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Solution 2.1:6c

From Förberedande kurs i matematik 1

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Current revision (11:39, 23 September 2008) (edit) (undo)
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Because the numerators are
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Because the denominators are <math>a^{2}-ab = a(a-b)</math> and <math>a-b</math>, both terms will have a common denominator <math>a(a-b)</math> if the top and bottom of the second term are multiplied by <math>a</math>,
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<math>a^{2}-ab=a\left( a-b \right)</math>
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and
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<math>a-b</math>, both terms will have a common denominator
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<math>a\left( a-b \right)</math>
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if the top and bottom of the second term are multiplied by
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<math>a</math>:
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{2a+b}{a^{2}-b}-\frac{2}{a-b} &= \frac{2a+b}{a(a-b)}-\frac{2}{a-b}\cdot\frac{a}{a}\\[5pt]
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& \frac{2a+b}{a^{2}-b}-\frac{2}{a-b}=\frac{2a+b}{a\left( a-b \right)}-\frac{2}{a-b}\centerdot \frac{a}{a} \\
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&= \frac{2a+b-2a}{a(a-b)}\\[5pt]
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& =\frac{2a+b-2a}{a\left( a-b \right)}=\frac{b}{a\left( a-b \right)} \\
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&= \frac{b}{a(a-b)}\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}

Current revision

Because the denominators are a2ab=a(ab) and ab, both terms will have a common denominator a(ab) if the top and bottom of the second term are multiplied by a,

a2b2a+b2ab=2a+ba(ab)2abaa=a(ab)2a+b2a=ba(ab).