Exercise 3 Page 389

By graphing the given equations, we can determine the solution to the System of Linear Equations. This will be the point at which the lines intersect. y=-2x-2 & (I) x+2y=2 & (II) We will start by finding the x- and y-intercepts for both equations.

Calculating the Intercepts

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 Equation (I), we will substitute 0 for y and solve for x. y=- 2x-2 Let's do it!

y=-2x-2

SubstituteIII

Substitute y= 0

0=-2x-2

▼

Solve for x

AddEqn

LHS+2=RHS+2

2=-2x

DivEqn

.LHS /(-2).=.RHS /(-2).

-1=x

RearrangeEqn

Rearrange equation

x=-1

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, just like with the x-intercept. Therefore, substituting 0 for x will give us the y-intercept.

y=-2x-2

Substitute

x= 0

y=-2( 0)-2

ZeroPropMult

Zero Property of Multiplication

y=-2

We can find the x- and y-intercepts of Equation (II) in the same way, by substituting 0 for each variable and solving for the other variable.

Information	y=-2x-2	x+2y=2
y= 0	0=-2x-2	x+2( 0)=2
Simplify	x=-1	x=2
x-intercept	(-1,0)	(2,0)
x= 0	y=-2( 0)-2	0+2y=2
Simplify	y=-2	y=1
y-intercept	(0,-2)	(0,1)

Graphing the System

To graph the two equations, we plot the intercepts and connect them with lines.

The solution to the system is the point where the lines intersect. The two lines appear to intersect at the point (-2,2).

Checking Our Answer

To check our answer, we need to substitute the solution into both of the equations to make sure that both create true statements.

y=-2x-2 & (I) x+2y=2 & (II)

(I), (II):x= -2, y= 2

2? =-2( -2)-2 -2+2( 2)? =2

(I), (II):Multiply

2? =4-2 -2+4? =2

(I), (II):Add and subtract terms

2=2 2=2

Because substituting (-2,2) into both equations produced true statements, we know that our solution is correct.