Exercise 8 Page 260

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. x+3 y=1 & (I) 5 x+6 y=14 & (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 Equation (I) by -2, y-terms will have opposite coefficients. -2( x+3 y)=-2(1) & (I) 5 x+6 y=14 & (II) ⇒ -2 x-6 y=-2 & (I) 5 x+6 y=14 & (II) We can see that the y-terms will eliminate each other if we add Equation (II) to Equation (I).

- 2x-6y=- 2 5x+6y=14

SysEqnAdd

(I): Add (II)

- 2x-6y+ 5x+6y=- 2+ 14 5x+6y=14

AddSubTerms

(I): Add and subtract terms

3x=12 5x+6y=14

DivEqn

(I): .LHS /3.=.RHS /3.

x=4 5x+6y=14

Now we can solve for y by substituting the value of x into either equation and simplifying. Let's use the second equation.

x=4 5x+6y=14

Substitute

(II): x= 4

x=4 5( 4)+6y=14

Multiply

(II): Multiply

x=4 20+6y=14

SubEqn

(II): LHS-20=RHS-20

x=4 6y=- 6

DivEqn

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

x=4 y=- 1

The solution, or intersection point, of the system of equations is (4,- 1).

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+3y=1 & (I) 5x+6y=14 & (II)

(I), (II): x= 4, y= - 1

4+3( - 1)? =1 5( 4)+6( - 1)? =14

Multiply

(II): Multiply

4+3(- 1)? =1 20+6(- 1)? =14

(I), (II): Multiplication Property of -1

4-3? =1 20-6? =14

(I), (II): Subtract terms

1=1 14=14

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