequalities can be either numerical inequality or algebraic inequality or a combination of both. Linear inequalities are inequalities that involve at least one linear algebraic expression, that is, a polynomial of degree 1 is compared with another algebraic expression of degree less than or equal to 1. There are several ways to represent various kinds of linear inequalities.

In this article, let us learn about linear inequalities, solving linear inequalities, graphing linear inequalities.

What Are Linear Inequalities?

Linear inequalities are defined as expressions in which two linear expressions are compared using the inequality symbols. The five symbols that are used to represent the linear inequalities are listed below:

We need to note that if, p < q, then p is some number that is strictly less than q. If p ≤ q, then it means that p is some number that is either strictly less than q or is exactly equal to q. Likewise, the same applies to the remaining two inequalities > (greater than) and ≥ (greater than or equal to).

Now, Let's say we have a linear inequality, 3x - 4 < 20. In this case LHS < RHS. We can see that the expression on the left-hand side, that is, 3x - 4 is in fact lesser than the number on the right-hand side, which is 20. We can represent this inequality pictorially on a weighing scale as:

Rules of Linear Inequalities

The 4 types of operations that are done on linear inequalities are addition, subtraction, multiplication, and division. Linear inequalities with the same solution are called equivalent inequality. There are rules for both equality and inequality. All the rules mentioned below are also true for inequalities involving lesser than or equal to (≤), and greater than or equal to (≥). Before learning how to solve linear inequalities, let's look at some of the important rules of inequality for all these operations.

Addition Rule of Linear Inequalities:

As per the addition rule of linear inequalities, adding the same number to each side of the inequality produces an equivalent inequality, that is the inequality symbol does not change.

If x > y, then x + a > y + a and if x < y, then x + a < y + a.

Subtraction Rule of Linear Inequalities:

As per the subtraction rule of linear inequalities, subtracting the same number from each side of the inequality produces an equivalent inequality, that is the inequality symbol does not change.

If x > y, then x − a > y − a and if x < y, then x − a < y − a.

Multiplication Rule of Linear Inequalities:

As per the multiplication rule of linear inequalities, multiplication on both sides of an inequality with a positive number always produces an equivalent inequality, that is the inequality symbol does not change.

If x > y and a > 0, then x × a > y × a and if x < y and a > 0, then x × a < y × a, Here, × is used as the multiplication symbol.

On the other hand, multiplication on both sides of the inequality with a negative number does not produce an equivalent inequality unless we also reverse the direction of the inequality symbol.

If x > y and a < 0, then x × a < y × a and if x < y and a < 0, then x × a > y × a.

Division Rule of Linear Inequalities:

As per the division rule of linear inequalities, division of both sides of an inequality with a positive number produces an equivalent inequality, that is the inequality symbol does not change.

If x > y and a > 0, then (x/a) > (y/a) and if x < y and a > 0, then (x/a) < (y/a).

On the other hand, the division of both sides of an inequality with a negative number produces an equivalent inequality if the inequality symbol is reversed.

If x > y and a < 0, then (x/a) < (y/a) and if x < y and a < 0, then (x/a) > (y/a)

Solving System Of Linear Inequalities

Solving linear inequalities in multi-step one variable is the same as solving multi-step linear equations ; begin by isolating the variable from the constants. As per the rules of inequalities, while we are solving multi-step linear inequalities, it is important for us to not forget to reverse the inequality sign when multiplying or dividing with negative numbers.

• Step 1: Simplify the inequality on both sides - on LHS as well as RHS as per the rules of inequality .
• Step 2: When the value is obtained, if the inequality is a strict inequality, the solution for x is less than or greater than the value obtained as defined in the question. And, if the inequality is not a strict inequality, then the solution for x is less than or equal to or greater than or equal to the value obtained as defined in the question.

Now, let's try solving linear inequalities with an example, to understand the concept.

2x + 3 > 7

To solve this linear inequality, we would follow the below-given steps:

2x > 7 - 3 ⇒ 2x > 4 ⇒ x > 2

The solution to this inequality will be the set of all values of x for which this inequality x > 2 is satisfied, that is, all real numbers strictly greater than 2.

Solving Linear Inequalities with Variable on Both Sides

Let us try solving linear inequalities with one variable by applying the concept we learned. Consider the following example.

3x - 15 > 2x + 11

We proceed as follows:

-15 - 11 > 2x - 3x ⇒ - 26 > - x ⇒ x > 26

Solving System of Linear Inequalities by Graphing

The system of two-variable linear inequalities is of the form ax + by > c or ax + by ≤ c. The signs of inequalities can change as per the set of inequalities given. To solve a system of two-variable linear inequalities, we must have at least two inequalities. Now, to solve a system of linear inequalities in two variables, let us consider an example.

2y - x > 1 and y - 2x < -1

First, we will plot the given inequalities on the graph. To do that, follow the given steps:

• Replace the inequality sign with equal to =, that is, we have 2y - x = 1 and y - 2x = -1. Since the linear inequality is strict, we draw dotted lines on the graph.
• Check if the origin (0, 0) satisfies the given linear inequalities. If it does, then shade the region on one side of the line which includes the origin. If the origin does not satisfy the linear inequality, shade the region on one side side of the line which does not include the origin. For 2y - x > 1, substitute (0, 0) we have: 2 × 0 - 0 > 1 ⇒ 0 > 1 which is not true. Hence, shade the side of the line 2y - x = 1 which does not include origin. Simillarly, for y - 2x < -1, substituting (0, 0), we have: 0 - 2 × 0 < -1 ⇒ 0 < -1 which is not true. Hence, hade the side of the line y - 2x = -1 which does not include origin.
• The common shaded will be the feasible region that forms the solution of the system of linear inequalities. If there is no common shaded region, then the solution does not exist. The violet-colored region in the graph given below shows the solution of the given system of linear inequalities.

Graphing Linear Inequalities

Linear inequalities with one variable are plotted on a number line, as the output gives the solution of one variable. Hence, graphing linear inequalities in one variable is done using a number line only. On the contrary, linear inequalities with two variables are plotted on a two-dimensional x and y axis graph, as the output gives the solution of two variables. Hence, graphing of two-variable linear inequalities is done using a graph.

Graphing Linear Inequalities - One Variable

Let's consider the below example.

4x > -3x + 21

The solution in this case is simple.

4x + 3x > 21 ⇒ 7x > 21 ⇒ x > 3

This can be plotted on a number line as:

Any point lying on the blue part of the number line will satisfy this inequality. Note that in this case, we have drawn a hollow dot at point 3. This is to indicate that 3 is not a part of the solution set (this is because the given inequality has a strict inequality). As per the solution obtained, the blue part of the number line satisfies the inequality. Let's take another example of linear inequalities:

3x ≤ 7 - 1 ⇒ 3x ≤ 6 ⇒ x ≤ 2

We want to represent this solution set on a number line. Thus, we highlight that part of the number line lying to the left of 2

We see that any number lying on the red part of the number line will satisfy this inequality and so it is a part of the solution set for this inequality. Note that we have drawn a solid dot exactly at point 2. This is to indicate that 2 is also a part of the solution set.

☛ Related Topics:

Check out the following pages related to linear inequalities

• Linear Inequalities with Two Variables
• Inequalities Involving Absolute Values
• Multiplying Polynomials
• Special Cases in Linear Equations
• One Variable Linear Equations and Inequations

Important Notes on Linear Inequalities

Here is a list of a few points that should be remembered while studying linear inequalities:

• In the case of linear inequalities, some other relationship like less than or greater than exists between LHS and RHS.
• A linear inequality is called so due to the highest power( exponents ) of the variable being 1.
• "Less than" and "greater than" are strict inequalities while "less than or equal to" and "greater than or equal to" are not strict linear inequalities.
• For every linear inequality which uses strict linear inequality, the value obtained for x is shown by a hollow dot. It shows that the value obtained is excluded.
• For every linear inequality which is not strict inequality, the value obtained for x is shown by a solid dot. It shows that the value obtained is included.

Examples on Solving Linear Inequalities

Example 1.  Find the solution to the linear inequality, -2x - 39 ≥ -15, and plot it on a number line.

Solution: We will solve the problem in the following way:

- 2x - 39 ≥ - 15 ⇒ - 2x ≥ 24

⇒ 2x ≤ - 24 ⇒ x ≤ - 12

The linear inequality will be plotted on a number line in the following way:

The solution set is plotted above. We can say that any number lying on the red part of the number line will satisfy this linear inequality and so it is a part of the solution set for this inequality. Hence, we have drawn a solid dot exactly at point -12. This is to indicate that -12 is also a part of the solution set.

Answer: Therefore, the solution is x ≤ -12

Example 2. Solve the linear inequalities in this linear inequality 2x - 5 > 3 - 7x

Solution: Let us proceed in the following way to solve the given linear inequality:

2x + 7x > 3 + 5 ⇒ 9x > 8 ⇒ x > 8/9

Answer: x > 8/9 is the solution for the linear inequality 2x - 5 > 3 - 7x

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Linear Inequalities Questions

Faqs on linear inequalities, what are linear inequalities in algebra.

Linear inequalities are defined as expressions in which two linear expressions are compared using the inequality symbols. These expressions could be numerical or algebraic or a combination of both.

What Is an Example of Linear Inequality?

An example of linear inequality is x - 5 > 3x - 10. Here, the LHS is strictly greater than the RHS since greater than symbol is used in this inequality. After solving, the inequality looks like this: 2x < 5 ⇒ x < (5/2).

What Are the Real-Life Uses of Linear Inequalities?

Inequalities are most often used in many real-life problems than equalities to determine the best solution to a problem. This solution can be as simple as finding how many of a product should be produced in order to maximize a profit or it can be as complicated as finding the correct combination of drugs to be given to a patient.

What Are the Uses of Linear Inequalities in Businesses?

Businesses use inequalities to create pricing models, plan their production lines, and control their inventory. They are also used for shipping or warehousing materials, and goods.

What Are the Symbols Used in Linear Inequalities?

The symbols used in linear inequalities are:

• Not equal (≠)
• Less than (<)
• Greater than (>)
• Less than or equal to (≤)
• Greater than or equal to (≥)

What Are the Two Similarities Between Linear Inequalities and Equations?

The similarities between linear inequalities and equations are:

• Both the mathematical statements relate two expressions to each other.
• Both are solved in the same way.

How to Solve Linear Inequalities in Two Variables?

To solve a system of two-variable linear inequalities, we must have at least two inequalities. We will plot the given inequalities on the graph and check for the common shaded region to determine the solution.

How to Solve Systems of Linear Inequalities by Graphing?

We plot the given inequalities on the graph similar to the linear equations, but with dotted lines, because of the inequality. Further, we check for the common shaded region to determine the solution. If there is no common shaded region, then the solution does not exist. The shaded region can be bounded or unbounded.

How are Quadratic Inequalities Different From Linear Inequalities?

Quadratic inequalities consist of algebraic expressions of degree 2 whereas linear inequalities consist of algebraic expressions of degree 1.

Solving Inequalities

Sometimes we need to solve Inequalities like these:

Our aim is to have x (or whatever the variable is) on its own on the left of the inequality sign:

We call that "solved".

Example: x + 2 > 12

Subtract 2 from both sides:

x + 2 − 2 > 12 − 2

x > 10

How to Solve

Solving inequalities is very like solving equations , we do most of the same things ...

... but we must also pay attention to the direction of the inequality .

Some things can change the direction !

< becomes >

> becomes <

≤ becomes ≥

≥ becomes ≤

Safe Things To Do

These things do not affect the direction of the inequality:

• Add (or subtract) a number from both sides
• Multiply (or divide) both sides by a positive number
• Simplify a side

Example: 3x < 7+3

We can simplify 7+3 without affecting the inequality:

But these things do change the direction of the inequality ("<" becomes ">" for example):

• Multiply (or divide) both sides by a negative number
• Swapping left and right hand sides

Example: 2y+7 < 12

When we swap the left and right hand sides, we must also change the direction of the inequality :

12 > 2y+7

Here are the details:

Adding or Subtracting a Value

We can often solve inequalities by adding (or subtracting) a number from both sides (just as in Introduction to Algebra ), like this:

Example: x + 3 < 7

If we subtract 3 from both sides, we get:

x + 3 − 3 < 7 − 3    

And that is our solution: x < 4

In other words, x can be any value less than 4.

What did we do?

And that works well for adding and subtracting , because if we add (or subtract) the same amount from both sides, it does not affect the inequality

Example: Alex has more coins than Billy. If both Alex and Billy get three more coins each, Alex will still have more coins than Billy.

What If I Solve It, But "x" Is On The Right?

No matter, just swap sides, but reverse the sign so it still "points at" the correct value!

Example: 12 < x + 5

If we subtract 5 from both sides, we get:

12 − 5 < x + 5 − 5    

That is a solution!

But it is normal to put "x" on the left hand side ...

... so let us flip sides (and the inequality sign!):

Do you see how the inequality sign still "points at" the smaller value (7) ?

And that is our solution: x > 7

Note: "x" can be on the right, but people usually like to see it on the left hand side.

Multiplying or Dividing by a Value

Another thing we do is multiply or divide both sides by a value (just as in Algebra - Multiplying ).

But we need to be a bit more careful (as you will see).

Positive Values

Everything is fine if we want to multiply or divide by a positive number :

Example: 3y < 15

If we divide both sides by 3 we get:

3y /3 < 15 /3

And that is our solution: y < 5

Negative Values

Well, just look at the number line!

For example, from 3 to 7 is an increase , but from −3 to −7 is a decrease.

See how the inequality sign reverses (from < to >) ?

Let us try an example:

Example: −2y < −8

Let us divide both sides by −2 ... and reverse the inequality !

−2y < −8

−2y /−2 > −8 /−2

And that is the correct solution: y > 4

(Note that I reversed the inequality on the same line I divided by the negative number.)

So, just remember:

When multiplying or dividing by a negative number, reverse the inequality

Multiplying or Dividing by Variables

Here is another (tricky!) example:

Example: bx < 3b

It seems easy just to divide both sides by b , which gives us:

... but wait ... if b is negative we need to reverse the inequality like this:

But we don't know if b is positive or negative, so we can't answer this one !

To help you understand, imagine replacing b with 1 or −1 in the example of bx < 3b :

• if b is 1 , then the answer is x < 3
• but if b is −1 , then we are solving −x < −3 , and the answer is x > 3

The answer could be x < 3 or x > 3 and we can't choose because we don't know b .

Do not try dividing by a variable to solve an inequality (unless you know the variable is always positive, or always negative).

A Bigger Example

Example: x−3 2 < −5.

First, let us clear out the "/2" by multiplying both sides by 2.

Because we are multiplying by a positive number, the inequalities will not change.

x−3 2 ×2 < −5  ×2  

x−3 < −10

Now add 3 to both sides:

x−3 + 3 < −10 + 3    

And that is our solution: x < −7

Two Inequalities At Once!

How do we solve something with two inequalities at once?

Example: −2 < 6−2x 3 < 4

First, let us clear out the "/3" by multiplying each part by 3.

Because we are multiplying by a positive number, the inequalities don't change:

−6 < 6−2x < 12

−12 < −2x < 6

Now divide each part by 2 (a positive number, so again the inequalities don't change):

−6 < −x < 3

Now multiply each part by −1. Because we are multiplying by a negative number, the inequalities change direction .

6 > x > −3

And that is the solution!

But to be neat it is better to have the smaller number on the left, larger on the right. So let us swap them over (and make sure the inequalities point correctly):

−3 < x < 6

• Many simple inequalities can be solved by adding, subtracting, multiplying or dividing both sides until you are left with the variable on its own.
• Multiplying or dividing both sides by a negative number
• Don't multiply or divide by a variable (unless you know it is always positive or always negative)

Solving Inequalities

Related Pages Solving Equations Algebraic Expressions More Algebra Lessons

In these lessons, we will look at the rules, approaches, and techniques for solving inequalities.

The following figure shows how to solve two-step inequalities. Scroll down the page for more examples and solutions.

The rules for solving inequalities are similar to those for solving linear equations. However, there is one exception when multiplying or dividing by a negative number.

To solve an inequality, we can:

• Add the same number to both sides.
• Subtract the same number from both sides.
• Multiply both sides by the same positive number.
• Divide both sides by the same positive number.
• Multiply both sides by the same negative number and reverse the sign.
• Divide both sides by the same negative number and reverse the sign.

Inequalities Of The Form “x + a > b” or “x + a < b”

Example: Solve x + 7 < 15

Solution: x + 7 < 15 x + 7 – 7 < 15 – 7 x < 8

Inequalities Of The Form “x – a < b” or “x – a > b”

Example: Solve x – 6 > 14

Solution: x – 6 > 14 x – 6+ 6 > 14 + 6 x > 20

Example: Solve the inequality x – 3 + 2 < 10

Solution: x – 3 + 2 < 10 x – 1 < 10 x – 1 + 1 < 10 + 1 x < 11

Inequalities Of The Form “a – x < b” or “a – x > b”

Example: Solve the inequality 7 – x < 9

Solution: 7 – x < 9 7 – x – 7 < 9 – 7 – x < 2 x > –2 (remember to reverse the symbol when multiplying by –1)

Example: Solve the inequality 12 > 18 – y

Solution: 12 > 18 – y 18 – y < 12 18 – y – 18 < 12 –18 – y < –6 y > 6 (remember to reverse the symbol when multiplying by –1)

Inequalities Of The Form “ < b” or “ > b”

Solving linear inequalities with like terms.

If an equation has like terms, we simplify the equation and then solve it. We do the same when solving inequalities with like terms.

Example: Evaluate 3x – 8 + 2x < 12

Solution: 3x – 8 + 2x < 12 3x + 2x < 12 + 8 5x < 20 x < 4

Example: Evaluate 6x – 8 > x + 7

Solution: 6x – 8 > x + 7 6x – x > 7 + 8 5x > 15 x > 3

Example: Evaluate 2(8 – p) ≤ 3(p + 7)

Solution: 2(8 – p) ≤ 3(p + 7) 16 – 2p ≤ 3p + 21 16 – 21 ≤ 3p + 2p –5 ≤ 5p –1 ≤ p p ≥ –1 (a < b is equivalent to b > a)

An Introduction To Solving Inequalities

Solving One-Step Linear Inequalities In One Variable

The solutions to linear inequalities can be expressed several ways: using inequalities, using a graph, or using interval notation.

The steps to solve linear inequalities are the same as linear equations, except if you multiply or divide by a negative when solving for the variable, you must reverse the inequality symbol.

Example: Solve. Express the solution as an inequality, graph and interval notation. x + 4 > 7 -2x > 8 x/-2 > -1 x - 9 ≥ -12 7x > -7 x - 9 ≤ -12

Solving Two-Step Linear Inequalities In One Variable

Example: Solve. Express the solution as an inequality, graph and interval notation. 3x + 4 ≥ 10 -2x - 1 > 9 10 ≥ -3x - 2 -8 > 5x + 12

Solving Linear Inequalities

Main rule to remember: If you multiply or divide by a negative number, the inequality flips direction.

Examples of how to solve linear inequalities are shown:

Example: Solve: 3x - 6 > 8x - 7

Students learn that when solving an inequality, such as -3x is less than 12, the goal is the same as when solving an equation: to get the variable by itself on one side.

Note that when multiplying or dividing both sides of an inequality by a negative number, the direction of the inequality sign must be switched.

For example, to solve -3x is less than 12, divide both sides by -3, to get x is greater than -4.

And when graphing an inequality on a number line, less than or greater than is shown with an open dot, and less than or equal to or greater than or equal to is shown with a closed dot.

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Solving inequalities

Here you will learn about solving inequalities, including how to solve linear inequalities, identify integers in the solution set, and represent solutions on a number line.

Students will first learn about solving simple inequalities as part of expressions and equations in 6th grade math and expand that knowledge in 7th grade math.

What is solving inequalities?

Solving inequalities allows you to calculate the values of an unknown variable in an inequality.

Solving inequalities is similar to solving equations, but where an equation has one unique solution, an inequality has a range of solutions.

In order to solve an inequality, you need to balance the inequality on each side of the inequality sign in the same way as you would balance an equation on each side of the equal sign. Solutions can be integers, decimals, positive numbers, or negative numbers.

For example,

Common Core State Standards

How does this relate to 6th grade and 7th grade math?

• Grade 6 – Expressions and Equations (6.EE.B.8) Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams.
• Grade 7 – Expressions and Equations (7.EE.4b) Solve word problems leading to inequalities of the form px + q > r or px + q < r, where p, q, and r are specific rational numbers. Graph the solution set of the inequality and interpret it in the context of the problem. For example: As a salesperson, you are paid \$50 per week plus \$3 per sale. This week you want your pay to be at least \$100. Write an inequality for the number of sales you need to make, and describe the solutions.

How to solve inequalities

In order to solve inequalities:

Rearrange the inequality so that all the unknowns are on one side of the inequality sign.

Rearrange the inequality by dividing by the \textbf{x} coefficient so that \textbf{‘x’} is isolated.

Write your solution with the inequality symbol.

[FREE] Solving Inequalities Worksheet (Grade 6 to 8)

Use this worksheet to check your grade 6 to 8 students’ understanding of solving inequalities. 15 questions with answers to identify areas of strength and support!

Solving linear inequalities examples

Example 1: solving linear inequalities.

4 x+6<26

In this case you are subtracting 6 from both sides.

\begin{aligned} 4x+6&<26\\ 4x&<20 \end{aligned}

This leaves 4x on the left side of the inequality sign and 20 on the right side.

2 Rearrange the inequality by dividing by the \textbf{x} coefficient so that \textbf{‘x’} is isolated.

In this case you need to divide both sides by 4.

\begin{aligned} 4x&<20\\ x&<5 \end{aligned}

This leaves x on the left side of the inequality sign and 5 on the right side.

3 Write your solution with the inequality symbol.

Any value less than 5 satisfies the inequality.

Example 2: solving linear inequalities

5x-4 \geq 26

In this case you need to add 4 to both sides.

\begin{aligned} 5x-4&\geq26\\ 5x&\geq30 \end{aligned}

This leaves 5x on the left side of the inequality sign and 30 on the right side.

In this case you need to divide both sides by 5.

\begin{aligned} 5x&\geq30\\ x&\geq6 \end{aligned}

This leaves x on the left side of the inequality sign and 6 on the right side.

Any value greater than or equal to 6 satisfies the inequality.

Example 3: solving linear inequalities with parentheses

3(x-4)\leq12

Let’s start by expanding the parentheses.

3x-12\leq12

Then you need to add 12 to both sides.

\begin{aligned} 3x-12&\leq12\\ 3x&\leq24 \end{aligned}

This leaves 3x on the left side of the inequality sign and 24 on the right side.

In this case you need to divide both sides by 3.

\begin{aligned} 3x&\leq24\\ x&\leq8 \end{aligned}

This leaves x on the left side of the inequality sign and 8 on the right side.

Any value less than or equal to 8 satisfies the inequality.

Example 4: solving linear inequalities with unknowns on both sides

5x-6 > 2x + 15

In this case you need to subtract 2x from both sides.

\begin{aligned} 5x-6&>2x+15\\ 3x-6&>15 \end{aligned}

This leaves 3x-6 on the left side of the inequality sign and 15 on the right side.

Rearrange the inequality so that \textbf{‘x’} s are on one side of the inequality sign and numbers on the other.

In this case you need to add 6 to both sides.

\begin{aligned} 3x-6&>15\\ 3x&>21 \end{aligned}

This leaves 3x on the left side of the inequality sign and 21 on the right side.

\begin{aligned} 3x&>21\\ x&>7 \end{aligned}

This leaves x on the left side of the inequality sign and 7 on the right side.

Any value greater than 7 satisfies the inequality.

Example 5: solving linear inequalities with fractions

\cfrac{x+3}{5}<2

Rearrange the inequality to eliminate the denominator.

In this case you need to multiply both sides by 5.

\begin{aligned} \cfrac{x+3}{5}&<2\\ x+3&<10 \end{aligned}

In this case you need to subtract 3 from both sides.

\begin{aligned} \cfrac{x+3}{5}&<2\\ x+3&<10\\ x&<7 \end{aligned}

Any value less than 7 satisfies the inequality.

Example 6: solving linear inequalities with non-integer solutions

In this case you need to subtract 6 from both sides.

\begin{aligned} 6x+1&\geq4\\ 6x&\geq3 \end{aligned}

In this case you need to divide both sides by 6.

\begin{aligned} 6x+1&\geq4\\ 6x&\geq3\\ x&\geq\cfrac{3}{6} \end{aligned}

This can be simplified to \, \cfrac{1}{2} \, or the decimal equivalent.

x\geq\cfrac{1}{2}

Any value greater than or equal to \, \cfrac{1}{2} \, satisfies the inequality.

Example 7: solving linear inequalities and representing solutions on a number line

Represent the solution on a number line:

2x-7 < 5

In this case you need to add 7 to both sides.

\begin{aligned} 2x-7&<5\\ 2x&<12 \end{aligned}

In this case you need to divide both sides by 2.

\begin{aligned} 2x-7& <5\\ 2x& <12\\ x& < 6 \end{aligned}

Represent your solution on a number line.

Any value less than 6 satisfies the inequality. An open circle is required at 6 and the values lower than 6 indicated with an arrow.

Example 8: solving linear inequalities with negative x coefficients

In this case you need to subtract 1 from both sides.

\begin{aligned} 1-2x & <7 \\ -2x & <6 \end{aligned}

In this case you need to divide both sides by negative 2.

6 \div-2=-3

Change the direction of the inequality sign.

Because you divided by a negative number, you also need to change the direction of the inequality sign.

\begin{aligned} 1-2x & <7 \\ -2x & <6 \\ x &>-3 \end{aligned}

Example 9: solving linear inequalities and listing integer values that satisfy the inequality

List the integer values that satisfy:

3 < x+1\leq8

In this case you need to subtract 1 from each part.

\begin{aligned} 3&<x+1\leq8\\ 2&<x\leq7\\ \end{aligned}

List the integer values satisfied by the inequality.

2<x\leq7

2 is not included in the solution set. 7 is included in the solution set. The integers that satisfy this inequality are:

3, 4, 5, 6, 7

Example 10: solving linear inequalities and listing integer values that satisfy the inequality

7\leq4x\leq20

In this case you need to divide each part by 4.

\begin{aligned} 7\leq \, & 4x\leq20\\ \cfrac{7}{4}\leq & \; x \leq5 \end{aligned}

\cfrac{7}{4} \leq x \leq 5

\cfrac{7}{4} \, is included in the solution set but it is not an integer.

The first integer higher is 2.

5 is also included in the solution set.

The integers that satisfy this inequality are:

Example 11: solving linear inequalities and representing the solution on a number line

-3<2x+5\leq7

In this case you need to subtract 5 from each part.

\begin{aligned} -3<2x+5&\leq7\\ -8<2x&\leq2 \end{aligned}

Rearrange the inequality so that \textbf{‘x’} is isolated. In this case you need to divide each part by \bf{2} .

\begin{aligned} -3<2x+5&\leq7\\ -8<2x&\leq2\\ -4<x&\leq1 \end{aligned}

Represent the solution set on a the number line

-4<x\leq1

-4 is not included in the solution set so requires an open circle. 1 is included in the solution set so requires a closed circle.

Put a solid line between the circles to indicate all the values that satisfy the solution set.

Teaching tips for solving inequalities

• Students should master solving simple inequalities (one-step inequalities) before moving on to two-step inequalities (or multi-step inequalities).
• Foster student engagement by practicing solving inequalities through classroom games such as BINGO rather than daily worksheets.
• Student practice problems should have a variety of inequalities such as inequalities with fractions, negative numbers, and parentheses. (see examples above)

Easy mistakes to make

• Solutions as inequalities Not including the inequality symbol in the solution is a common mistake. An inequality has a range of values that satisfy it rather than a unique solution so the inequality symbol is essential. For example, when solving x + 3 < 7 giving a solution of 4 or x = 4 is incorrect, the answer must be written as an inequality x < 4 .
• Balancing inequalities Errors can be made with solving equations and inequalities by not applying inverse operations or not balancing the inequalities. Work should be shown step-by-step with the inverse operations applied to both sides of the inequality. For example, when solving x + 3 < 7 , adding 3 to both sides rather than subtracting 3 from both sides.

Related inequalities lessons

• Inequalities
• Linear inequalities
• Inequalities on a number line
• Graphic inequalities
• Quadratic inequalities
• Greater than sign
• Less than sign

Practice solving inequalities questions

3x+7 < 31

\begin{aligned} 3x+7&<31\\ 3x&<24\\ x&<8 \end{aligned}

\begin{aligned} 4x-3&\geq25\\ 4x&\geq28\\ x&\geq7 \end{aligned}

2(x-5)\leq8

\begin{aligned} 2(x-5)&\leq8\\ 2x-10&\leq8\\ 2x&\leq18\\ x&\leq9 \end{aligned}

6x-5 > 4x + 1

\begin{aligned} 6x-5&>4x+1\\ 2x-5&>1\\ 2x&>6\\ x&>3 \end{aligned}

\cfrac{x-4}{2}>6

\begin{aligned} \cfrac{x-4}{2}&>6\\ x-4&>12\\ x&>16 \end{aligned}

\begin{aligned} 8x+1&\geq3\\ 8x&\geq2\\ x&\geq\cfrac{2}{8}\\ x&\geq\cfrac{1}{4} \end{aligned}

7. Represent the solution on a number line.

5x-2 < 28

\begin{aligned} 5x-2&<28\\ 5x&<30\\ x&<6 \end{aligned}

An open circle is required and all values less than 6 indicated.

2-3x > 14

\begin{aligned} 2-3x &>14 \\ -3x &>12 \\ x &< -4 \end{aligned}

Change the direction of the inequality sign as you have divided by a negative number.

9. List the integer values that satisfy:

2<x+3\leq5

\begin{aligned} 2<x&+3\leq5\\ -1< \, &x\leq2 \end{aligned}

-1 is not included in the solution set as is greater than -1.

2 is included in the solution set as x is less than or equal to 2.

10. List the integer values that satisfy:

4\leq3x\leq21

\begin{aligned} 4\leq3&x\leq21\\ \cfrac{4}{3} \, \leq \, &x\leq7 \end{aligned}

The first integer greater than \, \cfrac{4}{3} \, is 2.

7 is included in the solution set as x is less than or equal to 7.

11. List the integer values that satisfy:

-4<3x+2\leq5

\begin{aligned} -4<3x&+2\leq5\\ -6<3&x\leq3\\ -2< \, &x\leq1 \end{aligned}

-2 is not included in the solution set as x is greater than -2.

1 is included in the solution set as x is less than or equal to 1.

Solving inequalities FAQs

Solving inequalities is where you calculate the values that an unknown variable can be in an inequality.

To solve an inequality, you need to balance the inequality on each side of the inequality sign in the same way as you would balance an equation on each side of the equal sign. Solutions can be integers, decimals, positive numbers, or negative numbers.

The next lessons are

• Types of graphs
• Math formulas
• Coordinate plane

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Unit 2: Solving equations & inequalities

About this unit.

There are lots of strategies we can use to solve equations. Let's explore some different ways to solve equations and inequalities. We'll also see what it takes for an equation to have no solution, or infinite solutions.

Linear equations with variables on both sides

• Why we do the same thing to both sides: Variable on both sides (Opens a modal)
• Intro to equations with variables on both sides (Opens a modal)
• Equations with variables on both sides: 20-7x=6x-6 (Opens a modal)
• Equation with variables on both sides: fractions (Opens a modal)
• Equation with the variable in the denominator (Opens a modal)
• Equations with variables on both sides Get 3 of 4 questions to level up!
• Equations with variables on both sides: decimals & fractions Get 3 of 4 questions to level up!

Linear equations with parentheses

• Equations with parentheses (Opens a modal)
• Reasoning with linear equations (Opens a modal)
• Multi-step equations review (Opens a modal)
• Equations with parentheses Get 3 of 4 questions to level up!
• Equations with parentheses: decimals & fractions Get 3 of 4 questions to level up!
• Reasoning with linear equations Get 3 of 4 questions to level up!

Analyzing the number of solutions to linear equations

• Number of solutions to equations (Opens a modal)
• Worked example: number of solutions to equations (Opens a modal)
• Creating an equation with no solutions (Opens a modal)
• Creating an equation with infinitely many solutions (Opens a modal)
• Number of solutions to equations Get 3 of 4 questions to level up!
• Number of solutions to equations challenge Get 3 of 4 questions to level up!

Linear equations with unknown coefficients

• Linear equations with unknown coefficients (Opens a modal)
• Why is algebra important to learn? (Opens a modal)
• Linear equations with unknown coefficients Get 3 of 4 questions to level up!

Multi-step inequalities

• Inequalities with variables on both sides (Opens a modal)
• Inequalities with variables on both sides (with parentheses) (Opens a modal)
• Multi-step inequalities (Opens a modal)
• Using inequalities to solve problems (Opens a modal)
• Multi-step linear inequalities Get 3 of 4 questions to level up!
• Using inequalities to solve problems Get 3 of 4 questions to level up!

Compound inequalities

• Compound inequalities: OR (Opens a modal)
• Compound inequalities: AND (Opens a modal)
• A compound inequality with no solution (Opens a modal)
• Double inequalities (Opens a modal)
• Compound inequalities examples (Opens a modal)
• Compound inequalities review (Opens a modal)
• Solving equations & inequalities: FAQ (Opens a modal)
• Compound inequalities Get 3 of 4 questions to level up!

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5.7: Linear Inequalities in One Variable

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• Page ID 49372

• Denny Burzynski & Wade Ellis, Jr.
• College of Southern Nevada via OpenStax CNX

Inequalities

Relationships of Inequality

We have discovered that an equation is a mathematical way of expressing the relationship of equality between quantities. Not all relationships need be relationships of equality, however. Certainly the number of human beings on earth is greater than 20. Also, the average American consumes less than 10 grams of vitamin C every day. These types of relationships are not relationships of equality, but rather, relationships of inequality .

Linear Inequalities

Linear Inequality

A linear inequality is a mathematical statement that one linear expression is greater than or less than another linear expression.

Inequality Notation

The following notation is used to express relationships of inequality:

\(>\) Strictly Greater Than

\(<\) Strictly Less Than

\(\ge\) Greater than or equal to

\(\le\) Less than or equal to

Note that the expression \(x > 12\) has infinitely many solutions. Any number strictly greater than 12 will satisfy the statement. Some solutions are \(13, 15, 90, 12.1, 16.3\) and \(102.51\).

Sample Set A

The following are linear inequalities in one variable.

Example \(\PageIndex{1}\)

1. \(x \leq 12\) 2. \(x+7>4\) 3. \(y+3 \geq 2 y-7\) 4. \(P+26<10(4 P-6)\) 5. \(\dfrac{2 r-9}{5}>15\)

The following are not linear inequalities in one variable.

Example \(\PageIndex{2}\)

1. \(x^{2}<4\) The term \(x^{2}\) is quadratic, not linear. 2. \(x \leq 5 y+3\) There are two variables. This is a linear inequality in two variables. 3. \(y+1 \neq 5\) Although the symbol \(\neq\) certainly expresses an inequality, it is customary to use only the symbols \(<,>, \leq, \geq\).

A linear equation, we know, may have exactly one solution, infinitely many solutions, or no solution. Speculate on the number of solutions of a linear inequality. ( Hint: Consider the inequalities \(x<x−6\) and \(x\ge9\).)

A linear inequality may have infinitely many solutions, or no solutions.

The Algebra of Linear Inequalities

Inequalities can be solved by basically the same methods as linear equations. There is one important exception that we will discuss in item 3 of the algebra of linear inequalities.

Let \(a, b\), and \(c\) represent real numbers and assume that:

\(a < b\) (or \(a > b\))

Then, if \(a < b\):

• \(a+c < b+c\) and \(a-c < b-c\). If any real number is added to or subtracted from both sides of an inequality, the sense of the inequality remains unchanged.
• If \(c\) is a positive real number, then if \(a < b\), \(ac < bc\) and \(\dfrac{a}{c} < \dfrac{b}{c}\). If both sides of an inequality are multiplied or divided by the same positive number the sense of the inequality remains unchanged.
• If \(c\) is a negative real number, then if \(a < b\), \(ac > bc\) and \(\dfrac{a}{c} > \dfrac{b}{c}\) If both sides of an inequality are multiplied or divided by the same negative number, the inequality sign must be reversed (change direction) in order for the resulting inequality to be equivalent to the original inequality. (See problem 4 in the next set of examples.)

For example, consider the inequality \(3 < 7\).

Example \(\PageIndex{3}\)

For \(3<7,\) if 8 is added to both sides, we get \(3+8<7+8\) \( 11<15 \) True

Example \(\PageIndex{4}\)

For \(3 < 7\), if 8 is subtracted from both sides, we get:

\(3-8 < 7-8\)

\(-5 < -1\)

Example \(\PageIndex{5}\)

For \(3 < 7\), if both sides are multiplied by 8 (a positive number), we get:

\(8(3) > 8(7)\)

\(24 < 56\)

Example \(\PageIndex{6}\)

For \(3<7\), if both sides are multiplied by −8 (a negative number), we get

\((−8)3>(−8)7\)

Notice the change in direction of the inequality sign.

\(−24>−56\)

If we had forgotten to reverse the direction of the inequality sign we would have obtained the incorrect statement \(−24<−56\).

Example \(\PageIndex{7}\)

For \(3<7\), if both sides are divided by 8 (a positive number), we get

\(\dfrac{3}{8} < \dfrac{7}{8}\)

Example \(\PageIndex{8}\)

For \(3 < 7\), if both sides are divided by -8 (a negative number), we get:

\(\dfrac{3}{-8} > frac{7}{-8}\)

True, since \(-.375 > .875\)

Sample Set B

Solve the following linear inequalities. Draw a number line and place a point at each solution.

Example \(\PageIndex{9}\)

\(3x > 15\) Divide both sides by 3. The 3 is a positive number, so we need not reverse the sense of the inequality.

\(x > 5\)

Thus, all numbers strictly greater than 5 are solutions to the inequality \(3x > 15\)

Example \(\PageIndex{10}\)

\(2y-1 \le 16\) Add 1 to both sides.

\(2y \le 17\) Divide both sides 2.

\(y \le \dfrac{17}{2}\)

Example \(\PageIndex{11}\)

\(-8x + 5 < 14\) Subtract 5 both from both sides.

\(-8x < 9\) Divide both sides by -8. We must reverse the sense of the inequality since we are divide by a neagtive number.

\(x > -\dfrac{9}{8}\)

Example \(\PageIndex{12}\)

\(5-3(y+2) < 6y - 10\) \(5-3y-6 < 6y-10\) \(-3y-1 < 6y-10\) \(-9y < -9\) \(y > 1\)

Example \(\PageIndex{13}\)

\(\dfrac{2z+7}{-4} \ge -6\) Multiply by -4 \(2x+7 \le 24\) Notice the change in the sense of the inequality. \(2z \le 17\) \(z \le \dfrac{17}{2}\)

Practice Set B

Solve the following linear inequalities.

Practice Problem \(\PageIndex{1}\)

\(y−6≤5\)

\(y≤11\)

Practice Problem \(\PageIndex{2}\)

\(x+4>9\)

Practice Problem \(\PageIndex{3}\)

\(4x−1≥15\)

Practice Problem \(\PageIndex{4}\)

\(−5y+16≤7\)

\(y \ge \dfrac{9}{5}\)

Practice Problem \(\PageIndex{5}\)

\(7(4s−3)<2s+8\)

\(s < \dfrac{29}{2}\)

Practice Problem \(\PageIndex{6}\)

\(5(1−4h)+4<(1−h)2+6\)

\(h > \dfrac{1}{18}\)

Practice Problem \(\PageIndex{7}\)

\(18≥4(2x−3)−9x\)

\(x≥−30\)

Practice Problem \(\PageIndex{8}\)

\(-\dfrac{3b}{16} \le 4\)

\(b \ge \dfrac{-64}{3}\)

Practice Problem \(\PageIndex{9}\)

\(\dfrac{-7z+10}{-12} < -1\)

\(z < -\dfrac{2}{7}\)

Practice Problem \(\PageIndex{10}\)

\(-x -\dfrac{2}{3} \le \dfrac{5}{6}\)

\(x \ge \dfrac{-3}{2}

Compound Inequalities

Compound Inequality

Another type of inequality is the compound inequality . A compound inequality is of the form:

\(a < x < b\)

There are actually two statements here. The first statement is \(a<x\). The next statement is \(x<b\). When we read this statement we say "\(a\) is less than \(x\)," then continue saying "and \(x\) is less than \(b\)."

Just by looking at the inequality we can see that the number \(x\) is between the numbers \(a\) and \(b\). The compound inequality \(a<x<b\) indicates "betweenness." Without changing the meaning, the statement \(a<x\) can be read \(x>a\). (Surely, if the number \(a\) is less than the number \(x\), the number \(x\) must be greater than the number \(a\).) Thus, we can read \(a<x<b\) as "\(x\) is greater than a and at the same time is less than \(b\)." For example:

\(4 < x < 9\).

The letter \(x\) is some number strictly between \(4\) and \(9\). Hence, \(x\) is greater than \(4\) and, at the same time, less than \(9\). The numbers \(4\) and \(9\) are not included so we use open circles at these points.

\(-2 < z < 0\).

The \(z\) stands for some number between \(-2\) and \(0\). Hence, \(z\) is greater than \(-2\) but also less than \(0\).

\(1 < x + 6 < 8\).

The expression \(x + 6\) represents some number strictly between \(1\) and \(8\). Hence, \(x + 6\) represents some number strictly greater than \(1\), but less than \(8\).

\(\dfrac{1}{4} \le \dfrac{5x-2}{6} \le \dfrac{7}{9}\).

The term \(\dfrac{5x-2}{6}\) represents some number between and including \(\dfrac{1}{4}\) and \(\dfrac{7}{9}\). Hence, \(\dfrac{5x-2}{6}\) represents some number greater than or equal to \(\dfrac{1}{4}\) but less than or equal to \(\dfrac{7}{9}\).

Consider problem 3 above, \(1<x+6<8\). The statement says that the quantity \(x+6\) is between \(1\) and \(8\). This statement will be true for only certain values of \(x\). For example, if \(x=1\), the statement is true since \(1<1+6<8\). However, if \(x=4.9\), the statement is false since \(1<4.9+6<8\) is clearly not true. The first of the inequalities is satisfied since \(1\) is less than \(10.9\), but the second inequality is not satisfied since \(10.9\) is not less than \(8\).

We would like to know for exactly which values of \(x\) the statement \(1<x+6<8\) is true. We proceed by using the properties discussed earlier in this section, but now we must apply the rules to all three parts rather than just the two parts in a regular inequality.

Sample Set C

Example \(\pageindex{14}\).

Solve \(1 < x + 6 < 8\).

\(1−6<x+6−6<8−6\) Subtract \(6\) from all three parts.

\(-5 < x < 2\)

Thus, if \(x\) is any number strictly between \(-5\) and \(2\), the statement \(1 < x+6 < 8\) will be true.

Solve \(-3 < \dfrac{-2x-7}{5} < 8\)

\(-3 < \dfrac{-2x-7}{5}(5) < 8(5)\) Multiply each part by \(5\).

\(-15 < -2x-7 < 40\). Add \(7\) to all three parts.

\(-8 < -2x < 47\) Divide all three parts by \(-2\).

\(4 > x > -\dfrac{47}{2}\) Remember to reverse the direction of the inequality signs.

\(-\dfrac{47}{2} < x < 4\). It is customary (but not necessary) to write the inequality so that inequality arrows point to the left.

Thus, if \(x\) is any number between \(-\dfrac{47}{2}\) and \(4\), the original inequality will be satisfied.

Practice Set C

Find the values of x that satisfy the given continued inequality.

Practice Problem \(\PageIndex{11}\)

\(4<x−5<12\)

\(9<x<17\)

\(−3<7y+1<18\)

\(-\dfrac{4}{7} < y < \dfrac{17}{7}\)

\(0≤1−6x≤7\)

\(-1 \le x \le \dfrac{1}{6}\)

\(-5 \le \dfrac{2x+1}{3} \le 10\)

\(-8 \le x \le \dfrac{29}{2}\)

\(9 < \dfrac{-4x+5}{-2} < 14\)

\(\dfrac{23}{4} < x < \dfrac{33}{4}\)

Does \(4<x<−1\) have a solution?

For the following problems, solve the inequalities.

Exercise \(\PageIndex{1}\)

\(x+7<12\)

Exercise \(\PageIndex{2}\)

\(y−5≤8\)

Exercise \(\PageIndex{3}\)

\(y+19≥2\)

\(y≥−17\)

Exercise \(\PageIndex{4}\)

\(x−5>16\)

Exercise \(\PageIndex{5}\)

\(3x−7≤8\)

Exercise \(\PageIndex{6}\)

\(9y−12≤6\)

Exercise \(\PageIndex{7}\)

\(2z+8<7\)

\(z < -\dfrac{1}{2}\)

Exercise \(\PageIndex{8}\)

\(4x−14>21\)

Exercise \(\PageIndex{9}\)

\(−5x≤20\)

\(x≥−4\)

Exercise \(\PageIndex{10}\)

\(−8x<40\)

Exercise \(\PageIndex{11}\)

\(−7z<77\)

\(z>−11\)

Exercise \(\PageIndex{12}\)

\(−3y>39\)

Exercise \(\PageIndex{13}\)

\(\dfrac{x}{4} \ge 12\)

\(x≥48\)

Exercise \(\PageIndex{14}\)

\(\dfrac{y}{7} > 3\)

Exercise \(\PageIndex{15}\)

\(\dfrac{2x}{9} \ge 4\)

\(x≥18\)

Exercise \(\PageIndex{16}\)

\(\dfrac{5y}{2} \ge 15\)

Exercise \(\PageIndex{17}\)

\(\dfrac{10x}{3} \le 4\)

\(x \le \dfrac{6}{5}\)

Exercise \(\PageIndex{18}\)

\(-\dfrac{5y}{4} < 8\)

Exercise \(\PageIndex{19}\)

\(\dfrac{-12b}{5} < 24\)

\(b>−10\)

Exercise \(\PageIndex{20}\)

\(\dfrac{-6a}{7} \le -24\)

Exercise \(\PageIndex{21}\)

\(\dfrac{8x}{-5} > 6\)

\(x < -\dfrac{15}{4}\)

Exercise \(\PageIndex{22}\)

\(\dfrac{14y}{-3} \ge -18\)

Exercise \(\PageIndex{23}\)

\(\dfrac{21y}{-8} < -2\)

\(y < \dfrac{16}{21}\)

Exercise \(\PageIndex{24}\)

\(−3x+7≤−5\)

Exercise \(\PageIndex{25}\)

\(−7y+10≤−4\)

Exercise \(\PageIndex{26}\)

\(6x−11<31\)

Exercise \(\PageIndex{27}\)

\(3x−15≤30\)

\(x≤15\)

Exercise \(\PageIndex{28}\)

\(-2y + \dfrac{4}{3} \le -\dfrac{2}{3}\)

Exercise \(\PageIndex{29}\)

\(5(2x−5)≥15\)

Exercise \(\PageIndex{30}\)

\(4(x+1)>−12\)

Exercise \(\PageIndex{31}\)

\(6(3x−7)≥48\)

Exercise \(\PageIndex{32}\)

\(3(−x+3)>−27\)

Exercise \(\PageIndex{33}\)

\(−4(y+3)>0\)

\(y<−3\)

Exercise \(\PageIndex{34}\)

\(−7(x−77)≤0\)

Exercise \(\PageIndex{35}\)

\(2x−1<x+5\)

Exercise \(\PageIndex{36}\)

\(6y+12≤5y−1\)

Exercise \(\PageIndex{37}\)

\(3x+2≤2x−5\)

\(x≤−7\)

Exercise \(\PageIndex{38}\)

\(4x+5>5x−11\)

Exercise \(\PageIndex{39}\)

\(3x−12≥7x+4\)

\(x≤−4\)

Exercise \(\PageIndex{40}\)

\(−2x−7>5x\)

Exercise \(\PageIndex{41}\)

\(−x−4>−3x+12\)

Exercise \(\PageIndex{42}\)

\(3−x≥4\)

Exercise \(\PageIndex{43}\)

\(5−y≤14\)

\(y≥−9\)

Exercise \(\PageIndex{44}\)

\(2−4x≤−3+x\)

Exercise \(\PageIndex{45}\)

\(3[4+5(x+1)]<−3\)

\(x<−2\)

Exercise \(\PageIndex{46}\)

\(2[6+2(3x−7)]≥4\)

Exercise \(\PageIndex{47}\)

\(7[−3−4(x−1)]≤91\)

\(x≥−3\)

Exercise \(\PageIndex{48}\)

\(−2(4x−1)<3(5x+8)\)

Exercise \(\PageIndex{49}\)

\(−5(3x−2)>−3(−x−15)+1\)

Exercise \(\PageIndex{50}\)

\(−.0091x≥2.885x−12.014\)

Exercise \(\PageIndex{51}\)

What numbers satisfy the condition: twice a number plus one is greater than negative three?

\(x>−2\)

Exercise \(\PageIndex{52}\)

What numbers satisfy the condition: eight more than three times a number is less than or equal to fourteen?

Exercise \(\PageIndex{53}\)

One number is five times larger than another number. The difference between these two numbers is less than twenty-four. What are the largest possible values for the two numbers? Is there a smallest possible value for either number?

First number: any number strictly smaller that 6. Second number: any number strictly smaller than 30. No smallest possible value for either number. No largest possible value for either number.

Exercise \(\PageIndex{54}\)

The area of a rectangle is found by multiplying the length of the rectangle by the width of the rectangle. If the length of a rectangle is 8 feet, what is the largest possible measure for the width if it must be an integer (positive whole number) and the area must be less than 48 square feet?

Exercises for Review

Exercise \(\pageindex{55}\).

Simplify\((x^2y^3z^2)^5\)

\(x^{10}y^{15}z^{10}\)

Exercise \(\PageIndex{56}\)

Simplify \(−[−(−|−8|)]\).

Exercise \(\PageIndex{57}\)

Find the product. \((2x−7) (x+4)\).

\(2x^2 + x - 28\)

Exercise \(\PageIndex{58}\)

Twenty-five percent of a number is \(12.32\). What is the number?

Exercise \(\PageIndex{59}\)

The perimeter of a triangle is 40 inches. If the length of each of the two legs is exactly twice the length of the base, how long is each leg?

Free Linear Inequalities Word Problems Worksheet

While teaching high school, one of the biggest struggles I have seen students face is applying math concepts to the real-world. This is true when it comes to solving linear inequalities word problems too. Students often find it challenging to grasp how these mathematical principles translate into practical, everyday life scenarios.

That’s why I have put together this linear inequalities word problems worksheet! My goal is to help you learn a few tips and tricks and practice applying linear inequalities to the real-world! 

What are Linear Inequality Word Problems?

Lesson plans that focus on linear inequality word problems typically show students how to apply the skills they developed while solving inequalities to the real-world. There are a wide variety of inequality applications, ranging from social studies to physical science. Regardless of the application, the idea is that you will be faced with a word problem that requires you to model the scenario using a linear inequality.

In general, linear inequality word problems describe how one quantity has to be less than or greater than another. Your goal is then to use inequality symbols and  algebraic expressions  to represent the scenario algebraically.

For most, the concept of a linear inequality is first introduced in middle school (although this will vary by curriculum). My daughter, for example, is in 2nd grade math and is just starting to explore these problems. Others may not see this concept until later in high school (in some cases not until the 12th grade).

How to Solve a Linear Inequality Word Problem

The best way to solve any math word problem is to start by reading the question very carefully, and linear inequality word problems are no different!

I always encourage my students to underline or highlight any key words and important information. When it comes to how to solve a linear inequality word problem, the key words usually help you understand:

• which quantities you are working with
• whether you are working with  less than  or  greater than symbols

Once you have identified this important information, your goal is to write an algebraic expression using an inequality symbol that models the scenario. You can then solve the inequality using a similar process to what you would apply when solving one-step equations or  two step equations .

There are many different ways to represent the solution to an inequality problem. Sometimes you will be asked to use a number line, which shows all the negative values or positive values that belong to a solution set. The worksheet attached below will provide you with some practice using number lines to communicate your answers to inequality word problems.

​Solving a Linear Inequality Word Problem Example

Age problems are common applications that you will see when solving linear inequality word problems. These types of problems can be simple or complex, but I wanted to start by sharing a simple one here so that you can understand the basics of how to solve linear inequality word problems.

A father is 3 times as old as his son, but three times his son’s age is less than 30. What is the oldest the son can be?

We can begin by calling out the key words that give us important information. In this case, the following two pieces of information are considered important to the problem:

• “3 times as old as”
• “less than 30”

This tells us that we will be working with a “<” symbol, and multiplication of a quantity by 3. If we let  n  represent the age of the son, we can set up a linear inequality as follows:

$$3n<30$$

Reading this statement in English tells us that “3 times the son’s age is less than 30”. If you head back to the original problem, that seems to match the scenario given, doesn’t it? Great! That tells us that we have a good algebraic model for our real-world problem! Let’s move on and start solving!

Remember that we can solve a linear inequality using algebra in a similar way to solving a linear equation. This means that we can add or subtract terms on both sides of the inequality symbol, and we can also multiply and divide terms on both sides of the inequality symbol.

In this case, since we are multiplying  n  by 3, we divide both sides by 3 to isolate  n.

$$\frac{3n}{3}< \frac{30}{3}$$

$$n<10$$

This tells us that the son’s age must be less than 10 in order for the father to be three times his age but still less than 30 years old himself.

We can test this by multiplying a number greater than 10 by 3. For example, \(11 \times 3 = 33\). Notice that 33 is  not  less than 30. Therefore the son cannot be 11. The only values that will make this inequality statement true are values that are less than (not including) 10.

We can represent this solution on a number line by placing a hollow circle at 10 and drawing an arrow to the left toward the negative values. However, we should stop at zero since the son’s age cannot be less than zero. Note that the son must also not be equal to zero. If he were, the father would be \(3 \times 0 = 0\) as well!

If you need more practice with the algebra strategies that can be used to solve inequalities, check out this collection of  solving linear inequalities worksheets .

Linear Inequalities Word Problems Worksheet

Now that you have had some practice applying your understanding of linear inequalities to solve a real-world problem, you are ready to practice! Below I have included a linear inequalities word problem worksheet that covers a variety of problem types, ranging from age problems, to a bake sale problem with pink cupcakes!

As promised, this worksheet will also provide you with practice representing the solution to a linear inequality word problem on a number line. While you should be sure to attempt every problem as independent work first, make sure that you also check the answer key! This is an important step to make sure that you fully understand each problem.

My hope is that you find this linear inequalities word problems worksheet helpful as independent work whether you are in 1st grade math, 7th grade math, or anything in between!

Download the PDF worksheet by clicking below!

Practice Solving Linear Inequalities Word Problems

I hope this linear inequalities word problems worksheet has provided you with some practice with applying this important math concept to the real-world! My goal was to share examples that cover a variety of areas of life. Hopefully solving these problems also gave you an appreciation for how linear inequalities can be used in everyday life.

Whether you first learned about linear inequality word problems in the 6th grade or are just experiencing them for the first time in the 12th grade, the most important thing you can do is practice. Solving many different types of linear inequalities word problems will help you start to recognize patterns. This will help you with the first initial step of writing the inequality as an algebraic statement, something many students find challenging! 

With consistent practice, you’ll develop a strong foundation in solving linear inequalities, enhancing your problem-solving skills and confidence in applying mathematical concepts to diverse real-world scenarios. Keep exploring and practicing, and you’ll find that handling linear inequalities becomes more intuitive over time!

If you are looking for more word problem resources, check out this linear equations word problems worksheet !

Did you find this linear inequalities word problems worksheet helpful? Share this post and subscribe to Math By The Pixel on YouTube for more helpful mathematics content!

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Linear Inequalities In Two Variables

Linear inequalities in two variables represent the inequalities between two algebraic expressions where two distinct variables are included. In linear inequalities in two variables, we use greater than (>), less than (<), greater than or equal (≥) and less than or equal (≤) symbols, instead of using equal to a symbol (=).

What is Linear Inequalities?

Any two real numbers or two algebraic expressions associated with the symbol ‘<’, ‘>’, ‘≤’ or ‘≥’ form a linear inequality. For example, 9<11, 18>17 are examples of numerical inequalities and x+7>y, y<10-x, x ≥ y > 11 are examples of algebraic inequalities.

The symbols ‘<‘ and ‘>’ represent the strict inequalities and the symbols ‘≤’ and ‘≥’ represent slack inequalities. To represent linear inequalities in one variable in a number line is a visual representation and is a convenient way to represent the solutions of the inequality. Now, we will discuss the graph of a linear inequality in two variables.

What are Linear Inequalities in Two Variables?

Linear inequalities in two variables represent the unequal relation between two algebraic expressions that includes two distinct variables. Hence, the symbols used between the expression in two variables will be ‘<’, ‘>’, ‘≤’ or ‘≥’, but we cannot use equal to ‘=’ symbol here.

The examples of linear inequalities in two variables are:

• 3x < 2y + 5
• 8y – 9x > 10

Note:   4x 2 + 2x + 5 < 0 is not an example of linear inequality in one variable, because the exponent of x is 2 in the first term. It is a quadratic inequality.

How to Solve Linear Inequalities in Two Variables?

The solution for linear inequalities in two variables is an ordered pair that is true for the inequality statement. Let us say if Ax + By > C is a linear inequality where x and y are two variables, then an ordered pair (x, y) satisfying the statement will be the required solution.

The method of solving linear inequalities in two variables is the same as solving linear equations .

For example, if 2x + 3y > 4 is a linear inequality, then we can check the solution, by putting the values of x and y here.

Let x = 1 and y = 2

Taking LHS, we have;

2 (1) + 3 (2) = 2 + 6 = 8

Since, 8 > 4, therefore, the ordered pair (1, 2) satisfy the inequality 2x + 3y > 4. Hence, (1, 2) is the solution.

We can also put different values of x and y to find different solutions here.

Graphical Solution of Linear Inequalities in Two Variables

The statements involving symbols like ‘<’(less than), ‘>’ (greater than), ‘≤’’(less than or equal to), ‘≥’ (greater than or equal to) and two distinct variables are called linear inequalities in two variables. Let us see here, how to find the solution of such expressions, graphically.

Below are the two examples of linear inequalities shown in the figure. The graph of y > x – 2 and y ≤ 2x + 2 are:

Real-life Examples:

Following example validates the difference between equation and inequality:

Statement 1: The distance between your house and school is exactly 4.5 kilometres,

The mathematical expression of the above statement is,

x = 4.5 km, where ‘x’ is the distance between house and the school.

Statement 2: The distance between your house and the school is at least 4.5 kilometers.

Here, the distance can be 4.5 km or more than that. Therefore the mathematical expression for the above statement is,

x ≥ 4.5 km, where ‘x’ is a variable that is equal to the distance between the house and the school.

Important Facts

• We can add, subtract, multiply and divide by the same number to solve the inequalities
• While multiplying and dividing by negative number, the inequality sign get reversed
• In graphical solution, the ordered pair outside the shaded portion does not solve the inequality
• Numerical inequalities: If only numbers are involved in the expression, then it is a numerical inequality. Example:  10 > 8, 5 < 7
• Literal inequalities: x < 2, y > 5, z < 10 are the examples for literal inequalities.
• Double inequalities: 5 < 7 < 9 read as 7 less than 9 and greater than 5 is an example of double inequality.
• Strict inequality: Mathematical expressions involve only ‘<‘ or ‘>’  are called strict inequalities. Example: 2x + 3 < 6, 2x + 3y > 6
• Slack inequality: Mathematical expressions involve only ‘≤′ or ‘≥’ are called slack inequalities. Example: 2x + 3 ≤ 6, 2x + 3y ≥ 6

Related Articles

• Linear Inequalities
• Solving Linear Inequalities
• Represent Linear Inequalities In One Variable On Number Line
• Linear Equations In Two Variables
• Cross Multiplication- Pair Of Linear Equations In Two Variables

Solved Examples on Linear Inequalities in Two Variables

1.) Classify the following expressions into:

• Linear inequality in one variable.
• Linear inequality in two variables.
• Slack inequality.

5x < 6, 8x + 3y ≤ 5, 2x – 5 < 9 , 2x ≤ 9 , 2x + 3y < 10.

2.) Solve y < 2 graphically.

Solution: Graph of y = 2. So we can show it graphically as given below:

Let us select a point, (0, 0) in the lower half-plane I and putting y = 0 in the given inequality, we see that: 1 × 0 < 2 or 0 < 2 which is true. Thus, the solution region is the shaded region below the line y = 2.

Hence, every point below the line (excluding all the points on the line) determines the solution of the given inequality.

Linear Inequalities in Two Variables Word Problem

In an experiment, a solution of hydrochloric acid is to be kept between 25° and 30° Celsius. What is the range of temperature in degree Fahrenheit if the conversion formula is given by C = 5/9 (F – 32), where C and F represent the temperature in degree Celsius and degree Fahrenheit, respectively.

Solution: As per the question it is given:

25<C<30

Now if we put C = 5/9 (F – 32), we get;

25 < 5/9 (F – 32) < 30

9/5 x 25 < F – 32 < 30 x 9/5

45 < F -32< 54

77 < F < 86

Thus, the required range of temperature is between 77° F and 86° F.

Frequently Asked Questions – FAQs

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Systems of Linear Inequalities, Word Problems - Examples - Expii

Systems of linear inequalities, word problems - examples, explanations (3), word problems with systems of linear inequalities.

Solving word problems with systems of linear inequalities involves 2 steps:

Translate the word problem into mathematical expressions. We're finding a system of linear inequalities .

Solve! One strategy you can use to find solutions is to graph each inequality . Then, use the overlapping areas of the graphs to identify the solutions to the word problem.

Let's look at an example:

Image source: by Hannah Bonville

STEP 1: Translate the word problem into inequalities.

We're going to use x for the number of pizzas and y for the number of pounds of cookies.

Which inequality represents the number of pizzas Pilar needs?

Related Lessons

Systems of Linear Inequalities: Word Problems

When you are presented with a word problem that deals with inequalities, it may seem daunting at first to solve the problem. The good news is that there are two simple steps to follow in order to solve systems of linear inequalities!

Image source: by Anusha Rahman

Let's walk through an example problem together.

Jenn and John want to go on a road trip together. Jenn drives an average of 50mph, and John drives an average of 45mph. They want to drive less than 20 hours, but at least 500 miles. Write a system of linear inequalities to figure out how many miles John and Jenn must drive to make that happen.

Step 1: Translate the word problem into inequalities.

Let's start with establishing our variables. x=Jenn y=John

We know that Jenn and John want to drive at least 500 miles. At least generally means greater than or equal to . So if we incorporate the speed at which the two of them drive, we can say that:

50x+45y≥500

We also know that, combined, John and Jenn want to drive less than 20 hours. So, we can say that: x+y=20

Therefore, our system of linear equations is:

50x+45y≥500 x+y=20

Step 2: Solve the inequalities.

Let's solve these inequalities by graphing. Here is a graph of the two inequalities.

Made using Desmos

50x+45y≥500 is represented by the red shading. x+y=20 is represented by the blue shading. The overlap, or the solution, is represented by the purple shading.

(Video) Solving Inequalities: Word Problems

by Anusha Rahman

This video by Anusha Rahman walks you through an example word problem of how to solve a system of inequalities.

Example: Jackson is having a party for his friends on Saturday. Jackson expects at least 7 people to show up on Saturday. He is trying to figure out the total costs of all of the pizzas. He calls his local pizza place, Meredith’s pizza, who says that the cost of his pizzas will be less than 5 more than 2 times the number of people he needs pizza for.

Step 1: Establish variables x=the number of people y=the cost of all pizzas

Step 2: Set up the inequalities x≥7 y<5+2x

Step 3: Graph the inequalities to find the solution

Graph of: x≥7

Graph of y=5+2x

This is the graph of the line, y=5+2x. In order to graph the inequality, y<5+2x, we have to figure out what area on the graph to shade. To do that, we test out some points and see whether the inequality remains true!

Let's test the point (10,2). 2<2(10)+5 2<20+5 2<25 This statement is true ! So we will probably shade in this area, the bottom area of the graph. But, let's just check one more point to make sure.

Let's test the point (−10,2). 2<2(−10)+5 2<−20+5 2<−15 This statement is false .

So, we can see in the above graph that the area under the dotted line is shaded. This is the graph of y<5+2x.

Now, let's put both inequalities on the same graph to see the area of intersection!

The area shaded in purple is the solution to the system of inequalities! Because Jackson probably won't have negative friends or pay negative dollars, we can assume 0">x>0 and 0">y>0.

Jackson won’t know for certain what the price of his pizzas will be. Why? Because Jackson doesn’t know how many of his friends will be coming to his party. Especially when solving real-world problems, a system of inequalities is really helpful when we’re looking at something in the future, and we don’t have exact numbers, but we have a general idea, like Jackson does.