Insights on Graphing a Linear Equation Using a Table

Equation	Solution	Point
$3 = 2 x + 1$	$x = 1$	$(1, 3)$
$5 = 2 x + 1$	$x = 2$	$(2, 5)$
$7 = 2 x + 1$	$x = 3$	$(3, 7)$

Equation

Solution

Point

3 = 2 x + 1

x = 1

(1, 3)

5 = 2 x + 1

x = 2

(2, 5)

7 = 2 x + 1

x = 3

(3, 7)

$x$	$y$
$0$	$0$
$1$	$3$
$2$	$6$
$3$	$9$
$4$	$12$
$5$	$15$

x

y

0

0

1

3

2

6

3

9

4

12

5

15

x

2 x + 4

y = 2 x + 4

- 2

0

0.5

1

$x$	$2 x + 4$	$y = 2 x + 4$
$- 2$	$2 (- 2) + 4$
$0$	$2 (0) + 4$
$0.5$	$2 (0.5) + 4$
$1$	$2 (1) + 4$

x

2 x + 4

y = 2 x + 4

- 2

2 (- 2) + 4

0

2 (0) + 4

0.5

2 (0.5) + 4

1

2 (1) + 4

$x$	$2 x + 4$	$y = 2 x + 4$
$- 2$	$2 (- 2) + 4$	$0$
$0$	$2 (0) + 4$	$4$
$0.5$	$2 (0.5) + 4$	$5$
$1$	$2 (1) + 4$	$6$

x

2 x + 4

y = 2 x + 4

- 2

2 (- 2) + 4

0

0

2 (0) + 4

4

0.5

2 (0.5) + 4

5

1

2 (1) + 4

6

$x$	$2 x + 4$	Simplify	$y = 2 x + 4$
$- 2$	$2 (- 2) + 4$	$- 4 + 4$	$0$
$0$	$2 (0) + 4$	$0 + 4$	$4$
$0.5$	$2 (0.5) + 4$	$1 + 4$	$5$
$1$	$2 (1) + 4$	$2 + 4$	$6$

x

2 x + 4

Simplify

y = 2 x + 4

- 2

2 (- 2) + 4

- 4 + 4

0

0

2 (0) + 4

0 + 4

4

0.5

2 (0.5) + 4

1 + 4

5

1

2 (1) + 4

2 + 4

6

$6 x - 3 y = - 12$
$x$	$y$
$- 2$	$0$
$0$	$4$
$0.5$	$5$
$1$	$6$

6 x - 3 y = - 12

x

y

- 2

0

0

4

0.5

5

1

6

$x$	$y = 0.01 x + 6$	$y$
$350$	$y = 0.01 (350) + 6$	$9.5$
$410$	$y = 0.01 (410) + 6$	$10.1$
$550$	$y = 0.01 (550) + 6$	$11.5$
$600$	$y = 0.01 (600) + 6$	$12$

x

y = 0.01 x + 6

y

350

y = 0.01 (350) + 6

9.5

410

y = 0.01 (410) + 6

10.1

550

y = 0.01 (550) + 6

11.5

600

y = 0.01 (600) + 6

12

Number of Minutes	Total Cost
$200$	$$ 9$
$300$	$$ 9$
$350$	$$ 9.5$
$410$	$$ 10.1$
$550$	$$ 11.5$
$600$	$$ 12$

Number of Minutes

Total Cost

200

$ 9

300

$ 9

350

$ 9.5

410

$ 10.1

550

$ 11.5

600

$ 12

$x$	$y = 10 x + 50$	$y$
$35$	$y = 10 (35) + 50$	$400$
$95$	$y = 10 (95) + 50$	$1000$
$125$	$y = 10 (125) + 50$	$1300$

x

y = 10 x + 50

y

35

y = 10 (35) + 50

400

95

y = 10 (95) + 50

1000

125

y = 10 (125) + 50

1300

First, the given equation will be graphed. Then the obtained graph will be used to determine the maximum number of books that Maya can purchase.

Graphing the Equation

Before making the table of values for the given equation, one of the variables has to be isolated. It is more common to isolate

y,

so let it also be the case here.

3 x + 2 y = 46

▼

Solve for

y

SubEqn

$LHS - 3 x = RHS - 3 x$

2 y = - 3 x + 46

DivEqn

$LHS / 2 = RHS / 2$

y = \frac{- 3 x + 4 6}{2}

WriteSumFrac

Write as a sum of fractions

y = \frac{- 3 x}{2} + \frac{4 6}{2}

CalcQuot

Calculate quotient

y = - 1.5 x + 23

Now that

y

is isolated, the table of values can be constructed. The graph of a linear equation in two variables is a line. Through any two different points, there is exactly one line. Therefore, at least two different values of

x

need to be evaluated in the table of values.

$x$	$y = - 1.5 x + 23$	$y$
$0$	$y = - 1.5 (0) + 23$	$23$
$2$	$y = - 1.5 (2) + 23$	$20$
$4$	$y = - 1.5 (4) + 23$	$17$
$10$	$y = - 1.5 (10) + 23$	$8$

Note that in order to simplify the graphing process, the chosen $x -$ values are all even numbers. Each pair of $x$ and $y$ corresponds to a coordinate pair $(x, y) .$ Now $(0, 23),$ $(2, 20),$ $(4, 17),$ and $(10, 8)$ will be plotted on the same coordinate plane.

Finally, the four points can be connected using a straightedge.

Since the amount $y$ of money left and the number $x$ of books purchased cannot be negative, the graph is bounded only to the first quadrant. This also means that only part of the line is actually considered, which makes it a segment.

Finding the Maximum Number of Books

Note that the number of books has to be a whole number, as Maya cannot buy part of a book. Also, she cannot spend more money than she has, which means that the amount of money left has to be non-negative. A point which meets these conditions, the maximum value of the $x -$ coordinate, can be identified on the graph.

The $x -$ coordinate of the point corresponds to the maximum number of books that Maya can buy. As seen on the graph, its value is $15 .$ Therefore, Maya can purchase at most $15$ books.

$x$	$y = 2 x + 1$	$y$
$0$	$y = 2 (0) + 1$	$1$
$1$	$y = 2 (1) + 1$	$3$

x

y = 2 x + 1

y

0

y = 2 (0) + 1

1

1

y = 2 (1) + 1

3

$x$	$y = 5 x - 5$	$y$
$0$	$y = 5 (0) - 5$	$- 5$
$1$	$y = 5 (1) - 5$	$0$

x

y = 5 x - 5

y

0

y = 5 (0) - 5

- 5

1

y = 5 (1) - 5

0

Expression	$x = 2$	Simplify
$2 x + 1$	$2 (2) + 1$	$5$
$5 x - 5$	$5 (2) - 5$	$5$

Expression

x = 2

Simplify

2 x + 1

2 (2) + 1

5

5 x - 5

5 (2) - 5

5

{{ article.displayTitle }}

Catch-Up and Review

Plotting Points Corresponding to Linear Equations

A Table of Values and How to Make It

Phone Plan

Hint

Solution

Linear Equation and Graph of Its Solutions

Identifying Linear Equations

December Pay

Answer

Hint

Solution

Graphing a Linear Equation in Standard Form

Answer

Hint

Solution

Graphing the Equation

Finding the Maximum Number of Books

Matching a Line to a Table of Values

Solving a Linear Equation by Graphing

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}