Question

Solve and see whether or not it is extraneous
4/x^2+3x-10 -1/x^2+x-6 = 3/x^2-x-12

Answer 1

@satellite73

Answer 2

now it gets real annoying real fast
factor the denominators first
i suck at factoring so let me know what you get

Answer 3

Alright hold on lol:)

Answer 4

4/(x-2)(x+5) -1/(x-2)(x+3)= 3(x-4)(x+3)
@satellite73 done :)

Answer 5

whew ok

Answer 6

then you can clear the fractions as before but this time you have to multiply by the least common multiple of the denominators which is \[(x-2)(x+5)(x+3)(x-4)\] ugh

Answer 7

oh..oh gosh...headachee

Answer 8

actually it is not that bad, since the numerators are all numbers

Answer 9

\[4(x+3)(x-4)-1(x+5)(x-4)=3(x+5)(x-2)\] i think
check it

Answer 10

ok hold on:)

Answer 11

it is really just cancelling what you need to cancel

Answer 12

yes, thats true

Answer 13

then we have a raft of algebra to do
multiply all this muck out, combine like terms etc etc
i refuse to do it, i would cheat

Answer 14

Haha! I totally agree:) Ok, lets move onto another one

Answer 15

lol actually all the \(x^2\) terms go bye bye and you end up with only \(14x=2\) so \(x=\frac{1}{7}\)

Answer 16

here is how to cheat
http://www.wolframalpha.com/input/?i=4%28x%2B3%29%28x-4%29-1%28x%2B5%29%28x-4%29%3D3%28x%2B5%29%28x-2%29

Answer 17

and since \(\frac{1}{7}\) does not make any denominator zero, it is NOT extraneous

Answer 18

Marry me? Lol jk! Thanks so much :)

Answer 19

yw

Answer 20

and sure why not? it can be an OS wedding

Answer 21

Perfect! I'll start planning :D