(4x/x^2-18x+72)+(4/x-6)?

Question

I keep working this problem out and coming up with different answers each time. Can someone help me figure out what the real answer is please?

Answer 1

(4x/(x²-18x+72))+(4/(x-6))

=(4x/[(x-12)(x-6)])+(4/(x-6))

=(4/(x-6))[(x/(x-12))+1]

=(4/(x-6))[(x+x-12)/(x-12)]

=(4*(2x-12))/[(x-6)(x-12)]

=(4*2(x-6))/[(x-6)(x-12)]

=(8)/(x-12) .............. provided x is not equals to 6

Answer 2

First simplify the equation to the excellent of your ability: 4x/ (x-6)(x-12) + 4/(x-6) = 0 (The above (x-6)(x-12) is simplified from x^2-18x+seventy two (Use FOIL technique) then you definately do away with the fraction by applying multiplying the full equation by applying (x-6)(x-12): 4x + 4(x-12) = 0 then you definately distribute: 4x+4x-40 8=0 circulate over the 40 8 by applying including: 8x=40 8 Divide: very final answer: x=6 yet via fact in case you enter 6 decrease back into the equation you won't be able to do 4/0 subsequently that's no answer. (desire this helps you out)

Answer 3

Now, I am assuming that this is equal to zero. First, the 4x/xsquared can be simplified to 4/x, so your equation looks like this:
4/x -18x + 72 + 4/x -6 = 0
combine the two 4/x and subtract 6 from 72:
8/x -18x + 66 = 0
from there, i used a graphing calculator to find where the graph of this equation = 0. When i did this, x = 3.784 . . . Hope this helps.

Answer 4

x^2-18x+72 is
(x-12)(x-6) on factorisation
so the eqn becomes
(4x/(x-12)(x-6))+(4/(x-6))
4/(x-6) is the common term.so the eqn is
4/(x-6)(x/(x-12)+1)
4/(x-6)((x+x-12)/(x-12))
4/(x-6)((2x-12)/(x-12))
8/(x-6)((x-6)/(x-12))
since x-6 is in both the numerator and denomonator we can cancel the two
so the answer is
8/(x-12).

Answer 5

(x-6)(x-6)(x-12)
are the denominators

(4x(x-6)+4(x-6)(x-12))/
(x-6)(x-6)(x-12)

(4x^2-24x+4x^2-72x+288)/
(x-6)(x-6)(x-12)

(8x^2-96x+288)/(x-6)(x-6)
(x-12)
(8(x^2-12x+36))/(x-6)(x-6)
(x-12)
(8(x-6)(x-6))/(x-6)(x-6)(x-12)

Cancel out the x-6:
8/(x-12)