Houghton Mifflin Harcourt Algebra 1, 2015
HM
Houghton Mifflin Harcourt Algebra 1, 2015 View details
2. Using Intercepts
Continue to next subchapter

Exercise 6 Page 177

Substitute 0 for each of the variables and then solve the equation for the other variable.

x-intercept: -8
y-intercept: 12

Practice makes perfect

To determine the x-intercept for the given equation, we will substitute 0 for y and solve for x. To determine the y-intercept, we will substitute 0 for x and solve for y.

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. Consider the given equation. -9x+6y=72Let's substitute 0 for y and solve for x.
-9x+6y=72
-9x+6( 0)=72
-9x=72
x=-8
The x-intercept of the equation is -8.

Finding the y-intercept

In the same way as above, now 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.
-9x+6y=72
-9( 0)+6y=72
6y=72
y=12
The y-intercept of the equation is 12.