Solution 2.2:3b
From Förberedande kurs i matematik 1
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| - | {{ | + | First, we move all the terms over to the left-hand side: | 
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| - | {{ | + | <math>\frac{4x}{4x-7}-\frac{1}{2x-3}-1=0</math> | 
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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  | ||
| + | <math>x={7}/{5}\;</math>. | ||
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| + | It can easily happen that we calculate incorrectly, so we must check that the answer  | ||
| + | <math>x={7}/{5}\;</math> | ||
| + | satisfies the equation: | ||
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| + | <math>\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}</math> | ||
Revision as of 14:45, 17 September 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}
 
		  