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Solution 2.1:6b

From Förberedande kurs i matematik 1

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Current revision (11:35, 23 September 2008) (edit) (undo)
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The lowest common denominator for the three terms is <math>(x-2)(x+3)</math> and we expand each term so that all terms have the same denominator
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<center> [[Bild:2_1_6b.gif]] </center>
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{{Displayed math||<math>\begin{align}
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\frac{x}{x-2}+\frac{x}{x+3}-2
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&= \frac{x}{x-2}\cdot\frac{x+3}{x+3} + \frac{x}{x+3}\cdot\frac{x-2}{x-2} - 2\cdot\frac{(x-2)(x+3)}{(x-2)(x+3)}\\[5pt]
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&= \frac{x(x+3)+x(x-2)-2(x-2)(x+3)}{(x-2)(x+3)}\\[5pt]
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&= \frac{x^{2}+3x+x^{2}-2x-2(x^{2}+3x-2x-6)}{(x-2)(x+3)}\\[5pt]
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&= \frac{x^{2}+3x+x^{2}-2x-2x^{2}-6x+4x+12}{(x-2)(x+3)}\,\textrm{.}
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\end{align}</math>}}
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Now, collect the terms in the numerator
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{{Displayed math||<math>\begin{align}
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\frac{x}{x-2}+\frac{x}{x+3}-2 &= \frac{(x^{2}+x^{2}-2x^{2})+(3x-2x-6x+4x)+12}{(x-2)(x+3)}\\[5pt]
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&= \frac{-x+12}{(x-2)(x+3)}\,\textrm{.}
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\end{align}</math>}}
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Note: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.

Current revision

The lowest common denominator for the three terms is (x2)(x+3) and we expand each term so that all terms have the same denominator

xx2+xx+32=xx2x+3x+3+xx+3x2x22(x2)(x+3)(x2)(x+3)=(x2)(x+3)x(x+3)+x(x2)2(x2)(x+3)=(x2)(x+3)x2+3x+x22x2(x2+3x2x6)=(x2)(x+3)x2+3x+x22x2x26x+4x+12.

Now, collect the terms in the numerator

xx2+xx+32=(x2)(x+3)(x2+x22x2)+(3x2x6x+4x)+12=x+12(x2)(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.