Exercise 25 Page 98

We will graph this equation by finding and plotting its intercepts, then connecting them with a line. To find the x- and y-intercepts, we will 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 equal to 0, and the x-value is the x-intercept. To find the x-intercept of the given equation, we should substitute 0 for y and solve for x.

An x-intercept of -45 means that the graph passes through the x-axis at the point ( -45,0).

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.

-2/9x-5/9y=10

Substitute

x= 0

-2/9( 0)-5/9y=10

ZeroPropMult

Zero Property of Multiplication

-5/9y=10

MultEqn

LHS * 9=RHS* 9

-5 y =90

DivEqn

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

y=-18

A y-intercept of -18 means that the graph passes through the y-axis at the point (0, -18).

Graphing the equation

We can now graph the equation by plotting the intercepts and connecting them with a line.