Exercise 60 Page 431

Does either of the equations have an isolated variable in it?

(-4,-5)

Practice makes perfect

When solving a system of equations using substitution, there are three steps.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
Substitute this solution into one of the equations and solve for the value of the other variable.

Observing the given equations, it looks like it will be simplest to isolate x in the second equation.

2x-y=-3 & (I) x-y=1 & (II)

(II):LHS+y=RHS+y

2x-y=-3 x=1+y

Now that we've isolated x, we can solve the system by substitution.

2x-y=-3 & (I) x=1+y & (II)

(I):x= 1+y

2( 1+y)-y=-3 x=1+y

(I):Distribute 2

2+2y-y=-3 x=1+y

(I): Subtract terms

2+y=-3 x=1+y

(I):LHS-2=RHS-2

y=-5 x=1+y

Great! Now, to find the value of x, we need to substitute y=-5 into the second equation.

y=-5 x=1+y

(II):y= -5

y=-5 x=1 -5

(II): Subtract terms

y=-5 x=-4

The solution, or point of intersection, to this system of equations is the point (-4,-5).

Lesson Check p.429

Practice and Problem-Solving Exercises p.429-431

Standardized Test Prep p.431

Mixed Review p.431

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