anonymous
  • anonymous
Solve and see whether or not it is extraneous 4/x^2+3x-10 -1/x^2+x-6 = 3/x^2-x-12
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
now it gets real annoying real fast factor the denominators first i suck at factoring so let me know what you get
anonymous
  • anonymous
Alright hold on lol:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
4/(x-2)(x+5) -1/(x-2)(x+3)= 3(x-4)(x+3) @satellite73 done :)
anonymous
  • anonymous
whew ok
anonymous
  • 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
anonymous
  • anonymous
oh..oh gosh...headachee
anonymous
  • anonymous
actually it is not that bad, since the numerators are all numbers
anonymous
  • anonymous
\[4(x+3)(x-4)-1(x+5)(x-4)=3(x+5)(x-2)\] i think check it
anonymous
  • anonymous
ok hold on:)
anonymous
  • anonymous
it is really just cancelling what you need to cancel
anonymous
  • anonymous
yes, thats true
anonymous
  • 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
anonymous
  • anonymous
Haha! I totally agree:) Ok, lets move onto another one
anonymous
  • anonymous
lol actually all the \(x^2\) terms go bye bye and you end up with only \(14x=2\) so \(x=\frac{1}{7}\)
anonymous
  • anonymous
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
anonymous
  • anonymous
and since \(\frac{1}{7}\) does not make any denominator zero, it is NOT extraneous
anonymous
  • anonymous
Marry me? Lol jk! Thanks so much :)
anonymous
  • anonymous
yw
anonymous
  • anonymous
and sure why not? it can be an OS wedding
anonymous
  • anonymous
Perfect! I'll start planning :D

Looking for something else?

Not the answer you are looking for? Search for more explanations.