Exercise 10 Page 265

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. 8 x-7 y=-3 & (I) 6 x-5 y=-1 & (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 -5 and multiply Equation (II) by 7, the y-terms will have opposite coefficients. -5(8 x-7 y)=-5(-3) & (I) 7(6 x-5 y) =7(-1) & (II) ⇓ -40 x+ 35y=15 & (I) 42 x- 35y=-7 & (II) We can see that the y-terms will eliminate each other if we add Equation (I) to Equation (II).

-40x+35y=15 42x-35y=-7

SysEqnAdd

(II): Add (I)

-40x+35y=15 42x-35y+ ( -40x+ 35y)=-7+ 15

AddSubTerms

(II): Add and subtract terms

-40x+35y=15 2x=8

DivEqn

(II): .LHS /2.=.RHS /2.

-40x+35y=15 x=4

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

-40x+35y=15 x=4

Substitute

(I): x= 4

-40( 4)+35y=15 x=4

MultPosNeg

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

-160+35y=15 x=4

AddEqn

(I): LHS+160=RHS+160

35y=175 x=4

DivEqn

(I): .LHS /35.=.RHS /35.

y=5 x=4

The solution, or point of intersection, of the system of equations is (4,5).

Checking Our Answer

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.

8x-7y=-3 & (I) 6x-5y=-1 & (II)

(I): (II): x= 4, y= 5

8( 4)-7( 5)? =-3 & (I) 6( 4)-5( 5)? =-1 & (II)

Multiply

Multiply

32-35? =-3 & (I) 24-25? =-1 & (II)

SubTerms

Subtract terms

- 3=-3 ✓ & (I) - 1=-1 ✓ & (II)

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