writing linear equations from a table practice and problem solving c answers

Linear Equation Table

Part i. how linear equations relate to tables of values, equations as relationships.

The equation of a line expresses a relationship between x and y values on the coordinate plane. For instance, the equation $$y = x$$ expresses a relationship where every x value has the exact same y value. The equation $$ y = 2x $$ expresses a relationship in which every y value is double the x value, and $$ y = x + 1 $$ expresses a relationship in which every y value is 1 greater than the x value.

So what about a Table Of Values?

Since, as we just wrote, every linear equation is a relationship of x and y values, we can create a table of values for any line. These are just the $$ x $$ and $$ y $$ values that are true for the given line. In other words, a table of values is simply some of the points that are on the line.

Equation: $$ \red y = \blue x + 1 $$

Table of Values

Equation: y = 3x + 2

So, to create a table of values for a line, just pick a set of x values, substitute them into the equation and evaluate to get the y values.

Practice Creating a Table of Values

• Original problem

Create a table of values of the equation y = 5x + 2 .

Create the table and choose a set of x values.

Substitute each x value (left side column) into the equation.

Evaluate the equation (middle column) to arrive at the y value.

An Optional step, if you want, you can omit the middle column from your table, since the table of values is really just the x and y pairs. (We used the middle column simply to help us get the y values)

Create a table of values of the equation y = −6x + 2 .

An Optional step, if you want, you can omit the middle column from your table, since the table of values is really just the x and y pairs . (We used the middle column simply to help us get the y values)

Create a table of values of the equation y = −6x − 4

Create the table and choose a set of x values

Part II. Writing Equation from Table of Values

Often, students are asked to write the equation of a line from a table of values. To solve this kind of problem, simply chose any 2 points on the table and follow the normal steps for writing the equation of a line from 2 points .

Write the equation of a line from the table of values below.

Choose any two x, y pairs from the table and calculate the slope . Since, I like to work with easy, small numbers I chose (0, 3) and (1, 7).

Substitute slope into the slope intercept form of a line .

y = mx + b y = 4x + b

Find the value of 'b' in the slope intercept equation .

Since our table gave us the point (0, 3) we know that 'b' is 3. Remember 'b' is the y-intercept which, luckily, was supplied to us in the table.

Answer: y = 4x + 3

If you'd like, you could check your answer by substituting the values from the table into your equation. Each and every x, y pair from the table should work with your answer.

Write the equation from the table of values provided below.

Choose any two x, y pairs from the table and calculate the slope . I chose (2, 8) and (4, 9).

y = mx + b y = ½x + b

Now that we know the value of b, we can substitute it into our equation.

Answer: y = ½x + 7

Why can you not write the equation of a line from the table of values below?

The reason that this table could not represent the equation of a line is because the slope is inconsistent. For instance the slope of the 2 points at the top of the table (0, 1) and (1, 3) is different from the slope at the bottom (2, 8) and (3, 11).

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8.E: Solving Linear Equations (Exercises)

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8.1 - Solve Equations using the Subtraction and Addition Properties of Equality

In the following exercises, determine whether the given number is a solution to the equation.

• x + 16 = 31, x = 15
• w − 8 = 5, w = 3
• −9n = 45, n = 54
• 4a = 72, a = 18

In the following exercises, solve the equation using the Subtraction Property of Equality.

• y + 2 = −6
• a + \(\dfrac{1}{3} = \dfrac{5}{3}\)
• n + 3.6 = 5.1

In the following exercises, solve the equation using the Addition Property of Equality.

• u − 7 = 10
• x − 9 = −4
• c − \(\dfrac{3}{11} = \dfrac{9}{11}\)
• p − 4.8 = 14

In the following exercises, solve the equation.

• n − 12 = 32
• y + 16 = −9
• f + \(\dfrac{2}{3}\) = 4
• d − 3.9 = 8.2
• y + 8 − 15 = −3
• 7x + 10 − 6x + 3 = 5
• 6(n − 1) − 5n = −14
• 8(3p + 5) − 23(p − 1) = 35

In the following exercises, translate each English sentence into an algebraic equation and then solve it.

• The sum of −6 and m is 25.
• Four less than n is 13.

In the following exercises, translate into an algebraic equation and solve.

• Rochelle’s daughter is 11 years old. Her son is 3 years younger. How old is her son?
• Tan weighs 146 pounds. Minh weighs 15 pounds more than Tan. How much does Minh weigh?
• Peter paid $9.75 to go to the movies, which was $46.25 less than he paid to go to a concert. How much did he pay for the concert?
• Elissa earned $152.84 this week, which was $21.65 more than she earned last week. How much did she earn last week?

8.2 - Solve Equations using the Division and Multiplication Properties of Equality

In the following exercises, solve each equation using the Division Property of Equality.

• 13a = −65
• 0.25p = 5.25
• −y = 4

In the following exercises, solve each equation using the Multiplication Property of Equality.

• \(\dfrac{n}{6}\) = 18
• y −10 = 30
• 36 = \(\dfrac{3}{4}\)x
• \(\dfrac{5}{8} u = \dfrac{15}{16}\)

In the following exercises, solve each equation.

• −18m = −72
• \(\dfrac{c}{9}\) = 36
• 0.45x = 6.75
• \(\dfrac{11}{12} = \dfrac{2}{3} y\)
• 5r − 3r + 9r = 35 − 2
• 24x + 8x − 11x = −7−14

8.3 - Solve Equations with Variables and Constants on Both Sides

In the following exercises, solve the equations with constants on both sides.

• 8p + 7 = 47
• 10w − 5 = 65
• 3x + 19 = −47
• 32 = −4 − 9n

In the following exercises, solve the equations with variables on both sides.

• 7y = 6y − 13
• 5a + 21 = 2a
• k = −6k − 35
• 4x − \(\dfrac{3}{8}\) = 3x

In the following exercises, solve the equations with constants and variables on both sides.

• 12x − 9 = 3x + 45
• 5n − 20 = −7n − 80
• 4u + 16 = −19 − u
• \(\dfrac{5}{8} c\) − 4 = \(\dfrac{3}{8} c\) + 4

In the following exercises, solve each linear equation using the general strategy.

• 6(x + 6) = 24
• 9(2p − 5) = 72
• −(s + 4) = 18
• 8 + 3(n − 9) = 17
• 23 − 3(y − 7) = 8
• \(\dfrac{1}{3}\)(6m + 21) = m − 7
• 8(r − 2) = 6(r + 10)
• 5 + 7(2 − 5x) = 2(9x + 1) − (13x − 57)
• 4(3.5y + 0.25) = 365
• 0.25(q − 8) = 0.1(q + 7)

8.4 - Solve Equations with Fraction or Decimal Coefficients

In the following exercises, solve each equation by clearing the fractions.

• \(\dfrac{2}{5} n − \dfrac{1}{10} = \dfrac{7}{10}\)
• \(\dfrac{1}{3} x + \dfrac{1}{5} x = 8\)
• \(\dfrac{3}{4} a − \dfrac{1}{3} = \dfrac{1}{2} a + \dfrac{5}{6}\)
• \(\dfrac{1}{2}\)(k + 3) = \(\dfrac{1}{3}\)(k + 16)

In the following exercises, solve each equation by clearing the decimals.

• 0.8x − 0.3 = 0.7x + 0.2
• 0.36u + 2.55 = 0.41u + 6.8
• 0.6p − 1.9 = 0.78p + 1.7
• 0.10d + 0.05(d − 4) = 2.05

PRACTICE TEST

• \(\dfrac{23}{5}\)
• n − 18 = 31
• 4y − 8 = 16
• −8x − 15 + 9x − 1 = −21
• −15a = 120
• \(\dfrac{2}{3}\)x = 6
• x + 3.8 = 8.2
• 10y = −5y + 60
• 8n + 2 = 6n + 12
• 9m − 2 − 4m + m = 42 − 8
• −5(2x + 1) = 45
• −(d + 9) = 23
• 2(6x + 5) − 8 = −22
• 8(3a + 5) − 7(4a − 3) = 20 − 3a
• \(\dfrac{1}{4} p + \dfrac{1}{3} = \dfrac{1}{2}\)
• 0.1d + 0.25(d + 8) = 4.1
• Translate and solve: The difference of twice x and 4 is 16.
• Samuel paid $25.82 for gas this week, which was $3.47 less than he paid last week. How much did he pay last week?

Contributors and Attributions

Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at http://cnx.org/contents/[email protected] ."

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Course: Algebra 1   >   Unit 5

• Slope from equation
• Writing linear equations in all forms

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• What is a linear equation?
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• 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.
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Writing Linear Equations from Tables Notes & Practice

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Description.

Writing Equations of the form y = mx + b from tables.

This is a one-sided notes page on how to write linear equations from tables. There are nine examples. In some of the examples, the y-intercept is missing, so the student needs to extend the pattern in the table to find it. In other examples, points are missing and so there seems to be no pattern. Students can either write a new table to find the missing values to verify that the table does represent a linear situation (this would be the preferred method) or can find the slope between two points from the table.

Two examples "don't work" as one is not a function and the other is not linear and in the final example, the points are out of order and some are missing.

Also included is an independent practice page with 9 more questions that can be assigned for homework. Key included.

Also available: Writing Equations of Lines from Tables Digital Drag & Drop Activity

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