Exercise 30 Page 217

Since no variable term is isolated or has a coefficient equal to 1, it is not convenient to use the Substitution Method. Therefore, we will use the Elimination Method. In order to eliminate the x-terms, let's multiply the first equation by -1.

8x+5y=6 8x=3-2y

MultEqn

(I): LHS * (-1)=RHS* (-1)

-1(8x+5y)=-1(6) 8x=3-2y

Distr

(I): Distribute (-1)

(-1)8x+(-1)5y=-1(6) 8x=3-2y

MultPropNegOne

(I): Multiplication Property of -1

-8x+(-5y)=-6 8x=3-2y

AddNeg

(I): a+(- b)=a-b

-8x-5y=-6 8x=3-2y

We can see that the x-terms will eliminate each other if we add Equation (II) to Equation (I).

-8x-5y=-6 8x=3-2y

SysEqnAdd

(I): Add (II)

-8x-5y+( 8x)=-6+( 3-2y) 8x=3-2y

▼

(I):Solve for y

RemovePar

(I): Remove parentheses

-8x-5y+8x=-6+3-2y 8x=3-2y

AddTerms

(I): Add terms

-5y=-3-2y 8x=3-2y

AddEqn

(I): LHS+2y=RHS+2y

-5y+2y=-3-2y+2y 8x=3-2y

AddTerms

(I): Add terms

-3y=-3 8x=3-2y

DivEqn

(I): .LHS /-3.=.RHS /-3.

-3y/-3=-3/-3 8x=3-2y

MovePartNumRight

(I): a* b/c=a/c* b

-3/-3 * y=-3/-3 8x=3-2y

DivNegNeg

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

3/3 * y=3/3 8x=3-2y

QuotOne

(I): a/a=1

1 * y=1 8x=3-2y

IdPropMult

(I): Identity Property of Multiplication

y=1 8x=3-2y

Now we can solve for x by substituting the value of y into either equation and simplifying.

y=1 8x=3-2y

Substitute

(II): y= 1

y=1 8x=3-2( 1)

▼

(II):Solve for x

IdPropMult

(II): Identity Property of Multiplication

y=1 8x=3-2

SubTerms

(II): Subtract terms

y=1 8x=1

AddEqn

(II): LHS+8=RHS+8

y=1 8x/8=1/8

MovePartNumRight

(II): a* b/c=a/c* b

y=1 8/8 * x=1/8

QuotOne

(II): a/a=1

y=1 1 * x=1/8

IdPropMult

(II): Identity Property of Multiplication

y=1 x=1/8

The solution, or point of intersection, of the system of equations is ( 18,1).

Hints

Check the answer

Practice exercises

Asses your skill