Exercise 16 Page 391

Even though one of the equations have a variable with a coefficient of 1, the Substitution Method may not be the easiest. Instead, we will use the Elimination Method. To solve a system using this method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. This means that either the x-terms or the y-terms must cancel each other out. 5 x+ 2y=17 & (I) x- 2y=8 & (II) We can see that the y-terms will eliminate each other if we add Equation (I) to Equation (II).

5x+2y=17 & (I) x-2y=8 & (II)

SysEqnAdd

(II):Add I

5x+2y=17 x-2y +5x+2y=8 +17

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(II): Solve for x

AddTerms

(II): Add terms

5x+2y=17 6x=25

DivEqn

(II): .LHS /6.=.RHS /6.

5x+2y=17 x= 256

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

5x+2y=17 x= 256

Substitute

(I): x= 25/6

5( 256)+2y=17 x= 256

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(I): Solve for y

MoveLeftFacToNum

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

1256+2y=17 x= 256

SubEqn

(I):LHS-125/6=RHS-125/6

2y=17- 1256 x= 256

NumberToFrac

(I):a = 6* a/6

2y= 1026- 1256 x= 256

SubTerm

(I):Subtract term

2y=- 236 x= 256

DivEqn

(I):.LHS /2.=.RHS /2.

y=- 2312 x= 256

The solution to this system of equations is ( 256,- 2312).