Exercise 6 Page 213

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. This means that either the x- or the y-terms must cancel each other out. 5 y=15-5 x & (I) y=-2 x+3 & (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 Equation (II) by -5 the y-terms will have opposite coefficients. 5 y=15-5 x -5( y)=-5(-2 x+3) ⇓ 5y=15-5 x -5y=10 x-15 We can see that the y-terms will eliminate each other if we add Equation (I) to Equation (II).

5y=15-5x -5y=10x-15

SysEqnAdd

(II): Add (I)

5y=15-5x -5y+( 5y)=10x-15+( 15-5x)

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(II):Solve for x

RemovePar

(II): Remove parentheses

5y=15-5x -5y+5y=10x-15+15-5x

AddSubTerms

(II): Add and subtract terms

5y=15-5x 0=5x+0

IdPropAdd

(II): Identity Property of Addition

5y=15-5x 0=5x

DivEqn

(II): .LHS /5.=.RHS /5.

5y=15-5x 0/5=x

(II): 0/a=0

5y=15-5x 0=x

RearrangeEqn

(II): Rearrange equation

5y=15-5x x=0

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

5y=15-5x x=0

Substitute

(I): x= 0

5y=15-5( 0) x=0

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(I):Solve for y

ZeroPropMult

(I): Zero Property of Multiplication

5y=15-0 x=0

SubTerms

(I): Subtract terms

5y=15 x=0

DivEqn

(I): .LHS /5.=.RHS /5.

5y/5=15/5 x=0

MovePartNumRight

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

5/5 * y=15/5 x=0

QuotOne

(I): a/a=1

1 * y=15/5 x=0

IdPropMult

(I): Identity Property of Multiplication

y=15/5 x=0

CalcQuot

(I): Calculate quotient

y=3 x=0

The solution, or point of intersection, of the system of equations is (0,3). To check this solution, we will substitute it back into the given system and simplify. If doing so results in true statements for every equation in the system, our solution is correct.

5y=15-5x & (I) y=-2x+3 & (II)

(I), (II): x= 0, y= 3

5( 3) ? = 15-5( 0) 3 ? = -2( 0)+3

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Simplify

(I), (II): Zero Property of Multiplication

5(3) ? = 15-0 3 ? = 0+3

SubTerms

(I): Subtract terms

5(3) ? = 15 3 ? = 0+3

Multiply

(I): Multiply

15 = 15 3 ? = 0+3

IdPropAdd

(II): Identity Property of Addition

15 = 15 ✓ 3 = 3 ✓