how to solve this equation 2x/x^2 - 16 = 2/x^2 - 16 - 1/x+4 ?

Answer 1

2x/(x+4)(x-4)=2/(x+4)(x-4)-1/(x+4)

Multiply (x+4)(x-4) on both sides:
2x=2-(x-4)

Distributive property:
2x=2-x+4

Combine like terms:
2x=-x+6

Add 1x to both sides:
3x=6
x=2

Check:
2(2)/(2)^2-16=2/(2)^2-16-1/2+4
-1/3=-2/12-1/6
-1/3=-1/6-1/6
-1/3=-2/6
-1/3=-1/3

I hope this helps!

Answer 2

You really should use parentheses. I think you mean this

2x/(x^2-16)=2/(x^2-16)-1/(x+4) factor all denominators, multiply by the LCD which is x^2-16, or in factored form (x+4)(x-4)

2x=2-1(x-4)
2x=2-x+4
3x=6
x=2

Hope this helped

Answer 3

(x/(4x-16))-3=1/(x-4) Multiply each side by x-4 (x(x-4)/(4x-16))-3x+12=1 (x(x-4)/(4x-16))=3x-11 Multiply by 4x-16 x(x-4)=(3x-11)(4x-16) Now foil x^2-4x=12x^2-92x+176 Subtract the first side, from the first and the second 0=11x-88x+176 Now you can use the quadratic formula (88±√(7744-7744)/22)=(88±0)/22=4 Voilà. We now know that x=4. However, this leads to a problem, since plugging that in at the beginning gives 1/(4-4), or 1/0, which is not possible.

Answer 4

I am assuming you mean

2x/(x² - 16) = 2/(x² - 16) - 1/(x + 4)

So 2x/[{x + 4)(x - 4)] = 2/[(x + 4)(x - 4)] - 1/(x + 4)
Multilpy both sides by the LCD ie (x + 4)(x - 4)

Thus 2x = 2 - (x - 4)
2x = 2 - x + 4
3x = 6
x = 2

Answer 5

((2x)/(x^2 - 16)) = (2/(x^2 - 16)) - (1/(x + 4))

assuming nothing is mistyped, like

((2x)/(x^2 - 16)) = (2/(x - 4)) - (1/(x + 4))

---------

((2x)/(x^2 - 16)) - (2/(x^2 - 16)) = (-1/(x + 4))

(2x - 2)/(x^2 - 16) = (-1/(x + 4))

cross multiply

(2x - 2)(x + 4) = -x^2 + 16

2x^2 + 8x - 2x - 8 = -x^2 + 16
2x^2 + 6x - 8 = -x^2 + 16
3x^2 + 6x - 24 = 0
3(x^2 + 2x - 8) = 0
x^2 + 2x - 8 = 0
(x + 4)(x - 2) = 0

x = 2 or -4

since pluging in -4 in (1/(x + 4)) would make this problem undefined

ANS : x = 2

Answer 6

sorry, you have to do your own homework.

this is cheating.... thought you could get away with it ????

Answer 7

WHAT????????????????????????