Exercise 7 Page 393

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

(0.7,-0.8)

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 y in the first equation.

4x+y=2

SubEqn

LHS-4x=RHS-4x

y=2-4x

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

y=2-4x & (I) 3y+2x=-1 & (II)

Substitute

(II):y= 2-4x

y=2-4x 3( 2-4x)+2x=-1

Distr

(II):Distribute 3

y=2-4x 6-12x+2x=-1

AddTerms

(II): Add terms

y=2-4x 6-10x=-1

SubEqn

(II):LHS-6=RHS-6

y=2-4x -10x=-7

▼

(II): .LHS /(-10).=.RHS /(-10).

DivEqn

(II): .LHS /(-10).=.RHS /(-10).

y=2-4x x= -7- 10

DivNegNeg

(II): - a/- b=a/b

y=2-4x x= 710

WriteDec

(II): Write as a decimal

y=2-4x x=0.7

Now, to find the value of y, we need to substitute x=0.7 into either one of the equations in the given system. Let's use the first equation.

y=2-4x x=0.7

Substitute

(I):x= 0.7

y=2-4( 0.7) x=0.7

Multiply

(I): Multiply

y=2-2.8 x=0.7

SubTerm

(I): Subtract term

y=-0.8 x=0.7

The solution, or point of intersection, to this system of equations is the point (0.7,-0.8).

Exercise 7 Page 393

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