Exercise 38 Page 214

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. Since here the x-variable has the same coefficient, we can subtract the equations to eliminate x. x-3y=2 & (I) x-2y=1 & (II)Let's subtract Equation (I) from Equation (II).

x-3y=2 x-2y=1

SysEqnSub

(II): Subtract (I)

x-3y=2 x-2y-( x-3y)=1- 2

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(II):Solve for y

Distr

(II): Distribute - 1

x-3y=2 x-2y-x+3y=1-2

AddSubTerms

(II): Add and subtract terms

x-3y=2 y=- 1

Now we can find the value of x by substituting - 1 for y into Equation (I), and simplifying.

x-3y=2 & (I) y=- 1 & (II)

Substitute

(I): y= - 1

x-3( - 1)=2 y=- 1

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(I):Solve for x

MultPosNeg

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

x-(- 3)=2 y=- 1

SubNeg

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

x+3=2 y=- 1

SubEqn

(I): LHS-3=RHS-3

x=- 1 y=- 1

The solution to the system, which is the point of intersection of the lines, is (- 1,- 1).