Solution 1.2:6 - Revision history

Tek at 12:35, 22 September 2008

2008-09-22T12:35:30Z

←Older revision		Revision as of 12:35, 22 September 2008
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	the answer can be simplified to		the answer can be simplified to

-	{{Displayed math\|\|<math>\frac{38\cdot 12}{15\cdot 7}=\frac{2\cdot 19\cdot 2\cdot 2\cdot{}\rlap{/}3}{\rlap{/}3\cdot 5\cdot 7}=\frac{~~153~~}{35}\,</math>.}}	+	{{Displayed math\|\|<math>\frac{38\cdot 12}{15\cdot 7}=\frac{2\cdot 19\cdot 2\cdot 2\cdot{}\rlap{/}3}{\rlap{/}3\cdot 5\cdot 7}=\frac{152}{35}\,</math>.}}

Tek at 14:22, 19 September 2008

2008-09-19T14:22:20Z

←Older revision		Revision as of 14:22, 19 September 2008
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-	When we work with large expressions, it is often best to proceed step by step. A first step on the way can be to simplify all the parts:	+	When we work with large expressions, it is often best to proceed step by step. A first step on the way can be to simplify all the parts

		+	{{Displayed math\|\|<math>\frac{2}{\,3+\dfrac{1}{2}\vphantom{\Biggl(}\,}\ ,\quad \frac{\dfrac{1}{2}\vphantom{\Biggl(}}{\,\dfrac{1}{4}-\dfrac{1}{3}\vphantom{\Biggl(}\,}\quad\text{and}\quad\frac{3}{\,2-\dfrac{2}{7}\vphantom{\Biggl(}\,}\,</math>.}}

-	~~<math>\frac{~~2~~}{3+\frac{1}{2}}\~~ ,~~\quad \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}}</math>~~	+	We can do this by multiplying the top and bottom of each fraction by 2, 12 and 7 respectively, so as to get rid of the partial fractions
-	and	+
-	~~<math>\frac{3}{2-\frac{2}{~~7~~}}</math>~~	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	~~We can do this by multiplying the top and bottom of each fraction by~~	+	\frac{2}{3+\dfrac{1}{2}\vphantom{\Biggl(}} &= \frac{2\cdot 2}{\left( 3+\dfrac{1}{2} \right)\cdot 2\vphantom{\Biggl(}} = \frac{4}{3\cdot 2+\dfrac{1}{2}\cdot 2\vphantom{\Biggl(}} = \frac{4}{6+1} = \frac{4}{7}\,,\\[5pt]
-	<math~~>2</math>~~	+	\frac{\dfrac{1}{2}\vphantom{\Biggl(}}{\dfrac{1}{4}-\dfrac{1}{3}\vphantom{\Biggl(}} &= \frac{\dfrac{1}{2}\cdot 12\vphantom{\Biggl(}}{\left( \dfrac{1}{4}-\dfrac{1}{3} \right)\cdot 12\vphantom{\Biggl(}} = \frac{6}{\dfrac{12}{4}-\dfrac{12}{3}\vphantom{\Biggl(}} = \frac{6}{3-4} = \frac{6}{-1} = -6\,,\\[10pt]
-	,	+	\frac{3}{2-\dfrac{2}{7}\vphantom{\Biggl(}} &= \frac{3\cdot 7}{\left( 2-\dfrac{2}{7} \right)\cdot 7\vphantom{\Biggl(}} = \frac{21}{2\cdot 7-\dfrac{2}{7}\cdot 7\vphantom{\Biggl(}} = \frac{21}{14-2} = \frac{21}{12}\,\textrm{.}
-	~~<math>12</math>~~	+	\end{align}</math>}}
-	~~and~~	+
-	~~<math>7</math>~~	+
-	~~respectively, so as to get rid of the partial fractions:~~	+
-		+
-		+
-	<math>\begin{align}	+
-	& \frac{2}{3+\~~frac~~{1}{2}}=\ \frac{2\~~centerdot~~ 2}{\left( 3+\~~frac~~{1}{2} \right)\~~centerdot~~ 2}=\frac{4}{3\~~centerdot~~ 2+\~~frac~~{1}{2}\~~centerdot~~ 2}=\frac{4}{6+1}=\frac{4}{7} \\	+
-	& \\	+
-	& \frac{\~~frac~~{1}{2}}{\~~frac~~{1}{4}-\~~frac~~{1}{3}}=\frac{\~~frac~~{1}{2}\~~centerdot~~ 12}{\left( \~~frac~~{1}{4}-\~~frac~~{1}{3} \right)\~~centerdot~~ 12}=\frac{6}{\~~frac~~{12}{4}-\~~frac~~{12}{3}}=\frac{6}{3-4}=\frac{6}{-1}=-6 \\	+
-	& \\	+
-	& \frac{3}{2-\~~frac~~{2}{7}}=\frac{3\~~centerdot~~ 7}{\left( 2-\~~frac~~{2}{7} \right)\~~centerdot~~ 7}=\frac{21}{2\~~centerdot~~ 7-\~~frac~~{2}{7}\~~centerdot~~ 7}=\frac{21}{14-2}=\frac{21}{12} \\	+
-	\end{align}</math>	+

	The whole expression therefore equals		The whole expression therefore equals

		+	{{Displayed math\|\|<math>\frac{\dfrac{4}{7}-6\vphantom{\Biggl(}}{\dfrac{1}{2}-\dfrac{21}{12}\vphantom{\Biggl(}}\,</math>.}}

-	~~<math>\frac{\frac{~~4}{7~~}-6}{\frac{~~1}{2~~}-\frac{~~21}{12}}</math>	+	If we multiply the tops and bottoms of the fractions 4/7, 1/2 and 21/12 in the main fraction by their lowest common denominator, <math>7\cdot 12</math>, we obtain integers in the numerator and denominator

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{\dfrac{4}{7}-6\vphantom{\Biggl(}}{\dfrac{1}{2}-\dfrac{21}{12}\vphantom{\Biggl(}} &= \frac{\left( \dfrac{4}{7}-6 \right)\cdot 7\cdot 12\vphantom{\Biggl(}}{\left( \dfrac{1}{2}-\dfrac{21}{12} \right)\cdot 7\cdot 12\vphantom{\Biggl(}} = \frac{4\cdot 12-6\cdot 7\cdot 12}{7\cdot 6-21\cdot 7}\\[10pt]
		+	& =\frac{( 4-6\cdot 7)\cdot 12}{( 6-21)\cdot 7} = \frac{-38\cdot 12}{-15\cdot 7} = \frac{38\cdot 12}{15\cdot 7}\,\textrm{.}
		+	\end{align}</math>}}

		+	By factorizing 12, 15 and 38,

-	~~If we multiply the tops and bottoms of the fractions~~	+	{{Displayed math\|\|<math>\begin{align}
-	~~<math></math>~~	+	12 &= 2\cdot 6 = 2\cdot 2\cdot 3\,,\\
-		+	15 &= 3\cdot 5\,,\\
-	~~<math>~~{~~4}/~~{~~7}\;</~~math>	+	38 &= 2\cdot 19\,,\\
-	,	+	\end{align}</math>}}
-	~~<math>{1}/{2}\;</math>~~	+
-	~~and~~	+
-	~~<math>{21}/{12}\;</math>~~	+
-	~~in the main fraction by~~	+
-	~~their lowest common denominator,~~	+
-	~~<math>7\centerdot 12</math>~~	+
-	~~, we obtain integers in the numerator and denominator:~~	+
-		+
-		+
-	<math>\begin{align}	+
-	~~& \frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{~~12}}=\frac{\left( \frac{4}{7}-6 \right)\centerdot 7\centerdot 12}{\left( \frac{1}{2}-\frac{21}{12} \right)\centerdot 7\centerdot 12}=\frac{4\centerdot 12-6\centerdot 7\centerdot 12}{7\centerdot 6-21\centerdot 7} \\	+
-	& \\	+
-	~~& =\frac{\left( 4-6\centerdot 7 \right)\centerdot 12}{\left( 6-21 \right)\centerdot 7}=\frac{-38\centerdot 12}{-15\centerdot 7}=\frac{38\centerdot 12}{15\centerdot 7} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	~~By factorizing~~	+
-	~~<math>12</math>~~	+
-	,	+
-	~~<math>15</math>~~	+
-	~~and~~	+
-	~~<math>38</math>~~	+
-	,	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& 12~~=2\~~centerdot~~ 6=2\~~centerdot~~ 2\~~centerdot~~ 3 \\	+
-	& 15=3\~~centerdot~~ 5 \\	+
-	& 38=2\~~centerdot~~ 19 \\	+
-	\end{align}</math>	+

	the answer can be simplified to		the answer can be simplified to

-		+	{{Displayed math\|\|<math>\frac{38\cdot 12}{15\cdot 7}=\frac{2\cdot 19\cdot 2\cdot 2\cdot{}\rlap{/}3}{\rlap{/}3\cdot 5\cdot 7}=\frac{153}{35}\,</math>.}}
-	<math>\frac{38\~~centerdot~~ 12}{15\~~centerdot~~ 7}=\frac{2\~~centerdot~~ 19\~~centerdot~~ 2\~~centerdot~~ 2\~~centerdot~~ 3}{3\~~centerdot~~ 5\~~centerdot~~ 7}=\frac{153}{35}</math>	+

Ian at 14:48, 11 September 2008

2008-09-11T14:48:18Z

←Older revision		Revision as of 14:48, 11 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	When we work with large expressions, it is often best to proceed step by step. A first step on the way can be to simplify all the parts:
-	<~~center~~> ~~[[Image~~:~~1_2_6~~-1(2)~~.gif]]~~ </~~center~~>	+
-	<~~center~~> ~~[[Image~~:~~1_2_6~~-2(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\frac{2}{3+\frac{1}{2}}\ ,\quad \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}}</math>
		+	and
		+	<math>\frac{3}{2-\frac{2}{7}}</math>
		+
		+
		+	We can do this by multiplying the top and bottom of each fraction by
		+	<math>2</math>
		+	,
		+	<math>12</math>
		+	and
		+	<math>7</math>
		+	respectively, so as to get rid of the partial fractions:
		+
		+
		+	<math>\begin{align}
		+	& \frac{2}{3+\frac{1}{2}}=\ \frac{2\centerdot 2}{\left( 3+\frac{1}{2} \right)\centerdot 2}=\frac{4}{3\centerdot 2+\frac{1}{2}\centerdot 2}=\frac{4}{6+1}=\frac{4}{7} \\
		+	& \\
		+	& \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}}=\frac{\frac{1}{2}\centerdot 12}{\left( \frac{1}{4}-\frac{1}{3} \right)\centerdot 12}=\frac{6}{\frac{12}{4}-\frac{12}{3}}=\frac{6}{3-4}=\frac{6}{-1}=-6 \\
		+	& \\
		+	& \frac{3}{2-\frac{2}{7}}=\frac{3\centerdot 7}{\left( 2-\frac{2}{7} \right)\centerdot 7}=\frac{21}{2\centerdot 7-\frac{2}{7}\centerdot 7}=\frac{21}{14-2}=\frac{21}{12} \\
		+	\end{align}</math>
		+
		+	The whole expression therefore equals
		+
		+
		+	<math>\frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}</math>
		+
		+
		+
		+	If we multiply the tops and bottoms of the fractions
		+	<math></math>
		+
		+	<math>{4}/{7}\;</math>
		+	,
		+	<math>{1}/{2}\;</math>
		+	and
		+	<math>{21}/{12}\;</math>
		+	in the main fraction by
		+	their lowest common denominator,
		+	<math>7\centerdot 12</math>
		+	, we obtain integers in the numerator and denominator:
		+
		+
		+	<math>\begin{align}
		+	& \frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}=\frac{\left( \frac{4}{7}-6 \right)\centerdot 7\centerdot 12}{\left( \frac{1}{2}-\frac{21}{12} \right)\centerdot 7\centerdot 12}=\frac{4\centerdot 12-6\centerdot 7\centerdot 12}{7\centerdot 6-21\centerdot 7} \\
		+	& \\
		+	& =\frac{\left( 4-6\centerdot 7 \right)\centerdot 12}{\left( 6-21 \right)\centerdot 7}=\frac{-38\centerdot 12}{-15\centerdot 7}=\frac{38\centerdot 12}{15\centerdot 7} \\
		+	\end{align}</math>
		+
		+
		+	By factorizing
		+	<math>12</math>
		+	,
		+	<math>15</math>
		+	and
		+	<math>38</math>
		+	,
		+
		+
		+	<math>\begin{align}
		+	& 12=2\centerdot 6=2\centerdot 2\centerdot 3 \\
		+	& 15=3\centerdot 5 \\
		+	& 38=2\centerdot 19 \\
		+	\end{align}</math>
		+
		+	the answer can be simplified to
		+
		+
		+	<math>\frac{38\centerdot 12}{15\centerdot 7}=\frac{2\centerdot 19\centerdot 2\centerdot 2\centerdot 3}{3\centerdot 5\centerdot 7}=\frac{153}{35}</math>

Tekbot: Lösning 1.2:6 moved to Solution 1.2:6: Robot: moved page

2008-09-09T08:20:23Z

Lösning 1.2:6 moved to Solution 1.2:6: Robot: moved page

←Older revision

Revision as of 08:20, 9 September 2008

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

2008-08-21T06:27:19Z

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

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2008-03-26T10:10:23Z

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