Standard Form: Rewriting and Graphing Lines | Grade 8 Math | LevelKite

The equations $2x + y = 6$ and $y = -2x + 6$ look different, yet they describe exactly the same line. This lesson shows you why, and how to use that fact to draw a graph.

A linear equation in $x$ and $y$ is an equation whose solutions form a straight line when you plot them. Here $x$ and $y$ stand for real numbers, and a solution is a pair $(x, y)$ that makes the equation true. The graph is the set of all solutions, drawn as points on a coordinate plane.

The first equation is in standard form, $Ax + By = C$ . The letters $A$ , $B$ and $C$ stand for fixed numbers (integers in this lesson), and $A$ and $B$ are not both $0$ . The numbers $A$ and $B$ are coefficients: they multiply $x$ and $y$ . Both variable terms sit on one side, and the constant $C$ sits on the other. In $2x + y = 6$ , you have $A = 2$ , $B = 1$ and $C = 6$ .

The second equation is in slope-intercept form, $y = mx + b$ , where $y$ stands alone. The number $m$ is the slope: the amount $y$ changes each time $x$ increases by $1$ . The number $b$ is the y-intercept: the $y$ -value where the line crosses the $y$ -axis, so the point $(0, b)$ is on the line. In $y = -2x + 6$ , the slope is $-2$ and the y-intercept is $6$ .

Standard form hides the slope and the y-intercept, and slope-intercept form shows them. Changing the form does not change the line, as long as every step keeps the same solutions. Adding or subtracting the same quantity on both sides does that. So does multiplying or dividing both sides by the same number, provided that number is not $0$ .

Subtracting $2x$ from both sides of $2x + y = 6$ gives $y = -2x + 6$ . The table tests four points in both equations.

Each point passes both tests or fails both tests.

Point $(x, y)$	Test in $2x + y = 6$	Test in $y = -2x + 6$	On the line?
$(0, 6)$	$2(0) + 6 = 6$ , true	$-2(0) + 6 = 6$ , matches $y = 6$	Yes
$(1, 4)$	$2(1) + 4 = 6$ , true	$-2(1) + 6 = 4$ , matches $y = 4$	Yes
$(3, 0)$	$2(3) + 0 = 6$ , true	$-2(3) + 6 = 0$ , matches $y = 0$	Yes
$(2, 5)$	$2(2) + 5 = 9$ , not $6$	$-2(2) + 6 = 2$ , but $y = 5$	No

Key concept

Solve for y to change forms

A standard-form equation and its slope-intercept form are two descriptions of the same line. To change forms, solve for $y$ : subtract the $x$ -term from both sides, then divide every term by $B$ , the coefficient of $y$ . Each step does the same thing to both sides, so the solutions and the graph stay the same.

Condition: you need $B \neq 0$ , because you cannot divide by $0$ . If $B = 0$ , the equation has no $y$ -term, as in $x = 3$ , and its graph is a vertical line. A vertical line has an undefined slope, because $x$ never changes along it. It is also not a function $y = f(x)$ : a function gives exactly one $y$ -value for each $x$ -value, but on $x = 3$ the single $x$ -value goes with every $y$ -value. So it has no slope-intercept form.

Ax + By = C \quad\Longrightarrow\quad y = -\dfrac{A}{B}x + \dfrac{C}{B} \qquad (B \neq 0)

Worked example

Convert when y has a positive coefficient

Write $6x + 2y = 10$ in slope-intercept form. Then name the slope and the y-intercept.

The goal is to get y alone on the left side. Subtract 6x from both sides. Taking the same quantity from both sides keeps exactly the same solutions.
$6x + 2y - 6x = 10 - 6x \quad\Rightarrow\quad 2y = 10 - 6x$
Divide both sides by 2, the coefficient of y. Because 2 is not 0, this keeps the same solutions. Divide every term on the right side, not only the first one.
$y = \dfrac{10}{2} - \dfrac{6x}{2} = 5 - 3x$
Write the x-term first to match the pattern y = mx + b. Changing the order of the two terms does not change the result, as long as the minus sign stays with 3x.
$y = -3x + 5$
Read the two numbers. The slope m is the number that multiplies x, and the y-intercept b is the constant term.
$m = -3, \qquad b = 5$
Check with two different points from the new equation: x = 0 gives y = 5, and x = 1 gives y = -3(1) + 5 = 2. Both equations graph as straight lines, and two different points fix exactly one straight line. So if both points also make the original equation true, the two equations describe the same line.
$6(0) + 2(5) = 0 + 10 = 10 \qquad 6(1) + 2(2) = 6 + 4 = 10$

Answer

In slope-intercept form, $6x + 2y = 10$ is $y = -3x + 5$ . The slope is $-3$ and the y-intercept is $5$ .

Worked example

Convert, then graph

Write $3x - 2y = 6$ in slope-intercept form. Then graph the line.

Subtract 3x from both sides so that only the y-term stays on the left. The same subtraction on both sides keeps the same solutions. Each minus sign stays with its term: the left side is -2y, and the right side is 6 - 3x, written here with the x-term first.
$3x - 2y - 3x = 6 - 3x \quad\Rightarrow\quad -2y = -3x + 6$
Divide both sides by -2, the coefficient of y. It is not 0, so the solutions stay the same. Dividing by a negative number changes the sign of each term here: -3x divided by -2 is positive, and 6 divided by -2 is negative.
$y = \dfrac{-3x}{-2} + \dfrac{6}{-2} = \dfrac{3}{2}x - 3$
Read the form. The y-intercept is -3, so the line crosses the y-axis at (0, -3). The slope is 3/2, so y rises by 3/2 for each 1 that x increases. That is the same as rising by 3 each time x increases by 2.
$m = \dfrac{3}{2}, \qquad b = -3$
Find two more points, as you would for a table of values. Choose x-values that are multiples of 2, here x = 2 and x = 4, so that the fraction gives whole-number y-values. The new points are (2, 0) and (4, 3).
$y = \dfrac{3}{2}(2) - 3 = 3 - 3 = 0 \qquad y = \dfrac{3}{2}(4) - 3 = 6 - 3 = 3$
Check the three points (0, -3), (2, 0) and (4, 3) in the original standard-form equation. Each one gives 6, so each one is a solution.
$3(0) - 2(-3) = 0 + 6 = 6 \qquad 3(2) - 2(0) = 6 - 0 = 6 \qquad 3(4) - 2(3) = 12 - 6 = 6$
Plot the three points and draw the straight line through them. Extend the line past the points in both directions, because x can be any real number.

Answer

The equation $3x - 2y = 6$ becomes $y = \frac{3}{2}x - 3$ . Its graph is the straight line through $(0, -3)$ , $(2, 0)$ and $(4, 3)$ , shown as Line A in the figure below.

Line A is

3x - 2y = 6

. Point A is

(0, -3)

, Point B is

(2, 0)

and Point C is

(4, 3)

.

In the figure, the horizontal axis shows $x$ and the vertical axis shows $y$ . Both are plain numbers with no units, and one grid step is $1$ on each axis. Line A is the graph of $3x - 2y = 6$ . Point A is $(0, -3)$ , Point B is $(2, 0)$ and Point C is $(4, 3)$ : the three points from the example.

Trust the coordinates, not the picture. A point can look as if it sits on a line and still miss it. A point is on the line only when its exact coordinates make the equation true, which is why you checked all three by substitution. The coordinates also show the slope. From Point A to Point B, $x$ goes from $0$ to $2$ and $y$ goes from $-3$ to $0$ . So $y$ increases by $3$ while $x$ increases by $2$ , which matches the slope $\frac{3}{2}$ . The same change takes you from Point B to Point C.

Worked example

Graph straight from standard form

Graph $2x + 3y = 12$ without converting first. Then use slope-intercept form to confirm your graph.

You need points that make the equation true, and 0 is the easiest value to substitute. Let x = 0. The x-term becomes 0, leaving 3y = 12. Divide both sides by 3, which is not 0. This gives the point (0, 4), on the y-axis.
$2(0) + 3y = 12 \quad\Rightarrow\quad 3y = 12 \quad\Rightarrow\quad y = 4$
Now let y = 0. The y-term becomes 0, leaving 2x = 12. Divide both sides by 2, which is not 0. This gives the point (6, 0), on the x-axis.
$2x + 3(0) = 12 \quad\Rightarrow\quad 2x = 12 \quad\Rightarrow\quad x = 6$
Two different points fix a line, but find a third to check your arithmetic. Let x = 3, so the x-term is 6. Subtract 6 from both sides, then divide both sides by 3. This gives the point (3, 2).
$2(3) + 3y = 12 \quad\Rightarrow\quad 3y = 6 \quad\Rightarrow\quad y = 2$
Plot (0, 4), (3, 2) and (6, 0). Draw the straight line through them and extend it in both directions.
Confirm with slope-intercept form. Subtract 2x from both sides, then divide every term by 3, which is not 0.
$3y = -2x + 12 \quad\Rightarrow\quad y = -\dfrac{2}{3}x + 4$
Compare. The y-intercept 4 matches the point (0, 4). The slope -2/3 means y falls by 2 each time x rises by 3, which is exactly the change from (0, 4) to (3, 2) and from (3, 2) to (6, 0). Keep -2/3 as an exact fraction; a rounded decimal such as -0.67 is only an approximation.

Answer

The graph of $2x + 3y = 12$ is the straight line through $(0, 4)$ , $(3, 2)$ and $(6, 0)$ , shown as Line B in the figure below. In slope-intercept form, the same line is $y = -\frac{2}{3}x + 4$ .

Line B is

2x + 3y = 12

. Point D is

(0, 4)

, Point E is

(3, 2)

and Point F is

(6, 0)

.

In this figure the horizontal axis again shows $x$ , the vertical axis shows $y$ , and one grid step is $1$ on each axis. Line B is the graph of $2x + 3y = 12$ . Point D is $(0, 4)$ , Point E is $(3, 2)$ and Point F is $(6, 0)$ . Point D is on the $y$ -axis at the y-intercept $4$ . From Point D to Point E, and again from Point E to Point F, $x$ increases by $3$ while $y$ decreases by $2$ . That matches the slope $-\frac{2}{3}$ .

You now have two routes from standard form to a graph. When $B \neq 0$ , you can solve for $y$ first and build points from $y = mx + b$ . Or you can substitute values straight into the standard form. Both routes give points on the same line, because the two equations have exactly the same solutions. If both substitutions of $0$ lead to the same point, such as $(0, 0)$ , choose another $x$ -value to get a second point. Either way, check your points in the original equation.

Takeaway: when $B \neq 0$ , the standard form $Ax + By = C$ and the slope-intercept form $y = mx + b$ are two ways to write the same line. To change forms, solve for $y$ : subtract the $x$ -term from both sides, then divide every term by $B$ , and watch the signs. To graph, plot points whose exact coordinates make the equation true and draw the straight line through them. Next, you will explore how changing the slope or the y-intercept changes the graph.