Core Connections Algebra 1, 2013
CC
Core Connections Algebra 1, 2013 View details
3. Section 10.3
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Exercise 109 Page 506

To find the x-intercepts, substitute 0 for y and solve for x.

Top line:
y-intercept: (0,10)
x-intercept: (40,0)
Bottom line:
y-intercept: (0,3)
x-intercept: (- 9,0)

Practice makes perfect

We want to find the x- and y-intercepts for each line from the previous exercise.

Top Line

In the previous exercise we have already found that the y-intercept of the top line is (0,10). We also found the equation of the line. y=- 1/4x+10We can use this equation to determine the x-intercept of the line. 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 equation, we should substitute 0 for y and solve for x.
y=- 1/4x+10
0=- 1/4x+10
â–Ľ
Solve for x
- 10=- 1/4x
10=1/4x
40=x
x=40
An x-intercept of 40 means that the graph passes through the x-axis at the point ( 40,0).

Bottom Line

In the previous exercise we also found that the y-intercept of the bottom line is (0,3). We have found the equation of the line as well. y=1/3x+3 We can use this equation to determine the x-intercept of the bottom line. Again, to find the x-intercept of the equation, we will substitute 0 for y and solve for x.
y=1/3x+3
0=1/3x+3
â–Ľ
Solve for x
- 3=1/3x
- 9=x
x=- 9
An x-intercept of - 9 means that the graph passes through the x-axis at the point ( - 9,0).