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

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

now it gets real annoying real fast factor the denominators first i suck at factoring so let me know what you get
Alright hold on lol:)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question