Lösung 2.2:2a

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

Version vom 08:26, 22. Okt. 2008

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

\displaystyle \begin{align}

& \rlap{/}2\cdot{}\rlap{/}3\cdot 3\cdot\frac{5x}{\rlap{/}6} - 2\cdot{}\rlap{/}3\cdot{}\rlap{/}3\cdot\frac{x+2}{\rlap{/}9} = \rlap{/}2\cdot 3\cdot 3\cdot \frac{1}{\rlap{/}2} \\[5pt] &\qquad\Leftrightarrow\quad 3\cdot 5x-2\cdot (x+2) = 3\cdot 3\,\textrm{.}\\ \end{align}

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

\displaystyle 13x-4=9\,\textrm{.}

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

  1. Add 4 to both sides, \displaystyle \vphantom{x_2}13x-4+4=9+4\,, which gives \displaystyle \ 13x=13\,\textrm{.}
  2. Divide both sides by 13, \displaystyle \frac{13x}{13}=\frac{13}{13}\,, which gives the answer \displaystyle \ x=1\,\textrm{.}

The equation has \displaystyle x=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)

\displaystyle \begin{align}

\text{LHS} &= \frac{5\cdot 1}{6}-\frac{1+2}{9} = \frac{5}{6}-\frac{3}{9} = \frac{5}{6}-\frac{1}{3}\\[5pt] &= \frac{5}{6}-\frac{1\cdot 2}{3\cdot 2} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2} = \text{RHS}\,\textrm{.} \end{align}

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