How do I solve this complex fraction???

Question

[x+1- (6/x)] / [x + 5 + (6/x)] . I already know the answer but when i tried doing it i got x( x+7)/ x(x+11) which is not even close to the answer. Please help me by explaining the steps/ process of working this problem.

Answer 1

[x+1- (6/x)] / [x + 5 + (6/x)]

multiply the numerator and denominator by x:

numerator becomes
= x^2 + x - 6
= (x-3)*(x+2)

denominator becomes
= x^2 + 5x + 6
= (x+3)*(x+2)

equation then becomes
= (x-3)*(x+2) / [ (x+3)*(x+2) ]
= (x-3) / (x+3)

Answer 2

Multiply both the numerator and denominator by x which is equivalent to multiplying the equation by 1. You now have

[x^2 + x - 6]/[x^2 + 5x +6] =
(x + 3)(x - 2) / (x + 3)(x + 2) =
(x - 2) / (x + 2)

Answer 3

numerator:
[x + 1- (6/x)] = [ x^2/x + x/x - 6/x] = [x^2 + x - 6]/x

denominator:
[x + 5 + (6/x)] = [x^2 + 5x + 6]/x

numerator over denominator (remember to flip the x in the denominator)

[x^2 + x - 6] x / x [x^2 + 5x + 6 ]

The x's cancel out, leaving

[x^2 + x - 6] / [x^2 + 5x + 6 ]

factor:
(x-2)(x+3) / (x+2)(x+3)

The (x+3) cancel out, leaving
(x-2)/(x+2)

Answer 4

First you need to get a common denominator on both sides (x), so multiply each term by x/x:

You get:
(x*(x/x) + 1*(x/x) -6/x)/(x*(x/x) +5*(x/x) +6/x)
You can rewrite it as:
[(x^2+x-6)/x]/[(x^2+5x+6)/x]
When you divide, x cancel and you get:
(x^2+x-6)/(x^2+5x+6)
[(x+3)(x-2)/(x+3)(x+2)] When you divide (x+3) cancel out and the answer is:
(x-2)/(x+2)

I hope this helps.

Answer 5

[x+1-(6/x)]/[x+5+(6/x)]

-----first, expand the equations.. thus we got:
={[x^2+x-6]/x}/{[x^2+5x+6]/x}

after expansion, cancel the denominators x
=[x^2+x-6]/[x^2+5x+6]

next get the factors of the denominator and the numerator so we got this
=[(x+3)(x-2)]/[(x+3)(x+2)]

since the numerator and the denominator have common (x+3), this would cancel
=(x-2)/(x+2)

may this one help you

Answer 6

i'm not sure but is it (back of the book)

(x-3)(x+2)/(x+3)(x+2)
x+2 cancel out
x-3/x+3