Exercise 54 Page 173

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. In this exercise, this means that either the x-terms or the y-terms must cancel each other out. x+4 y=12 & (I) 2 x-8 y=4 & (II) Currently, none of the terms in this system will cancel out. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply (I) by 2, the y-terms will be additive inverses. 2( x+4 y)=2(12) 2 x-8 y=4 ⇓ 2 x+ 8y=24 2 x- 8y=4We can see that the y-terms will eliminate each other if we add (I) to (II).

2x+8y=24 2x-8y=4

SysEqnAdd

(II): Add (I)

2x+8y=24 2x-8y+( 2x+8y)=4+( 24)

▼

(II):Solve for x

RemovePar

(II): Remove parentheses

2x+8y=24 2x-8y+2x+8y=4+24

AddTerms

(II): Add terms

2x+8y=24 4x=28

DivEqn

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

2x+8y=24 x=7

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

2x+8y=24 x=7

Substitute

(I): x= 7

2( 7)+8y=24 x=7

▼

(I):Solve for y

Multiply

(I): Multiply

14+8y=24 x=7

SubEqn

(I): LHS-14=RHS-14

8y=10 x=7

DivEqn

(I): .LHS /8.=.RHS /8.

y= 108 x=7

ReduceFrac

(I): a/b=.a /2./.b /2.

y= 54 x=7

The solution, or intersection point, of the system of equations is (7, 54).