Exercise 13 Page 326

To determine the x- and y-intercepts of a line, we need to substitute 0 for one variable, solve, then repeat for the other variable.

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 equation, we should substitute 0 for y and solve for x.

-5x+3y=-7.5

Substitute

y= 0

-5x+3( 0)=-7.5

▼

Solve for x

ZeroPropMult

Zero Property of Multiplication

-5x=-7.5

DivEqn

.LHS /-5.=.RHS /-5.

x=-7.5/-5

DivNegNeg

- a/- b=a/b

x=7.5/5

CalcQuot

Calculate quotient

x=1.5

An x-intercept of 1.5 means that the graph passes through the x-axis at the point ( 1.5,0).

Finding the y-intercept

Let's use the same concept to find the y-intercept. 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.

-5x+3y=-7.5

Substitute

x= 0

-5( 0)+3y=-7.5

▼

Solve for y

ZeroPropMult

Zero Property of Multiplication

3y=-7.5

DivEqn

.LHS /3.=.RHS /3.

y=-7.5/3

CalcQuot

Calculate quotient

y=-2.5

A y-intercept of -2.5 means that the graph passes through the y-axis at the point (0, -2.5).