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

1. anonymous

@satellite73

2. anonymous

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

3. anonymous

Alright hold on lol:)

4. anonymous

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

5. anonymous

whew ok

6. anonymous

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

7. anonymous

8. anonymous

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

9. anonymous

$4(x+3)(x-4)-1(x+5)(x-4)=3(x+5)(x-2)$ i think check it

10. anonymous

ok hold on:)

11. anonymous

it is really just cancelling what you need to cancel

12. anonymous

yes, thats true

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous
17. anonymous

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

18. anonymous

Marry me? Lol jk! Thanks so much :)

19. anonymous

yw

20. anonymous

and sure why not? it can be an OS wedding

21. anonymous

Perfect! I'll start planning :D