Solution 1.2:5b - Revision history

Tek at 11:48, 19 September 2008

2008-09-19T11:48:15Z

←Older revision		Revision as of 11:48, 19 September 2008
Line 1:		Line 1:
	Method 1		Method 1

-	One solution is to calculate the numerator and denominator in the main fraction individually:	+	One solution is to calculate the numerator and denominator in the main fraction individually

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{1}{2}+\frac{1}{3} &= \frac{1\cdot 3}{2\cdot 3}+\frac{1\cdot 2}{3\cdot 2} = \frac{3}{6}+\frac{2}{6} = \frac{5}{6}\,,\\[10pt]
		+	\frac{1}{3}-\frac{1}{2} &= \frac{1\cdot 2}{3\cdot 2}-\frac{1\cdot 3}{2\cdot 3} = \frac{2}{6}-\frac{3}{6} = -\frac{1}{6}\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	The whole expression then reduces to a double fraction which we calculate by multiplying top and bottom by the reciprocal of the denominator
-	~~& \frac{1}{2}+\frac{1}{3}=\frac{1\centerdot 3}{2\centerdot 3}+\frac{1\centerdot 2}{3\centerdot 2}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6} \\~~	+
-	~~& \\~~	+
-	~~& \frac{1}{3}-\frac{1}{2}=\frac{1\centerdot 2}{3\centerdot 2}-\frac{1\centerdot 3}{2\centerdot 3}=\frac{2}{6}-\frac{3}{6}=-\frac{1}{6} \\~~	+
-	~~\end{align}</math>~~	+

-		+	{{Displayed math\|\|<math>
-		+	\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{6}\vphantom{\Biggl(}\,}{\,-\dfrac{1}{6}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}{\,-\dfrac{1}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}=\frac{5}{-1}=-5\,</math>.}}
-	~~The whole expression then reduces to a double fraction which we calculate by multiplying top and bottom by the reciprocal of the denominator:~~	+
-		+
-		+
-	<math>~~\begin{align}~~	+
-	& \frac{~~1}{2}+~~\~~frac{1}{3}=~~\~~frac~~{1~~\centerdot 3~~}{2~~\centerdot 3~~}+\~~frac~~{1~~\centerdot 2~~}{3~~\centerdot 2~~}=\~~frac~~{~~3}{6}+~~\~~frac{2~~}~~{6}=~~\~~frac{5~~}{6} \\	+
-	~~& \\~~	+
-	~~& \frac~~{1}{3}-\~~frac~~{1}{2}=\~~frac~~{1\~~centerdot 2~~}{3\~~centerdot 2~~}-\frac{1\~~centerdot 3}{2~~\~~centerdot 3}=\frac~~{2}{6}-\~~frac~~{3}{~~6}=~~-\~~frac~~{1}{6} \\	+
-	~~& \frac~~{\~~frac{1~~}{2}+\frac{1}{3}}{\~~frac~~{1}{3}-\~~frac~~{1}{2}}=\~~frac~~{\~~frac{5}{6~~}}{-\~~frac~~{1}{6}}=\~~frac~~{\~~frac~~{5}{6}\~~centerdot 6}~~{-\~~frac{1}{6~~}\~~centerdot 6~~}=\frac{5}{-1}=-5 \\	+
-	~~\end{align}~~</math>	+


	Method 2		Method 2

-	Another way to solve the exercise is to ~~multiplying~~ the top and bottom of the main fraction by	+	Another way to solve the exercise is to multiply the top and bottom of the main fraction by <math>3\cdot 2=6</math>, so that all denominators in the partial fractions 1/2 and 1/3 can eliminated in one step
-	<math>3\~~centerdot~~ 2=6</math>	+
-	, so that all denominators in the partial fractions	+
-	~~<math>{~~1}/{2~~}\;</math>~~	+
-	and	+
-	~~<math>{~~1}/{3~~}\;</math>~~	+
-	can eliminated in one step:	+
-		+

-		+	{{Displayed math\|\|<math>
-	<math>~~\begin{align}~~	+	\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\left( \dfrac{1}{2}+\dfrac{1}{3} \right)\cdot 6\vphantom{\Biggl(}\,}{\,\left( \dfrac{1}{3}-\dfrac{1}{2} \right)\cdot 6\vphantom{\Biggl(}\,}=\frac{\,\dfrac{6}{2}+\dfrac{6}{3}\vphantom{\Biggl(}\,}{\,\dfrac{6}{3}-\dfrac{6}{2}\vphantom{\Biggl(}\,}=\frac{3+2}{2-3}=\frac{5}{-1}=-5\,\textrm{.}</math>}}
-	~~& \\~~	+
-	& \frac{\~~frac~~{1}{2}+\~~frac~~{1}{3}}{\~~frac~~{1}{3}-\~~frac~~{1}{2}}=\frac{\left( \~~frac~~{1}{2}+\~~frac~~{1}{3} \right)\~~centerdot~~ 6}{\left( \~~frac~~{1}{3}-\~~frac~~{1}{2} \right)\~~centerdot~~ 6}=\frac{\~~frac~~{6}{2}+\~~frac~~{6}{3}}{\~~frac~~{6}{3}-\~~frac~~{6}{2}}=\frac{3+2}{2-3}=\frac{5}{-1}=-5 \\	+
-	~~\end~~{~~align~~}</math>	+

Ian at 14:54, 11 September 2008

2008-09-11T14:54:44Z

←Older revision		Revision as of 14:54, 11 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	Method 1
-	<~~center~~> ~~[[Image~~:~~1_2_5b.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	One solution is to calculate the numerator and denominator in the main fraction individually:
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{2}+\frac{1}{3}=\frac{1\centerdot 3}{2\centerdot 3}+\frac{1\centerdot 2}{3\centerdot 2}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6} \\
		+	& \\
		+	& \frac{1}{3}-\frac{1}{2}=\frac{1\centerdot 2}{3\centerdot 2}-\frac{1\centerdot 3}{2\centerdot 3}=\frac{2}{6}-\frac{3}{6}=-\frac{1}{6} \\
		+	\end{align}</math>
		+
		+
		+
		+	The whole expression then reduces to a double fraction which we calculate by multiplying top and bottom by the reciprocal of the denominator:
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{2}+\frac{1}{3}=\frac{1\centerdot 3}{2\centerdot 3}+\frac{1\centerdot 2}{3\centerdot 2}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6} \\
		+	& \\
		+	& \frac{1}{3}-\frac{1}{2}=\frac{1\centerdot 2}{3\centerdot 2}-\frac{1\centerdot 3}{2\centerdot 3}=\frac{2}{6}-\frac{3}{6}=-\frac{1}{6} \\
		+	& \frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{3}-\frac{1}{2}}=\frac{\frac{5}{6}}{-\frac{1}{6}}=\frac{\frac{5}{6}\centerdot 6}{-\frac{1}{6}\centerdot 6}=\frac{5}{-1}=-5 \\
		+	\end{align}</math>
		+
		+
		+	Method 2
		+
		+	Another way to solve the exercise is to multiplying the top and bottom of the main fraction by
		+	<math>3\centerdot 2=6</math>
		+	, so that all denominators in the partial fractions
		+	<math>{1}/{2}\;</math>
		+	and
		+	<math>{1}/{3}\;</math>
		+	can eliminated in one step:
		+
		+
		+
		+	<math>\begin{align}
		+	& \\
		+	& \frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{3}-\frac{1}{2}}=\frac{\left( \frac{1}{2}+\frac{1}{3} \right)\centerdot 6}{\left( \frac{1}{3}-\frac{1}{2} \right)\centerdot 6}=\frac{\frac{6}{2}+\frac{6}{3}}{\frac{6}{3}-\frac{6}{2}}=\frac{3+2}{2-3}=\frac{5}{-1}=-5 \\
		+	\end{align}</math>

Tekbot: Lösning 1.2:5b moved to Solution 1.2:5b: Robot: moved page

2008-09-09T08:19:43Z

Lösning 1.2:5b moved to Solution 1.2:5b: Robot: moved page

←Older revision

Revision as of 08:19, 9 September 2008

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

2008-08-21T06:26:59Z

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

←Older revision		Revision as of 06:26, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_2_5b.gif]] </center>	+	<center> [[Image:1_2_5b.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Lina: Ny sida: {{NAVCONTENT_START}} Bild:1_2_5b.gif {{NAVCONTENT_STOP}}

2008-03-26T10:08:18Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_5b.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:1_2_5b.gif]] </center>
{{NAVCONTENT_STOP}}