Exercise 37 Page 214

To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. This does not happen in the given system. However, if we multiply Equation (II) by 3, the y-variable will have the same coefficient in both equations. 2x-3y=- 14 3(3x-y)= 3(7) ⇕ 2x- 3y=- 14 & (I) 9x- 3y=21 & (II)Let's now subtract Equation (I) from Equation (II).

2x-3y=- 14 9x-3y=21

SysEqnSub

(II): Subtract (I)

2x-3y=- 14 9x-3y-( 2x-3y)=21-( - 14)

▼

(II):Solve for x

Distr

(II): Distribute - 1

2x-3y=- 14 9x-3y-2x+3y=21-(- 14)

SubNeg

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

2x-3y=- 14 9x-3y-2x+3y=21+14

AddSubTerms

(II): Add and subtract terms

2x-3y=- 14 7x=35

DivEqn

(II): .LHS /7.=.RHS /7.

2x-3y=- 14 x=5

Now we can find the value of y by substituting 5 of x into Equation (I), and simplifying.

2x-3y=- 14 & (I) x=5 & (II)

Substitute

(I): x= 5

2( 5)-3y=- 14 x=5

▼

(I):Solve for y

Multiply

(I): Multiply

10-3y=- 14 x=5

SubEqn

(I): LHS-10=RHS-10

- 3y=- 24 x=5

DivEqn

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

y=8 x=5

The solution to the system, which is the point of intersection of the lines, is (5,8).