Exercise 2 Page 274

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. 2 x- y=4 & (I) 3 x+ y=1 & (II)We can see that the y-terms will eliminate each other if we add Equation (II) to Equation (I).

2x-y=4 3x+y=1

SysEqnAdd

(I): Add (II)

2x-y+ 3x+y=4+ 1 3x+y=1

▼

(I):Solve for x

AddTerms

(I): Add terms

5x=5 3x+y=1

DivEqn

(I): .LHS /5.=.RHS /5.

x=1 3x+y=1

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

x=1 3x+y=1

Substitute

(II): x= 1

x=1 3( 1)+y=1

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(II):Solve for y

IdPropMult

(II): Identity Property of Multiplication

x=1 3+y=1

SubEqn

(II): LHS-3=RHS-3

x=1 y=- 2

The solution, or point of intersection, of the system of equations is (1,- 2). This corresponds to answer G.