Solution 2.2:3c - Revision history

Tek at 08:11, 24 September 2008

Tek — Wed, 24 Sep 2008 08:11:04 GMT

←Older revision		Revision as of 08:11, 24 September 2008
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-	~~(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>}}

Ian at 15:12, 17 September 2008

Ian — Wed, 17 Sep 2008 15:12:55 GMT

←Older revision		Revision as of 15:12, 17 September 2008
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-	{{~~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>

Tekbot: Lösning 2.2:3c moved to Solution 2.2:3c: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:45:43 GMT

Lösning 2.2:3c moved to Solution 2.2:3c: Robot: moved page

←Older revision

Revision as of 08:45, 9 September 2008

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Tekbot — Thu, 21 Aug 2008 06:39:59 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Lina — Mon, 31 Mar 2008 14:39:26 GMT

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