Question

Find the following difference.
I NEED SOME DAMN HELP!!!!!

Answer 1

\[\frac{ 2x-3 }{ x^2+x-20 }-\frac{ x+4 }{ x^2+9x+20 }\]

Answer 2

Usually here we want to factor these first.
It will help find a common denominator.

Answer 3

She is offline ^^

Answer 4

(x+5)(x-4) & (x+5)(x+4)

Answer 5

Alright, so you factored out the denominator for both of the fractions correctly, it looks like this then,\[\frac{ 2x-3 }{ (x+5)(x-4) }-\frac{ (x+4) }{ (x+4)(x+5) }\]so by simplifying that you'll get\[\frac{ 2x-3 }{ (x+5)(x-4) }-\frac{ 1 }{ (x+5) }\]from there you need to multiply your second fraction by\[\frac{ (x-4) }{ (x-4) }\]in order to get your common denominator, and from there you should be able to finish the problem.

Answer 6

\[\frac{ 2x-3 }{ x^2+x-20 }-\frac{ x+4 }{ x^2+9x+20 }\]
\[\frac{ 2x-3 }{ (x+5)(x-4) }-\frac{ x+4 }{ (x+5)(x+4) }\]
\[\frac{ 2x-3 }{ (x+5)(x-4) }-\frac{ 1 }{ (x+5)(1) }\]
\[\frac{ 2x-3 }{ (x+5)(x-4) }-\frac{ x-4 }{ (x+5)(x-4) }\]
\[\frac{ (2x-3) -(x-4)}{ (x+5)(x-4) }\]

Answer 7

\[\frac{ 2x^2+12 }{ (x+5)(x-5) }\]

Answer 8

right?

Answer 9

@iambatman

Answer 10

\[(2x-3)-(x-4)=2x-x-3-(-4)=x-3+4=x+1\]\[\frac{x+1}{(x+5)(x-4)}=\frac{x+1}{x^2+x-20}\]