Solutions, Tables, and Lines | Grade 8 Math | LevelKite

You know slope-intercept form, $y = mx + b$ , where the slope $m$ and the $y$ -intercept $b$ are fixed numbers. In this lesson, a linear equation is any equation in $x$ and $y$ that can be written this way. The variables $x$ and $y$ each stand for a real number, and $x$ may be any real number.

What makes such an equation true? One number alone cannot do it. You need a value for $x$ and a value for $y$ together. You write them as an ordered pair $(x, y)$ , with the $x$ -value always first and the $y$ -value second. The two numbers are the pair's coordinates.

Try the pair $(1, 3)$ in $y = 2x + 1$ . Substitute it, which means replace each variable with its number: $3 = 2(1) + 1$ . The right side is $2 + 1 = 3$ , so the equation says $3 = 3$ , which is true. A pair that makes the equation true is a solution of the equation. Now try $(1, 5)$ . The right side is still $3$ , but the left side is $5$ . The equation says $5 = 3$ , which is false, so $(1, 5)$ is not a solution.

Worked example

Testing two ordered pairs

Decide whether each ordered pair is a solution of $y = 2x + 1$ : first $(3, 7)$ , then $(-2, -1)$ .

Read the pair (3, 7) in order. An ordered pair is always written (x, y), so the first number is the x-value and the second number is the y-value.
$x = 3, \quad y = 7$
Substitute 3 for x on the right side of the equation. This is allowed because x stands for exactly this number in the pair. Multiply first, then add.
$2(3) + 1 = 6 + 1 = 7$
Compare with the left side. The left side is y, and y is 7 for this pair. Both sides are the same number, so the equation is true and (3, 7) is a solution.
$7 = 7$
Now read (-2, -1): the x-value is -2 and the y-value is -1. Substitute -2 for x. A positive number times a negative number is negative, so 2 times -2 is -4.
$2(-2) + 1 = -4 + 1 = -3$
Compare with the left side, which is y = -1. The two sides are different numbers, so the equation is false and (-2, -1) is not a solution.
$-1 \neq -3$

Answer

The pair $(3, 7)$ is a solution of $y = 2x + 1$ , and the pair $(-2, -1)$ is not a solution.

Line A is the line

y = 2x + 1

. Point A is at

(3, 7)

, Point B is at

(7, 3)

and Point C is at

(-2, -1)

.

In this figure, and in every figure in this lesson, the horizontal axis shows $x$ and the vertical axis shows $y$ . Each grid square is 1 unit. The axes have no measurement unit, because $x$ and $y$ are plain numbers here. Line A is the line with equation $y = 2x + 1$ .

Point A is at $(3, 7)$ , the solution from the example, and it lies on Line A. Point B is at $(7, 3)$ and Point C is at $(-2, -1)$ . Neither pair is a solution, and neither point lies on Line A. Swapping the coordinates of Point A gives Point B, a different point in a different place.

Line A crosses the $y$ -axis at $(0, 1)$ , not at the origin. The equation agrees: $1 = 2(0) + 1$ is true, but $0 = 2(0) + 1$ says $0 = 1$ , which is false. Because Line A misses the origin, $y = 2x + 1$ is linear but not proportional. A proportional relationship has the form $y = kx$ for a fixed number $k$ , and its line passes through the origin.

Key concept

The line is every solution

A solution of a linear equation in $x$ and $y$ is an ordered pair $(x, y)$ that makes the equation true when you substitute the first number for $x$ and the second number for $y$ .

The graph of the equation is the set of all points whose coordinates are solutions. For an equation $y = mx + b$ , where $m$ and $b$ are fixed numbers and $x$ can be any real number, the graph is one straight line. Every solution is a point on that line, and every point on that line is a solution.

So you can test any point by substituting its exact coordinates, and you can draw the line by plotting a few solutions.

A table of values is an organized list of solutions. You choose some $x$ -values, substitute each one to compute its $y$ -value, and write each pair in its own row. Each row is a solution, because you computed $y$ to make the equation true. You may choose any $x$ -values, because $x$ can be any real number. Small integers, including a negative number and $0$ , keep the arithmetic simple.

How many rows do you need? Exactly one straight line passes through two different points, so two different solutions already fix the line. Use three. The third row is a check on your arithmetic.

Here is the check, and it uses slope. On a line $y = mx + b$ , two different points always have different $x$ -values, so the change in $x$ between them is not $0$ and you can divide by it. The change in $y$ divided by the change in $x$ equals the slope $m$ for every such pair of points. If your table gives any other number, a $y$ -value is wrong.

Worked example

From an equation to a table to a line

Make a table of values for $y = -2x + 3$ using $x = -1$ , $x = 0$ and $x = 2$ . Then graph the line.

Substitute -1 for x. The product of two negative numbers is positive, so -2 times -1 is 2. Then add 3. The first row of the table is the pair (-1, 5).
$y = -2(-1) + 3 = 2 + 3 = 5$
Substitute 0 for x. Any number times 0 is 0, so only the 3 is left. The second row is the pair (0, 3), the point where the line crosses the y-axis.
$y = -2(0) + 3 = 0 + 3 = 3$
Substitute 2 for x. A negative number times a positive number is negative, so -2 times 2 is -4. Then add 3. The third row is the pair (2, -1).
$y = -2(2) + 3 = -4 + 3 = -1$
Plot each pair, starting at the origin. The x-value tells you how far to move right (positive) or left (negative). The y-value tells you how far to move up (positive) or down (negative). So (-1, 5) is 1 unit left and 5 units up, (0, 3) is 3 units up on the y-axis, and (2, -1) is 2 units right and 1 unit down.
Check with slope before you draw. From (-1, 5) to (0, 3), x changes by 0 - (-1) = 1 and y changes by 3 - 5 = -2. From (0, 3) to (2, -1), x changes by 2 - 0 = 2 and y changes by -1 - 3 = -4. Neither change in x is 0, so you can divide. Both quotients equal the slope -2 from the equation, so the three points lie on one straight line.
$\frac{-2}{1} = -2 \qquad \frac{-4}{2} = -2$
Draw the straight line through the three points and extend it past them in both directions. The line keeps going because x can be any real number, not only the three values in the table.

Answer

The graph of $y = -2x + 3$ is the straight line through $(-1, 5)$ , $(0, 3)$ and $(2, -1)$ , extended in both directions.

Table of values for y = -2x + 3

x	Substitute into y = -2x + 3	y	Ordered pair (x, y)
$-1$	$-2(-1) + 3 = 2 + 3$	$5$	$(-1, 5)$
$0$	$-2(0) + 3 = 0 + 3$	$3$	$(0, 3)$
$2$	$-2(2) + 3 = -4 + 3$	$-1$	$(2, -1)$

Points A, B and C are the table pairs

(-1, 5)

,

(0, 3)

and

(2, -1)

. Point D at

(1, 1)

is another solution on Line A, the graph of

y = -2x + 3

.

This figure shows the finished graph. As before, the horizontal axis shows $x$ , the vertical axis shows $y$ , and each grid square is 1 unit. Each row of the table became one point: Point A is $(-1, 5)$ , Point B is $(0, 3)$ and Point C is $(2, -1)$ . Line A passes through all three, and it is the graph of $y = -2x + 3$ .

The table holds only three solutions, but the line holds every solution. Point D at $(1, 1)$ was never in the table, yet it lies on Line A. Substituting confirms it: $-2(1) + 3 = -2 + 3 = 1$ . The coordinates of a solution do not have to be integers either. For $x = 0.5$ , you get $y = -2(0.5) + 3 = -1 + 3 = 2$ , so $(0.5, 2)$ is a solution too.

A drawing can make a point look as if it is on a line when it is slightly off. To be sure, substitute the exact coordinates.

Worked example

A slope that is a fraction

Graph $y = \frac{1}{2}x - 2$ from a table of values. Then decide whether $(3, -1)$ is a solution.

You may use any x-values, but multiples of 2 are convenient here. The slope has denominator 2, and half of an even number is an integer, so each y-value will be an integer that sits exactly on a grid crossing. Use -6, 0 and 6.
Substitute -6 for x. Half of -6 is -3, and -3 minus 2 is -5. The first pair is (-6, -5).
$\frac{1}{2}(-6) - 2 = -3 - 2 = -5$
Substitute 0, then 6, in the same way. Half of 0 is 0 and half of 6 is 3. Subtract 2 each time. The other two pairs are (0, -2) and (6, 1).
$\frac{1}{2}(0) - 2 = 0 - 2 = -2 \qquad \frac{1}{2}(6) - 2 = 3 - 2 = 1$
Check with slope. From (-6, -5) to (0, -2), x changes by 0 - (-6) = 6 and y changes by -2 - (-5) = 3. From (0, -2) to (6, 1), x changes by 6 - 0 = 6 and y changes by 1 - (-2) = 3. Each quotient is 3 divided by 6, which equals the slope one half, so the points lie on one straight line. Plot them and draw the line through them, extended in both directions.
$\frac{3}{6} = \frac{1}{2}$
Test (3, -1) by substituting its exact coordinates, x = 3 and y = -1. Half of 3 is exactly 1.5, and 1.5 minus 2 is -0.5. The left side is y = -1, and -1 is not equal to -0.5, so the equation is false.
$\frac{1}{2}(3) - 2 = 1.5 - 2 = -0.5 \neq -1$
Compare with the figure that follows. Points A, B and C are the table pairs (-6, -5), (0, -2) and (6, 1), and Line A passes through all three. Point D is (3, -1). It sits 0.5 units directly below (3, -0.5), the point of Line A with x = 3. That gap is small on the grid, which is why you substitute instead of judging by eye.

Answer

The graph of $y = \frac{1}{2}x - 2$ is the line through $(-6, -5)$ , $(0, -2)$ and $(6, 1)$ . The pair $(3, -1)$ is not a solution; the solution with $x = 3$ is $(3, -0.5)$ .

Points A, B and C are the table pairs

(-6, -5)

,

(0, -2)

and

(6, 1)

on Line A, the graph of

y = \frac{1}{2}x - 2

. Point D at

(3, -1)

is not on Line A. The horizontal axis shows

x

; the vertical axis shows

y

.

Takeaway. A solution of a linear equation is an ordered pair $(x, y)$ , with $x$ first and $y$ second, that makes the equation true. For an equation $y = mx + b$ , all of the solutions together form one straight line. So a table of values gives you exact points to plot, the line through them is the graph, and substituting exact coordinates tells you whether any other point belongs on it.

Next, you will graph horizontal and vertical lines. Vertical lines have an undefined slope, so they cannot be written as $y = mx + b$ and need their own approach.