Lösung 2.2:3b

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K (Lösning 2.2:3b moved to Solution 2.2:3b: Robot: moved page)
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First, we move all the terms over to the left-hand side:
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<center> [[Image:2_2_3b-1(2).gif]] </center>
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<math>\frac{4x}{4x-7}-\frac{1}{2x-3}-1=0</math>
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<center> [[Image:2_2_3b-2(2).gif]] </center>
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Then, we multiply the top and bottom of all three terms by appropriate factors so that they have the same common denominator, in the following way,
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<math>\frac{4x}{4x-7}\centerdot \frac{2x-3}{2x-3}-\frac{1}{2x-3}\centerdot \frac{4x-7}{4x-7}-\frac{\left( 2x-3 \right)\left( 4x-7 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0</math>
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and so that we can rewrite the left-hand side giving
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<math>\frac{4x\left( 2x-3 \right)-\left( 4x-7 \right)-\left( 2x-3 \right)\left( 4x-7 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0</math>
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We expand the numerator
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<math>\frac{8x^{2}-12x-\left( 4x-7 \right)-\left( 8x^{2}-14x-12x+21 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0</math>
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and simplify
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<math>\frac{10x-14}{\left( 2x-3 \right)\left( 4x-7 \right)}=0</math>
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This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when
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<math>10x-14=0</math>
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which gives
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<math>x={7}/{5}\;</math>.
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It can easily happen that we calculate incorrectly, so we must check that the answer
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<math>x={7}/{5}\;</math>
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satisfies the equation:
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<math>\begin{align}
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& \text{LHS }~~~=\text{ }~~~\frac{4\centerdot \frac{7}{5}}{4\centerdot \frac{7}{5}-7}-\frac{1}{2\centerdot \frac{7}{5}-3}\text{ }=\text{ }\left\{ \text{ multiply top and bottom by 5} \right\} \\
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& \text{ }~~~ \\
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& =~\frac{4\centerdot \frac{7}{5}}{4\centerdot \frac{7}{5}-7}\centerdot \frac{5}{5}-\frac{1}{2\centerdot \frac{7}{5}-3}\centerdot \frac{5}{5}=\frac{4\centerdot 7}{4\centerdot 7-7\centerdot 5}-\frac{5}{2\centerdot 7-3\centerdot 5} \\
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& \\
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& =\frac{4}{4-5}-\frac{5}{14-15}=-4-\left( -5 \right)=1\text{ }~~~=\text{ RHS}~~~ \\
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\end{align}</math>

Version vom 14:45, 17. Sep. 2008

First, we move all the terms over to the left-hand side:


\displaystyle \frac{4x}{4x-7}-\frac{1}{2x-3}-1=0


Then, we multiply the top and bottom of all three terms by appropriate factors so that they have the same common denominator, in the following way,


\displaystyle \frac{4x}{4x-7}\centerdot \frac{2x-3}{2x-3}-\frac{1}{2x-3}\centerdot \frac{4x-7}{4x-7}-\frac{\left( 2x-3 \right)\left( 4x-7 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0


and so that we can rewrite the left-hand side giving


\displaystyle \frac{4x\left( 2x-3 \right)-\left( 4x-7 \right)-\left( 2x-3 \right)\left( 4x-7 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0


We expand the numerator


\displaystyle \frac{8x^{2}-12x-\left( 4x-7 \right)-\left( 8x^{2}-14x-12x+21 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0


and simplify


\displaystyle \frac{10x-14}{\left( 2x-3 \right)\left( 4x-7 \right)}=0


This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when


\displaystyle 10x-14=0


which gives \displaystyle x={7}/{5}\;.

It can easily happen that we calculate incorrectly, so we must check that the answer \displaystyle x={7}/{5}\; satisfies the equation:


\displaystyle \begin{align} & \text{LHS }~~~=\text{ }~~~\frac{4\centerdot \frac{7}{5}}{4\centerdot \frac{7}{5}-7}-\frac{1}{2\centerdot \frac{7}{5}-3}\text{ }=\text{ }\left\{ \text{ multiply top and bottom by 5} \right\} \\ & \text{ }~~~ \\ & =~\frac{4\centerdot \frac{7}{5}}{4\centerdot \frac{7}{5}-7}\centerdot \frac{5}{5}-\frac{1}{2\centerdot \frac{7}{5}-3}\centerdot \frac{5}{5}=\frac{4\centerdot 7}{4\centerdot 7-7\centerdot 5}-\frac{5}{2\centerdot 7-3\centerdot 5} \\ & \\ & =\frac{4}{4-5}-\frac{5}{14-15}=-4-\left( -5 \right)=1\text{ }~~~=\text{ RHS}~~~ \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:3b“