Exercise 35 Page 383

Since the y in the first equation is already isolated, the easiest way to solve the system is to use the Substitution Method. Let's substitute y= 23x+1 into the second equation.

y= 23x+1 & (I) 2x+3y=27 & (II)

Substitute

(II): y= 23x+1

y= 23x+1 2x+3( 23x+1)=27

Distr

(II): Distribute 23

y= 23x+1 2x+2x+3=27

AddTerms

(II): Add terms

y= 23x+1 4x+3=27

SubEqn

(II): LHS-3=RHS-3

y= 23x+1 4x=24

DivEqn

(II): .LHS /4.=.RHS /4.

y= 23x+1 x=6

Now that we have found a solution for x, we can substitute that into our first equation to solve for y.

y= 23x+1 x=6

Substitute

(I): x= 6

y= 23( 6)+1 x=6

Multiply

(I): Multiply

y=4+1 x=6

AddTerms

(I): Add terms

y=5 x=6

The solution to the system — the point of intersection — is (6,5).