Lösung 2.2:3a

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We multiply the top and bottom of the terms on the left-hand side by appropriate factors so that they have the same common denominator, in the following way,
We multiply the top and bottom of the terms on the left-hand side by appropriate factors so that they have the same common denominator, in the following way,
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{{Displayed math||<math>\frac{x+3}{x-3}\cdot \frac{x-2}{x-2}-\frac{x+5}{x-2}\cdot \frac{x-3}{x-3}=0\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{x+3}{x-3}\cdot \frac{x-2}{x-2}-\frac{x+5}{x-2}\cdot \frac{x-3}{x-3}=0\,\textrm{.}</math>}}
Now, the numerators can be subtracted,
Now, the numerators can be subtracted,
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{{Displayed math||<math>\frac{(x+3)(x-2)-(x+5)(x-3 )}{(x-2)(x-3)}=0\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{(x+3)(x-2)-(x+5)(x-3 )}{(x-2)(x-3)}=0\,\textrm{.}</math>}}
Expand the brackets in the numerator,
Expand the brackets in the numerator,
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{{Displayed math||<math>\frac{x^{2}-2x+3x-6-(x^{2}-3x+5x-15)}{(x-2)(x-3)}=0</math>}}
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{{Abgesetzte Formel||<math>\frac{x^{2}-2x+3x-6-(x^{2}-3x+5x-15)}{(x-2)(x-3)}=0</math>}}
and simplify
and simplify
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{{Displayed math||<math>\frac{-x+9}{(x-2)(x-3)}=0\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{-x+9}{(x-2)(x-3)}=0\,\textrm{.}</math>}}
The left-hand side will be zero only when its numerator is zero (provided the denominator is not also zero), which gives us that the equation's solutions are given by the solutions to
The left-hand side will be zero only when its numerator is zero (provided the denominator is not also zero), which gives us that the equation's solutions are given by the solutions to
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{{Displayed math||<math>-x+9=0\,</math>,}}
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{{Abgesetzte Formel||<math>-x+9=0\,</math>,}}
i.e. <math>x=9</math>.
i.e. <math>x=9</math>.
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Substituting <math>x=9</math> into the original equation shows that we have calculated correctly,
Substituting <math>x=9</math> into the original equation shows that we have calculated correctly,
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{{Displayed math||<math>\text{LHS}=\frac{9+3}{9-3}-\frac{9+5}{9-2}=\frac{12}{6}-\frac{14}{7}=2-2=0=\text{RHS.}</math>}}
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{{Abgesetzte Formel||<math>\text{LHS}=\frac{9+3}{9-3}-\frac{9+5}{9-2}=\frac{12}{6}-\frac{14}{7}=2-2=0=\text{RHS.}</math>}}

Version vom 08:27, 22. Okt. 2008

We multiply the top and bottom of the terms on the left-hand side by appropriate factors so that they have the same common denominator, in the following way,

\displaystyle \frac{x+3}{x-3}\cdot \frac{x-2}{x-2}-\frac{x+5}{x-2}\cdot \frac{x-3}{x-3}=0\,\textrm{.}

Now, the numerators can be subtracted,

\displaystyle \frac{(x+3)(x-2)-(x+5)(x-3 )}{(x-2)(x-3)}=0\,\textrm{.}

Expand the brackets in the numerator,

\displaystyle \frac{x^{2}-2x+3x-6-(x^{2}-3x+5x-15)}{(x-2)(x-3)}=0

and simplify

\displaystyle \frac{-x+9}{(x-2)(x-3)}=0\,\textrm{.}

The left-hand side will be zero only when its numerator is zero (provided the denominator is not also zero), which gives us that the equation's solutions are given by the solutions to

\displaystyle -x+9=0\,,

i.e. \displaystyle x=9.

Substituting \displaystyle x=9 into the original equation shows that we have calculated correctly,

\displaystyle \text{LHS}=\frac{9+3}{9-3}-\frac{9+5}{9-2}=\frac{12}{6}-\frac{14}{7}=2-2=0=\text{RHS.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:3a“