Lösung 2.2:1c

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Because there is an
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<center> [[Image:2_2_1c.gif]] </center>
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<math>x</math>
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on both the left- and right-hand sides,
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the first step is to subtract
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<math>{x}/{3}\;</math>
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from both sides,
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<math>\frac{1}{3}x-1-\frac{1}{3}x=x-\frac{1}{3}x</math>
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so as to collect
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<math>x</math>
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on the right-hand side
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<math>-1=\frac{2}{3}x.</math>
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Then, multiply both sides by
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<math>{3}/{2}\;</math>,
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<math>\frac{3}{2}\centerdot \left( -1 \right)=\frac{3}{2}\centerdot \frac{2}{3}x</math>
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so that
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<math>{2}/{3}\;</math>
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can be eliminated on the right-hand side to give us
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<math>-\frac{3}{2}=x</math>

Version vom 14:21, 16. Sep. 2008

Because there is an \displaystyle x on both the left- and right-hand sides,

the first step is to subtract \displaystyle {x}/{3}\; from both sides,


\displaystyle \frac{1}{3}x-1-\frac{1}{3}x=x-\frac{1}{3}x


so as to collect \displaystyle x on the right-hand side


\displaystyle -1=\frac{2}{3}x.


Then, multiply both sides by \displaystyle {3}/{2}\;,


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


so that \displaystyle {2}/{3}\; can be eliminated on the right-hand side to give us


\displaystyle -\frac{3}{2}=x

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:1c“