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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
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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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<math>\left( x-2 \right)\left( x+3 \right)</math>
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and we expand each term so that all terms have the same denominator:
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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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<math>\begin{align}
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Now, collect the terms in the numerator
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& \frac{x}{x-2}+\frac{x}{x+3}-2=\frac{x}{x-2}\centerdot \frac{x+3}{x+3}+\frac{x}{x+3}\centerdot \frac{x-2}{x-2}-2\centerdot \frac{\left( x-2 \right)\left( x+3 \right)}{\left( x-2 \right)\left( x+3 \right)} \\
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& =\frac{x\left( x+3 \right)+x\left( x-2 \right)-2\left( x-2 \right)\left( x+3 \right)}{\left( x-2 \right)\left( x+3 \right)} \\
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& =\frac{x^{2}+3x+x^{2}-2x-2\left( x^{2}+3x-2x-6 \right)}{\left( x-2 \right)\left( x+3 \right)} \\
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& =\frac{x^{2}+3x+x^{2}-2x-2x^{2}-6x+4x+12}{\left( x-2 \right)\left( x+3 \right)} \\
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\end{align}</math>
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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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Now, collect together the terms in the numerator:
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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.
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<math>\begin{align}
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& \frac{x}{x-2}+\frac{x}{x+3}-2=\frac{\left( x^{2}+x^{2}-2x^{2} \right)+\left( 3x-2x-6x+4x \right)+12}{\left( x-2 \right)\left( x+3 \right)} \\
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& =\frac{-x+12}{\left( x-2 \right)\left( x+3 \right)} \\
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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.
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Current revision

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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