How do you add #3/(x+2) + 6/(x-1)#?

Answer 1

Eli Assouline

#(9(x+1))/(x^2+x-2)#

Detail

#(3(x-1)+6(x+2))/((x+2)(x-1)#

#(3x-3+6x+12)/(x^2-x+2x-2)#

#(9x+9)/(x^2+x-2)#

#(9(x+1))/(x^2+x-2)#

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Answer 2

Isaiah Anthony

To add the fractions 3/(x+2) and 6/(x-1), you need to find a common denominator. The common denominator in this case is (x+2)(x-1).

To add the fractions, you multiply the numerator of each fraction by the denominator of the other fraction.

The resulting numerator is (3(x-1)) + (6(x+2)).

Simplifying the numerator gives 3x - 3 + 6x + 12.

Combining like terms, the numerator becomes 9x + 9.

The denominator remains (x+2)(x-1).

Therefore, the sum of 3/(x+2) + 6/(x-1) is (9x + 9)/((x+2)(x-1)).

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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.