Solve the following system by the elimination method. Check the solution.

5x-5=-7y, 6x+5y=-11

Hmm

add 7y+5 to both sides of 1st equation

5x+7y=5

eliminate x's

multiply first equation by - 6 and 2nd by 5 and add them

-30x-42y=-30

30x+25y=-55 +

0x-17y=-85

-17y=-85

divide both sides by - 17

y=5

sub back

5x-5=-7y

5x-5=-7 (5)

5x-5=-35

add 5

5x=-30

divide 5

x=-6

x=-6

y=5

subsitute to find out

5x-5=-7y

5 (-6) - 5=-7 (5)

-30-5=-35

-35=-35

yep

6x+5y=-11

6 (-6) + 5 (5) = - 11

-36+25=-11

-11=-11

yep

x=-6

y=5

(-6,5)

I assume elimination method is to eliminate one of the variables. in order to eliminate one then that variable must have the same coefficient in both equations. I look at what will be easier to eliminate. The x values look easier with 6 and 5. in order to make them equal I must multiply each equation by the other equations x coefficient. (remember that to multiply an equation by a number is to multiply both sides of the equal sign by that number.)

so:

6 (5x-5) = 6 (-7y)

30x - 30 = - 42y

5 (6x+5y) = 5 (-11)

30x + 25y = - 55

now I subtract the 2 equations.

I will subtract the left sides of equations and subtract the right sides of equations on other side

30x - 30 - (30x + 25y) = - 42y - (-55)

distribute the minus sign into the parenthesis

30x - 30 - 30x - 25y = - 42y + 55

cancel out the x values

- 30 - 25y = - 42y + 55

solve for t

17y = 85

Y = 5

plug y into either equation

6x + 5 (5) = - 11

6x + 25 = - 11

6x = - 36

x = - 6

answer (5,-6)

this can be checked by plugging the values into other equatiin