Lösung 2.2:3b

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First, we move all the terms over to the left-hand side:
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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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{{Displayed math||<math>\frac{4x}{4x-7}-\frac{1}{2x-3}-1=0\,\textrm{.}</math>}}
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,
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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{{Displayed math||<math>\frac{4x}{4x-7}\cdot\frac{2x-3}{2x-3} - \frac{1}{2x-3}\cdot\frac{4x-7}{4x-7} - \frac{(2x-3)(4x-7)}{(2x-3)(4x-7)} = 0</math>}}
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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
and so that we can rewrite the left-hand side giving
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{{Displayed math||<math>\frac{4x(2x-3) - (4x-7) - (2x-3)(4x-7)}{(2x-3)(4x-7)}=0\,\textrm{.}</math>}}
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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
We expand the numerator
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{{Displayed math||<math>\frac{8x^{2}-12x-(4x-7)-(8x^{2}-14x-12x+21)}{(2x-3)(4x-7)} = 0</math>}}
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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
and simplify
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{{Displayed math||<math>\frac{10x-14}{(2x-3)(4x-7)}=0\,\textrm{.}</math>}}
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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
This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when
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{{Displayed math||<math>10x-14=0\,,</math>}}
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<math>10x-14=0</math>
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which gives <math>x=7/5\,</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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It can easily happen that we calculate incorrectly, so we check that the answer <math>x=7/5</math> satisfies the equation,
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<math>\begin{align}
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{{Displayed math||<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{LHS } &= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7} - \frac{1}{2\cdot\frac{7}{5}-3}
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& \text{ }~~~ \\
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= \{\,\text{multiply top and bottom by 5}\,\}\\[5pt]
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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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&= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7}\cdot\frac{5}{5} - \frac{1}{2\cdot \frac{7}{5}-3}\cdot\frac{5}{5}
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& \\
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= \frac{4\cdot 7}{4\cdot 7-7\cdot 5}-\frac{5}{2\cdot 7-3\cdot 5}\\[5pt]
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& =\frac{4}{4-5}-\frac{5}{14-15}=-4-\left( -5 \right)=1\text{ }~~~=\text{ RHS}~~~ \\
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&= \frac{4}{4-5}-\frac{5}{14-15}
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\end{align}</math>
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= -4-(-5) = 1 = \text{RHS.}
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\end{align}</math>}}

Version vom 07:40, 24. Sep. 2008

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

Vorlage:Displayed math

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,

Vorlage:Displayed math

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

Vorlage:Displayed math

We expand the numerator

Vorlage:Displayed math

and simplify

Vorlage:Displayed math

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

Vorlage:Displayed math

which gives \displaystyle x=7/5\,.

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

Vorlage:Displayed math

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