Lösung 2.2:3c

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
Start by rewriting the terms on the left-hand side as one term having a common denominator
Start by rewriting the terms on the left-hand side as one term having a common denominator
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{1}{x-1}-\frac{1}{x+1} &= \frac{1}{x-1}\cdot\frac{x+1}{x+1} - \frac{1}{x+1}\cdot\frac{x-1}{x-1}\\[5pt]
\frac{1}{x-1}-\frac{1}{x+1} &= \frac{1}{x-1}\cdot\frac{x+1}{x+1} - \frac{1}{x+1}\cdot\frac{x-1}{x-1}\\[5pt]
&= \frac{x+1}{(x-1)(x+1)} - \frac{x-1}{(x-1)(x+1)}\\[5pt]
&= \frac{x+1}{(x-1)(x+1)} - \frac{x-1}{(x-1)(x+1)}\\[5pt]
Zeile 10: Zeile 10:
If we also write <math>3x-3=3(x-1)</math>, the equation can be rewritten as
If we also write <math>3x-3=3(x-1)</math>, the equation can be rewritten as
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{{Displayed math||<math>\frac{2}{(x-1)(x+1)}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3(x-1)}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{2}{(x-1)(x+1)}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3(x-1)}\,\textrm{.}</math>}}
Because <math>x=1</math> cannot be a solution to the equation, the factor
Because <math>x=1</math> cannot be a solution to the equation, the factor
<math>x-1</math> can be removed from the denominator of both sides (i.e. actually, we multiply both sides by <math>x-1</math> and then eliminate it)
<math>x-1</math> can be removed from the denominator of both sides (i.e. actually, we multiply both sides by <math>x-1</math> and then eliminate it)
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{{Displayed math||<math>\frac{2}{x+1}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{2}{x+1}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3}\,\textrm{.}</math>}}
Then, both sides are multiplied by 3 and <math>x+1</math>, so that we get an equation without any denominators
Then, both sides are multiplied by 3 and <math>x+1</math>, so that we get an equation without any denominators
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{{Displayed math||<math>6\bigl(x^{2}+\tfrac{1}{2}\bigr) = (6x-1)(x+1)\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>6\bigl(x^{2}+\tfrac{1}{2}\bigr) = (6x-1)(x+1)\,\textrm{.}</math>}}
Expanding both sides
Expanding both sides
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{{Displayed math||<math>6x^{2}+3=6x^{2}+5x-1\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>6x^{2}+3=6x^{2}+5x-1\,\textrm{.}</math>}}
The ''x''² terms cancel each other out and we obtain a first-order equation,
The ''x''² terms cancel each other out and we obtain a first-order equation,
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{{Displayed math||<math>3=5x-1\,,</math>}}
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{{Abgesetzte Formel||<math>3=5x-1\,,</math>}}
which has the solution
which has the solution
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{{Displayed math||<math>x=\frac{4}{5}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>x=\frac{4}{5}\,\textrm{.}</math>}}
We check whether we have calculated correctly by substituting <math>x=4/5</math> into the original equation,
We check whether we have calculated correctly by substituting <math>x=4/5</math> into the original equation,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\text{LHS} &= \biggl(\frac{1}{\frac{4}{5}-1}-\frac{1}{\frac{4}{5}+1}\biggr)\bigl( \bigl(\tfrac{4}{5}\bigr)^{2}+\tfrac{1}{2}\bigr)
\text{LHS} &= \biggl(\frac{1}{\frac{4}{5}-1}-\frac{1}{\frac{4}{5}+1}\biggr)\bigl( \bigl(\tfrac{4}{5}\bigr)^{2}+\tfrac{1}{2}\bigr)
= \biggl(\frac{1}{-\frac{1}{5}}-\frac{1}{\frac{9}{5}}\biggr)\bigl(
= \biggl(\frac{1}{-\frac{1}{5}}-\frac{1}{\frac{9}{5}}\biggr)\bigl(

Version vom 08:27, 22. Okt. 2008

Start by rewriting the terms on the left-hand side as one term having a common denominator

\displaystyle \begin{align}

\frac{1}{x-1}-\frac{1}{x+1} &= \frac{1}{x-1}\cdot\frac{x+1}{x+1} - \frac{1}{x+1}\cdot\frac{x-1}{x-1}\\[5pt] &= \frac{x+1}{(x-1)(x+1)} - \frac{x-1}{(x-1)(x+1)}\\[5pt] &= \frac{(x+1)-(x-1)}{(x-1)(x+1)}\\[5pt] &= \frac{2}{(x-1)(x+1)}\,\textrm{.} \end{align}

If we also write \displaystyle 3x-3=3(x-1), the equation can be rewritten as

\displaystyle \frac{2}{(x-1)(x+1)}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3(x-1)}\,\textrm{.}

Because \displaystyle x=1 cannot be a solution to the equation, the factor \displaystyle x-1 can be removed from the denominator of both sides (i.e. actually, we multiply both sides by \displaystyle x-1 and then eliminate it)

\displaystyle \frac{2}{x+1}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3}\,\textrm{.}

Then, both sides are multiplied by 3 and \displaystyle x+1, so that we get an equation without any denominators

\displaystyle 6\bigl(x^{2}+\tfrac{1}{2}\bigr) = (6x-1)(x+1)\,\textrm{.}

Expanding both sides

\displaystyle 6x^{2}+3=6x^{2}+5x-1\,\textrm{.}

The x² terms cancel each other out and we obtain a first-order equation,

\displaystyle 3=5x-1\,,

which has the solution

\displaystyle x=\frac{4}{5}\,\textrm{.}

We check whether we have calculated correctly by substituting \displaystyle x=4/5 into the original equation,

\displaystyle \begin{align}

\text{LHS} &= \biggl(\frac{1}{\frac{4}{5}-1}-\frac{1}{\frac{4}{5}+1}\biggr)\bigl( \bigl(\tfrac{4}{5}\bigr)^{2}+\tfrac{1}{2}\bigr) = \biggl(\frac{1}{-\frac{1}{5}}-\frac{1}{\frac{9}{5}}\biggr)\bigl( \tfrac{16}{25}+\tfrac{1}{2}\bigr)\\[5pt] &= \bigl(-5-\tfrac{5}{9}\bigr)\cdot\frac{16\cdot 2+25}{2\cdot 25} = -\frac{50}{9}\cdot\frac{57}{50} = -\frac{57}{9} = -\frac{19}{3}\,,\\[15pt] \text{RHS} &= \frac{6\cdot\frac{4}{5}-1}{3\cdot\frac{4}{5}-3} = \frac{\frac{24}{5}-\frac{5}{5}}{\frac{12}{5}-\frac{15}{5}} = \frac{\frac{1}{5}\cdot (24-5)}{\frac{1}{5}\cdot (12-15)} = \frac{19}{-3} = -\frac{19}{3}\,\textrm{.} \end{align}

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