Lösung 2.1:7a

Aus Online Mathematik Brückenkurs 1

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If we multiply the top and bottom of the first fraction by
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<center> [[Image:2_1_7a.gif]] </center>
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<math>x+5</math>
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and the second by
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<math>x+3</math>, then they will both have the same numerator and we can work out the expression by subtracting the numerators:
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<math>\begin{align}
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& \frac{2}{x+3}-\frac{2}{x+5}=\frac{2}{x+3}\centerdot \frac{x+5}{x+5}-\frac{2}{x+5}\centerdot \frac{x+3}{x+3} \\
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& =\frac{2\left( x+5 \right)-2\left( x+3 \right)}{\left( x+3 \right)\left( x+5 \right)}=\frac{2x+10-2x-6}{\left( x+3 \right)\left( x+5 \right)}=\frac{4}{\left( x+3 \right)\left( x+5 \right)} \\
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\end{align}</math>

Version vom 12:38, 16. Sep. 2008

If we multiply the top and bottom of the first fraction by \displaystyle x+5 and the second by \displaystyle x+3, then they will both have the same numerator and we can work out the expression by subtracting the numerators:


\displaystyle \begin{align} & \frac{2}{x+3}-\frac{2}{x+5}=\frac{2}{x+3}\centerdot \frac{x+5}{x+5}-\frac{2}{x+5}\centerdot \frac{x+3}{x+3} \\ & =\frac{2\left( x+5 \right)-2\left( x+3 \right)}{\left( x+3 \right)\left( x+5 \right)}=\frac{2x+10-2x-6}{\left( x+3 \right)\left( x+5 \right)}=\frac{4}{\left( x+3 \right)\left( x+5 \right)} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:7a“