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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.

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2x-2/x-1
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yea.
My teacher told me the answer was 2x+8, I was supposed to figure out how to get it
ok let me see I might have made a careless mistake on min please
ok
Factoring everything I get (((x-4)(x+4))/(5(x-3)))*(10(x-3)²/(x-4)(x-3))
after cancellations u get (x+4)2 => 2x+8
Factor all terms that can be factorized, you get: \[{x^2-16 \over 5x-15}.{10(x-3)^2 \over x^2-7x+12}={(x-4)(x+4) \over 5(x-3)}.{10(x-3)^2 \over (x-4)(x-3)}\] Now, Cancel any common factors, in the numerator and denominator, you get: \[2(x+4)=2x+8\]
Wow thank you!
darn it.
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