Lösung 2.3:3e

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In this case, we see that the left-hand side contains the factor
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<center> [[Image:2_3_3e.gif]] </center>
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<math>x+\text{3}</math>, which we can take out to obtain
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<math>\begin{align}
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& \left( x+\text{3} \right)\left( x-1 \right)-\left( x+\text{3} \right)\left( 2x-9 \right)=\left( x+\text{3} \right)\left( \left( x-1 \right)-\left( 2x-9 \right) \right) \\
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& =\left( x+\text{3} \right)\left( x-1-2x+9 \right)=\left( x+\text{3} \right)\left( -x+8 \right). \\
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\end{align}</math>
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This rewriting of the equation results in the new equation
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<math>\left( x+\text{3} \right)\left( -x+8 \right)=0</math>
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which has the solutions
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<math>x=-\text{3}</math>
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and
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<math>x=\text{8}</math>.
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We check the solution
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<math>x=\text{8 }</math>
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by substituting it into the equation:
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LHS
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<math>=\left( 8+3 \right)\centerdot \left( 8-1 \right)-\left( 8+3 \right)\centerdot \left( 2\centerdot 8-9 \right)=11\centerdot 7-11\centerdot 7=0=</math>
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RHS

Version vom 15:14, 20. Sep. 2008

In this case, we see that the left-hand side contains the factor \displaystyle x+\text{3}, which we can take out to obtain


\displaystyle \begin{align} & \left( x+\text{3} \right)\left( x-1 \right)-\left( x+\text{3} \right)\left( 2x-9 \right)=\left( x+\text{3} \right)\left( \left( x-1 \right)-\left( 2x-9 \right) \right) \\ & =\left( x+\text{3} \right)\left( x-1-2x+9 \right)=\left( x+\text{3} \right)\left( -x+8 \right). \\ \end{align}


This rewriting of the equation results in the new equation


\displaystyle \left( x+\text{3} \right)\left( -x+8 \right)=0


which has the solutions \displaystyle x=-\text{3} and \displaystyle x=\text{8}.

We check the solution \displaystyle x=\text{8 } by substituting it into the equation:

LHS \displaystyle =\left( 8+3 \right)\centerdot \left( 8-1 \right)-\left( 8+3 \right)\centerdot \left( 2\centerdot 8-9 \right)=11\centerdot 7-11\centerdot 7=0= RHS

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