Exercise 8 Page 159

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 12 means that the graph passes through the x-axis at the point ( 12,0).

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.

2x-5y=1

Substitute

x= 0

2( 0)-5y=1

ZeroPropMult

Zero Property of Multiplication

-5y=1

DivEqn

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

y=1/-5

MoveNegDenomToFrac

Put minus sign in front of fraction

y=-1/5

A y-intercept of - 15 means that the graph passes through the y-axis at the point (0, - 15).

Graphing the Equation

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