Lösung 2.2:3c

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
(The exercise is taken from an actual exam from Spring Term 1944!)
+
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:
+
{{Displayed math||<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{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}</math>}}
 +
If we also write <math>3x-3=3(x-1)</math>, the equation can be rewritten as
-
<math>\begin{align}
+
{{Displayed math||<math>\frac{2}{(x-1)(x+1)}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3(x-1)}\,\textrm{.}</math>}}
-
& \frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{x-1}\centerdot \frac{x+1}{x+1}-\frac{1}{x+1}\centerdot \frac{x-1}{x-1} \\
+
-
& =\frac{x+1}{\left( x-1 \right)\left( x+1 \right)}-\frac{x-1}{\left( x-1 \right)\left( x+1 \right)}=\frac{\left( x+1 \right)-\left( x-1 \right)}{\left( x-1 \right)\left( x+1 \right)}=\frac{2}{\left( x-1 \right)\left( x+1 \right)} \\
+
-
\end{align}</math>
+
 +
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)
-
If we also write
+
{{Displayed math||<math>\frac{2}{x+1}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3}\,\textrm{.}</math>}}
-
<math>3x-3=3\left( x-1 \right)</math>, the equation can be rewritten as
+
 +
Then, both sides are multiplied by 3 and <math>x+1</math>, so that we get an equation without any denominators
-
<math>\frac{2}{\left( x-1 \right)\left( x+1 \right)}\left( x^{2}+\frac{1}{2} \right)=\frac{6x-1}{3\left( x-1 \right)}</math>
+
{{Displayed math||<math>6\bigl(x^{2}+\tfrac{1}{2}\bigr) = (6x-1)(x+1)\,\textrm{.}</math>}}
-
 
+
-
 
+
-
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>\frac{2}{x+1}\left( x^{2}+\frac{1}{2} \right)=\frac{6x-1}{3}</math>
+
-
 
+
-
 
+
-
Then, both sides are multiplied by
+
-
<math>3</math>
+
-
and
+
-
<math>x+1</math>, so that we get an equation without any denominators.
+
-
 
+
-
 
+
-
<math>6\left( x^{2}+\frac{1}{2} \right)=\left( 6x-1 \right)\left( x+1 \right)</math>
+
-
 
+
Expanding both sides
Expanding both sides
 +
{{Displayed math||<math>6x^{2}+3=6x^{2}+5x-1\,\textrm{.}</math>}}
-
<math>6x^{2}+3=6x^{2}+5x-1</math>
+
The ''x''² terms cancel each other out and we obtain a first-order equation,
-
 
+
-
 
+
-
the x2 terms cancel each other out and we obtain a first-order equation,
+
-
 
+
-
 
+
-
<math>3=5x-1</math>
+
 +
{{Displayed math||<math>3=5x-1\,,</math>}}
which has the solution
which has the solution
 +
{{Displayed math||<math>x=\frac{4}{5}\,\textrm{.}</math>}}
-
<math>x=\frac{4}{5}</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 x=4/5 into the original equation:
+
-
 
+
-
 
+
-
<math>\begin{align}
+
-
& \text{LHS}=\left( \frac{1}{\frac{4}{5}-1}-\frac{1}{\frac{4}{5}+1} \right)\left( \left( \frac{4}{5} \right)^{2}+\frac{1}{2} \right)=\left( \frac{1}{-\frac{1}{5}}-\frac{1}{\frac{9}{5}} \right)\left( \frac{16}{25}+\frac{1}{2} \right) \\
+
-
& \\
+
-
& =\left( -5-\frac{5}{9} \right)\frac{16\centerdot 2+25}{2\centerdot 25}=-\frac{50}{9}\centerdot \frac{57}{50}=-\frac{57}{9}=-\frac{19}{3}, \\
+
-
\end{align}</math>
+
-
 
+
-
 
+
-
<math>\text{RHS}=\frac{6\centerdot \frac{4}{5}-1}{3\centerdot \frac{4}{5}-3}=\frac{\frac{24}{5}-\frac{5}{5}}{\frac{12}{5}-\frac{15}{5}}=\frac{\frac{1}{5}\centerdot \left( 24-5 \right)}{\frac{1}{5}\centerdot \left( 12-15 \right)}=\frac{19}{-3}=-\frac{19}{3}</math>
+
{{Displayed math||<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)
 +
= \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}</math>}}

Version vom 08:11, 24. Sep. 2008

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

Vorlage:Displayed math

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

Vorlage:Displayed math

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)

Vorlage:Displayed math

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

Vorlage:Displayed math

Expanding both sides

Vorlage:Displayed math

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

Vorlage:Displayed math

which has the solution

Vorlage:Displayed math

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

Vorlage:Displayed math

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