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How do you simplify #(a+2)/(a-4) + (a-2)/(a+3)#?

Answer 1

Nathan Axel

See below

#(a+2)/(a-4)+(a-2)/(a+3)=((a+2)(a+3)+(a-2)(a-4))/((a-4)(a+3))=#

#=(a^2+3a+cancel(2a)+6+a^2-4a-cancel(2a)+6)/(a^2+3a-4a-12)#

#=(2a^2-a+12)/(a^2-a-12)#

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Answer 2

Eva Abramson

To simplify the expression (a+2)/(a-4) + (a-2)/(a+3), we need to find a common denominator for the two fractions. The common denominator is (a-4)(a+3).

Next, we can rewrite the fractions with the common denominator:

[(a+2)(a+3)]/[(a-4)(a+3)] + [(a-2)(a-4)]/[(a-4)(a+3)]

Expanding the numerators, we get:

[(a^2 + 5a + 6)]/[(a-4)(a+3)] + [(a^2 - 6a + 8)]/[(a-4)(a+3)]

Combining the fractions, we have:

[(a^2 + 5a + 6) + (a^2 - 6a + 8)]/[(a-4)(a+3)]

Simplifying the numerator, we get:

(2a^2 - a + 14)/[(a-4)(a+3)]

Therefore, the simplified expression is (2a^2 - a + 14)/[(a-4)(a+3)].

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.