Exercise 7 Page 242

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. In this exercise, this means that either the x-terms or the y-terms must cancel each other out. x+ y=4 & (I) -3 x- y=-8 & (II) We can see that the y-terms will eliminate each other if we add (I) to (II). Now we can now solve for y by substituting the value of x into either equation and simplifying. Let's use the first equation.

x+y=4 x=2

Substitute

(I): x= 2

2+y=4 x=2

SubEqn

(I): LHS-2=RHS-2

y=2 x=2

The solution, or intersection point, of the system of equations is (2,2).

Checking Our Solution

To check this solution, we will substitute it back into the given system and simplify. If doing so results in true statements for every equation in the system, our solution is correct.

x+y=4 & (I) -3x-y=-8 & (II)

(I), (II): x= 2, y= 2

2+ 2? =4 & (I) -3( 2)- 2? =-8 & (II)

MultNegPos

(II):(- a)b = - ab

2+2? =4 & (I) -6-2? =-8 & (II)

(I), (II): Add and subtract terms

4=4 & (I) -8=-8 & (II)

Because both equations are true statements, we know our solution is correct.