McGraw Hill Glencoe Algebra 1, 2012
MH
McGraw Hill Glencoe Algebra 1, 2012 View details
1. Graphing Linear Equations
Continue to next subchapter

Exercise 24 Page 160

Substitute 0 for one variable and solve. Then, repeat for the other and graph.

Practice makes perfect

We will graph this equation by finding and plotting its intercepts, then connecting them with a line. To find the x- and y-intercepts, we will need to substitute 0 for one variable, solve, then repeat for the other variable.

Finding the x-intercept

Think of the point where the graph of an equation crosses the x-axis. The y-value of that ( x, y) coordinate pair is equal to 0, and the x-value is the x-intercept. To find the x-intercept of the given equation, we should substitute 0 for y and solve for x.
5-y=-3x
5-( 0)=-3x
5=-3 x
â–Ľ
Solve for x
5/-3=x
-5/3=x
x=-5/3
x=-1 23

An x-intercept of -1 23 means that the graph passes through the x-axis at the point ( -1 23,0).

Finding the y-intercept

Let's use the same concept to find the y-intercept. Consider the point where the graph of the equation crosses the y-axis. The x-value of the ( x, y) coordinate pair at the y-intercept is 0. Therefore, substituting 0 for x will give us the y-intercept.
5-y=-3 x
5-y=-3( 0)
5-y=0
â–Ľ
Solve for y
5=y
y=5
A y-intercept of 5 means that the graph passes through the y-axis at the point (0, 5).

Graphing the Equation

We can now graph the equation by plotting the intercepts and connecting them with a line.