Question

(x^2+x-30)/(x-5) =11

Answer 1

I would recommend you get rid of that fraction first of all by multiplying both sides of the equation by the denominator of the fraction (x-5) \[\frac{x^2+x-30}{x-5}\times (x-5)=11 \times (x-5)\]

Answer 2

If you will note by inspection \[x ^{2}+x-30 =(x+6)(x-5)\]therefore the x-5 can be cancelled giving you:
x+6=11

Answer 3

\[\frac{x^2+x-30}{x-5} =11\]\[x^2+x-30 =11(x-5)\]\[x^2+x-30 =11x-55\]\[x^2-10x+25 =0\]\[(x-5)^2 =0\] \[x= 5\]

Answer 4

making it simple to solve. x=5

Answer 5

thank you

Answer 6

Unfortunately, when x=5, the denominator x-5 is zero (since x-5=5-5=0), so x=5 cannot be a solution.


So there are no solutions.

Answer 7

This is why it's important to check any answers you get.

Answer 8

Excellent catch Jim.

Answer 9

Great spot jim, should of checked my answer.

Answer 10

awesome. cause i know the answer is

there is no solution because the origional equation if undefined at x=5

i just didnt know how to get there