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Solution 2.1:6b

From Förberedande kurs i matematik 1

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The lowest common denominator for the three terms is (x−2)(x+3) and we expand each term so that all terms have the same denominator

xx−2+xx+3−2=xx−2x+3x+3+xx+3x−2x−2−2(x−2)(x+3)(x−2)(x+3)=(x−2)(x+3)x(x+3)+x(x−2)−2(x−2)(x+3)=(x−2)(x+3)x2+3x+x2−2x−2(x2+3x−2x−6)=(x−2)(x+3)x2+3x+x2−2x−2x2−6x+4x+12.

Now, collect the terms in the numerator

xx−2+xx+3−2=(x−2)(x+3)(x2+x2−2x2)+(3x−2x−6x+4x)+12=−x+12(x−2)(x+3).

Note: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.

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