Solution 2.1:6c

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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 \displaystyle a^{2}-ab = a(a-b) and \displaystyle a-b, both terms will have a common denominator \displaystyle a(a-b) if the top and bottom of the second term are multiplied by \displaystyle a,

\displaystyle \begin{align}

\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] &= \frac{2a+b-2a}{a(a-b)}\\[5pt] &= \frac{b}{a(a-b)}\,\textrm{.} \end{align}

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