Exercise 19 Page 217

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. x+4 y=1 & (I) 3 x+5 y=10 & (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 (I) by -3 the x-terms will have opposite coefficients. -3( x+4 y)=-3(1) 3 x+5 y=10 ⇓ -3x-12 y=-3 3x+5 y=10 We can see that the x-terms will eliminate each other if we add Equation (II) to Equation (I).

-3x-12y=-3 3x+5y=10

SysEqnAdd

(I): Add (II)

-3x-12y+( 3x+5y)=-3+( 10) 3x+5y=10

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(I):Solve for y

RemovePar

(I): Remove parentheses

-3x-12y+3x+5y=-3+10 3x+5y=10

AddTerms

(I): Add terms

-7y=7 3x+5y=10

DivEqn

(I): .LHS /-7.=.RHS /-7.

-7y/-7=7/-7 3x+5y=10

MovePartNumRight

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

-7/-7 * y =7/-7 3x+5y=10

DivNegNeg

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

7/7 * y =7/-7 3x+5y=10

MoveNegDenomToFrac

(I): Put minus sign in front of fraction

7/7 * y =-7/7 3x+5y=10

QuotOne

(I): a/a=1

1 * y =-1 3x+5y=10

IdPropMult

(I): Identity Property of Multiplication

y =-1 3x+5y=10

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

y =-1 3x+5y=10

Substitute

(II): y= -1

y =-1 3x+5( -1)=10

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(II):Solve for x

MultPropNegOne

(II): Multiplication Property of -1

y =-1 3x+(-5)=10

AddNeg

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

y =-1 3x-5=10

AddEqn

(II): LHS+5=RHS+5

y =-1 3x-5+5=10+5

AddTerms

(II): Add terms

y =-1 3x=15

DivEqn

(II): .LHS /3.=.RHS /3.

y =-1 3x/3=15/3

MovePartNumRight

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

y =-1 3/3 * x=15/3

QuotOne

(II): a/a=1

y =-1 1 * x=15/3

IdPropMult

(II): Identity Property of Multiplication

y =-1 x=15/3

CalcQuot

(II): Calculate quotient

y =-1 x=5

The solution, or point of intersection, of the system of equations is (5,-1). 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.

x+4y=1 & (I) 3x+5y=10 & (II)

(I), (II): x= 5, y= -1

5+4( -1) ? = 1 3( 5)+5( -1) ? = 10

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Simplify

(I), (II): Multiplication Property of -1

5+(-4) ? = 1 3(5)+(-5) ? = 10

Multiply

(II): Multiply

5+(-4) ? = 1 15+(-5) ? = 10

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

5-4 ? = 1 15-5 ? = 10

(I), (II): Subtract terms

1 = 1 ✓ 10 = 10 ✓