Lösung 2.2:2a

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If we divide up the denominators that appear in the equation into small integer factors
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<center> [[Image:2_2_2a-1(2).gif]] </center>
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<math>6=2\centerdot 3</math>,
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<math>9=3\centerdot 3</math>
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and 2, we see that the lowest common denominator is
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<center> [[Image:2_2_2a-2(2).gif]] </center>
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<math>2\centerdot 3\centerdot 3=18</math>. Thus, we multiply both sides of the equation by
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<math>2\centerdot 3\centerdot 3</math>
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in order to avoid having denominators in the equation:
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<math>\begin{align}
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& 2\centerdot 3\centerdot 3\centerdot \frac{5x}{6}-2\centerdot 3\centerdot 3\centerdot \frac{x+2}{9}=2\centerdot 3\centerdot 3\centerdot \frac{1}{2} \\
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& \Leftrightarrow 3\centerdot 5x-2\centerdot \left( x+2 \right)=3\centerdot 3 \\
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\end{align}</math>
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We can rewrite the left-hand side as
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<math>3\centerdot 5x-2\centerdot \left( x+2 \right)=15x-2x-4=13x-4</math>, so that we get the equation
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<math>13x-4=9</math>
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We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get
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<math>x</math>
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by itself on one side:
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1. Add 4 to both sides:
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<math>13x-+4=9+4</math>
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which gives
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<math>13x=13</math>
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.
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2. Divide both sides by 13:
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<math>\frac{13x}{13}=\frac{13}{13}</math>
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which gives the answer
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<math>x=1</math>.
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The equation has
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<math>1</math>
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as the solution.
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When we have obtained an answer, it is important to go back to the original equation to check that
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<math>x=1</math>
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really is the correct answer( i.e. that we haven't calculated incorrectly):
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LHS =
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<math>\frac{5\centerdot 1}{6}-\frac{1\centerdot 2}{9}=\frac{5}{6}-\frac{3}{9}=\frac{5}{6}-\frac{1}{3}=\frac{5}{6}-\frac{1\centerdot 2}{3\centerdot 2}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2}</math>
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= RHS

Version vom 12:25, 17. Sep. 2008

If we divide up the denominators that appear in the equation into small integer factors \displaystyle 6=2\centerdot 3, \displaystyle 9=3\centerdot 3 and 2, we see that the lowest common denominator is \displaystyle 2\centerdot 3\centerdot 3=18. Thus, we multiply both sides of the equation by \displaystyle 2\centerdot 3\centerdot 3 in order to avoid having denominators in the equation:


\displaystyle \begin{align} & 2\centerdot 3\centerdot 3\centerdot \frac{5x}{6}-2\centerdot 3\centerdot 3\centerdot \frac{x+2}{9}=2\centerdot 3\centerdot 3\centerdot \frac{1}{2} \\ & \Leftrightarrow 3\centerdot 5x-2\centerdot \left( x+2 \right)=3\centerdot 3 \\ \end{align}


We can rewrite the left-hand side as \displaystyle 3\centerdot 5x-2\centerdot \left( x+2 \right)=15x-2x-4=13x-4, so that we get the equation


\displaystyle 13x-4=9


We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get \displaystyle x by itself on one side:

1. Add 4 to both sides:

\displaystyle 13x-+4=9+4

which gives

\displaystyle 13x=13 .

2. Divide both sides by 13:

\displaystyle \frac{13x}{13}=\frac{13}{13}

which gives the answer \displaystyle x=1.

The equation has \displaystyle 1 as the solution.

When we have obtained an answer, it is important to go back to the original equation to check that \displaystyle x=1 really is the correct answer( i.e. that we haven't calculated incorrectly):

LHS = \displaystyle \frac{5\centerdot 1}{6}-\frac{1\centerdot 2}{9}=\frac{5}{6}-\frac{3}{9}=\frac{5}{6}-\frac{1}{3}=\frac{5}{6}-\frac{1\centerdot 2}{3\centerdot 2}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2} = RHS

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