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

Exercise 16 Page 177

Find the intercepts, plot them on the coordinate plane, and connect them with a straight line.

Practice makes perfect

We will graph the following equation by finding its intercepts. -3y=7x-21

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 0, and the x-value is the x-intercept. To find the x-intercept of the given equation, we will substitute 0 for y and solve for x.

-3y=7x-21
-3( 0)=7x-21
0=7x-21
-7x=-21
x=3

The x-intercept is the point (3,0). Let's continue with the y-intercept.

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.

-3y=7x-21
-3y=7( 0)-21
-3y=-21
y=7

Thus, the point (0,7) is the y-intercept.

Graphing the line

Now, we will plot the intercepts on a coordinate plane and connect them with a straight line.