Aus Online Mathematik Brückenkurs 1

Version vom 08:45, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 2.2:3c moved to Solution 2.2:3c: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 15:12, 17. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	(The exercise is taken from an actual exam from Spring Term 1944!)
-	<center> [[Image:2_2_3c-1(3).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	Start by rewriting the terms on the left-hand side as one term having a common denominator:
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:2_2_3c-2(3).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	<math>\begin{align}
-	{{NAVCONTENT_START}}	+	& \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} \\
-	<center> [[Image:2_2_3c-3(3).gif]] </center>	+	& =\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)} \\
-	{{NAVCONTENT_STOP}}	+	\end{align}</math>
		+
		+
		+	If we also write
		+	<math>3x-3=3\left( x-1 \right)</math>, the equation can be rewritten as
		+
		+
		+	<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>
		+
		+
		+	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
		+
		+
		+	<math>6x^{2}+3=6x^{2}+5x-1</math>
		+
		+
		+	the x2 terms cancel each other out and we obtain a first-order equation,
		+
		+
		+	<math>3=5x-1</math>
		+
		+
		+	which has the solution
		+
		+
		+	<math>x=\frac{4}{5}</math>
		+
		+
		+	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>

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Version vom 15:12, 17. Sep. 2008

(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:

\displaystyle \begin{align} & \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}

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

\displaystyle \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)}

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}\left( x^{2}+\frac{1}{2} \right)=\frac{6x-1}{3}

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

\displaystyle 6\left( x^{2}+\frac{1}{2} \right)=\left( 6x-1 \right)\left( x+1 \right)

Expanding both sides

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

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}

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

\displaystyle \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}

\displaystyle \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}