Exercise 59 Page 208

Since neither equation has 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 of linear equations this way, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. 3 x-5 y=26 & (I) -2 x-3 y=-11 & (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 and multiply (II) by 3, the x-terms will have opposite coefficients. 2(3 x-5 y)=2(26) 3(-2 x-3 y)=3(-11) ⇓ 6x-10 y=52 -6x-9 y=-33 We can see that the x-terms will eliminate each other if we add (I) to (II).

6x-10y=52 -6x-9y=-33

SysEqnAdd

(II): Add (I)

6x-10y=52 -6x-9y+( 6x-10y)=-33+( 52)

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(II):Solve for y

RemovePar

(II): Remove parentheses

6x-10y=52 - 6x - 9y + 6x - 10y=- 33 + 52

AddSubTerms

(II): Add and subtract terms

6x-10y=52 -19y=19

DivEqn

(II): .LHS /(-19).=.RHS /(-19).

6x-10y=52 y=-1

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

6x-10y=52 y=-1

Substitute

(I): y= -1

6x-10( -1)=52 y=-1

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(I):Solve for x

MultNegNegOnePar

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

6x+10=52 y=-1

SubEqn

(I): LHS-10=RHS-10

6x=42 y=-1

DivEqn

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

x=7 y=-1

The solution to the system of equations is (7,-1).