Question

Help I will medal and fan. Kind of confusing

Answer 1

question

Answer 2

no need for that, just post it.

Answer 3

well, HINT: \(\large\color{slate}{ x^2-9=(x+3)(x-3) }\)

Answer 4

combine the fractions all into 1 common denominator

Answer 5

no you are wrong

Answer 6

I am wrong?

Answer 7

oh, he is.... I though... oops

Answer 8

no @SkepticGod i wrong

Answer 9

I solved it and got up to this but if i divide I only get 0=0\[\frac{ -x^3+9x^2+9x-81 }{ x^3+3x^2-9x-27 }\]

Answer 10

how do I find the value of x from this point

Answer 11

so solomon how do I find the value of x?

Answer 12

\(\large\color{slate}{ \displaystyle \frac{x }{x^2-9}-\frac{1}{x+3} =\frac{1}{4x-12} }\)

\(\large\color{slate}{ \displaystyle \frac{x }{x^2-9}-\frac{x-3}{(x+3)(x-3)} =\frac{1}{4x-12} }\)

\(\large\color{slate}{ \displaystyle \frac{x-(x-3) }{x^2-9} =\frac{1}{4x-12} }\)

\(\large\color{slate}{ \displaystyle \frac{3 }{x^2-9} =\frac{1}{4x-12} }\)

\(\large\color{slate}{ \displaystyle \frac{3\times 4 }{4(x^2-9)} =\frac{1\times(x+3)}{(4x-12)(x+3)} }\)

\(\large\color{slate}{ \displaystyle \frac{12 }{4(x^2-9)} =\frac{x+3}{4(x-3)(x+3)} }\)

\(\large\color{slate}{ \displaystyle 12 =x+3 }\)

Answer 13

Thank you so much I never thought about (x^2-9) is the same as (x-3)(x+3)