Solution 2.1:5b

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m (Lösning 2.1:5b moved to Solution 2.1:5b: Robot: moved page)
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We can factorize the denominators as
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<center> [[Image:2_1_5b.gif]] </center>
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<math>y^{2}-2y=y\left( y-2 \right)</math>
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<math>y^{2}-4=\left( y-2 \right)\left( y+2 \right)</math>
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[by the conjugate rule]
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and then we see that the terms' lowest common denominator is
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<math>y\left( y-2 \right)\left( y+2 \right)</math>
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because it is the product that contains the smallest number of factors which contain both
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<math>y\left( y-2 \right)</math>
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and
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<math>\left( y-2 \right)\left( y+2 \right)</math>
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.
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Now, we rewrite the fractions so that they have same denominators and start simplifying:
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<math>\begin{align}
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& y^{2}-4=\left( y-2 \right)\left( y+2 \right) \\
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& y^{2}-2y=y\left( y-2 \right) \\
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& \frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4}=\frac{1}{y\left( y-2 \right)}\centerdot \frac{y+2}{y+2}-\frac{2}{\left( y-2 \right)\left( y+2 \right)}\centerdot \frac{y}{y} \\
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& \\
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& =\frac{y+2}{y\left( y-2 \right)\left( y+2 \right)}-\frac{2y}{\left( y-2 \right)\left( y+2 \right)y} \\
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& \\
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& =\frac{y+2-2y}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-y+2}{y\left( y-2 \right)\left( y+2 \right)} \\
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\end{align}</math>
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The numerator can be rewritten as
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<math>-y+2=-\left( y-2 \right)</math>
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and we can eliminate the common factor
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<math>y-2</math>
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.
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<math>\frac{-y+2}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-\left( y-2 \right)}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-1}{y\left( y+2 \right)}=-\frac{1}{y\left( y+2 \right)}</math>

Revision as of 09:15, 16 September 2008

We can factorize the denominators as


\displaystyle y^{2}-2y=y\left( y-2 \right)



\displaystyle y^{2}-4=\left( y-2 \right)\left( y+2 \right) [by the conjugate rule]

and then we see that the terms' lowest common denominator is \displaystyle y\left( y-2 \right)\left( y+2 \right) because it is the product that contains the smallest number of factors which contain both \displaystyle y\left( y-2 \right) and \displaystyle \left( y-2 \right)\left( y+2 \right) .

Now, we rewrite the fractions so that they have same denominators and start simplifying:


\displaystyle \begin{align} & y^{2}-4=\left( y-2 \right)\left( y+2 \right) \\ & y^{2}-2y=y\left( y-2 \right) \\ & \frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4}=\frac{1}{y\left( y-2 \right)}\centerdot \frac{y+2}{y+2}-\frac{2}{\left( y-2 \right)\left( y+2 \right)}\centerdot \frac{y}{y} \\ & \\ & =\frac{y+2}{y\left( y-2 \right)\left( y+2 \right)}-\frac{2y}{\left( y-2 \right)\left( y+2 \right)y} \\ & \\ & =\frac{y+2-2y}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-y+2}{y\left( y-2 \right)\left( y+2 \right)} \\ \end{align}


The numerator can be rewritten as \displaystyle -y+2=-\left( y-2 \right) and we can eliminate the common factor \displaystyle y-2 .


\displaystyle \frac{-y+2}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-\left( y-2 \right)}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-1}{y\left( y+2 \right)}=-\frac{1}{y\left( y+2 \right)}

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