Solve. (x/x -8) + 7= 8/x - 8?

Question

Solve. (x/x -8) + 7= 8/x - 8

Does anyone know how to do this kind of problem because i am trying to learn the steps

Answer 1

x/(x-8) + 7 = 8/(x-8)
Multiply everything by (x-8)
x + 7(x-8) = 8
Distribute the 7
x + 7x - 56 = 8
Combine x and 7x
8x - 56 = 8
Add 56 to each side
8x = 64
Divide both sides by 8.
x = 8

Note that x-8 cannot equal zero, but we just found x to be 8.
That means x-8 = 8-8 = 0
Thus, we have no solution.

Answer 2

x/x-8 +7x-56/x-8 = 8/x-8 -- I have multiplied 7/1(x-8/x-8)
x + 7x -56 -8/x-8 = 0 -- we have common denominators now
8x-64/x-8 = 0

There is no such thing as a zero demoninator so we can safely say that X cannot equal 8

Answer 3

ok i thnk i can help you out... what you need to do is cross multiply, since you have x/x - 8 you need to make it x/x -8/1 and do the same for the other side.... 8/x - 8/1... then solve it, so all the math and i think the answer is x= 1/15

Answer 4

(x/(x - 8)) + 7 = (8/(x - 8))

(8/(x - 8)) - (x/(x - 8)) = 7
(8 - x)/(x - 8) = 7
((-x + 8)/(x - 8)) = 7
(-(x - 8)/(x - 8)) = 7
-1 = 7

So this would be No Solution, but if you mean

(x/(x - 8)) + 7 = (8/x) - 8

Multiply everything by x(x - 8)

x^2 + 7x(x - 8) = 8(x - 8) - 8x(x - 8)
x^2 + (7 + 8)x(x - 8) - 8(x - 8) = 0
x^2 + 15x(x - 8) - 8(x - 8) = 0
x^2 + 15x^2 - 120x - 8x + 64 = 0
16x^2 - 128x + 64 = 0
16(x^2 - 8x + 4) = 0
x^2 - 8x + 4 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-8) ± sqrt((-8)^2 - 4(1)(4)))/(2(1))
x = (8 ± sqrt(64 - 16))/2
x = (8 ± sqrt(48))/2
x = (8 ± sqrt(16 * 3))/2
x = (8 ± 4sqrt(3))/2
x = 4 ± 2sqrt(3)

ANS : 4 ± 2sqrt(3)

Answer 5

x/x - 8 + 7 = 8/x - 8
move all x's to one side...and all numbers to the other changing the sign (+/-) as you move them.
x/x - 8x = 8 - 7 - 8
reduce that to:
x/x - 8x = - 7
realize that anything over itself = 1
1 - 8x = - 7
move the one to the number side
- 8x = -8
divide both sides by -8
x = 1

Answer 6

It is hard to tell from your question whether second part is (8/x)-8 or 8/(x-8). that is why there are 2 different answers above.