Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
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Exercise 4 Page 190

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

x-intercept: -4
y-intercept: 3

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. -6x+8y=24As we have already said, let's substitute 0 for y and solve for x.
-6x+8y=24
-6x+8( 0)=24
-6x=24
x=-4
The x-intercept of the equation is -4.

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.
-6x+8y=24
-6( 0)+8y=24
8y=24
y=3
As a result, 3 is the y-intercept of the equation.