Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
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Exercise 3 Page 190

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

x-intercept: 6
y-intercept: -4

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. 2x-3y=12As we have already said, let's substitute 0 for y and solve for x.
2x-3y=12
2x-3( 0)=12
2x=12
x=6
The x-intercept of the equation is 6.

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.
2x-3y=12
2( 0)-3y=12
-3y=12
y=-4
As a result, -4 is the y-intercept of the equation.