Exercise 1 Page 267

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. 8x+3 y=-9 & (I) -8x+ y=29 & (II) 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. This means that either the x-terms or the y-terms must cancel each other out. We can see that the x-terms will eliminate each other if we add Equation (I) to Equation (II).

8x+3y=-9 -8x+ y=29

SysEqnAdd

(II): Add (I)

8x+3y=-9 -8x+y+( 8x+3y)=29+( - 9)

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(II):Solve for y

AddSubTerms

Add and subtract terms

8x+3y=-9 4y=20

DivEqn

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

8x+3y=-9 y=5

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

8x+3y=-9 y=5

Substitute

(I): y= 5

8x+3 ( 5)=- 9 y=5

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(I):Solve for x

Multiply

(I): Multiply

8x+15=-9 y=5

SubEqn

(I): LHS-15=RHS-15

8x=-24 y=5

DivEqn

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

x=-3 y=5

The solution, or point of intersection, of the system of equations is (-3,5).