Solution 2.2:3b - Revision history

Tek at 07:40, 24 September 2008

2008-09-24T07:40:38Z

←Older revision		Revision as of 07:40, 24 September 2008
Line 1:		Line 1:
-	First, we move all the terms over to the left-hand side:	+	First, we move all the terms over to the left-hand side,
-		+
-		+
-	~~<math>\frac{4x}{4x-7}-\frac{1}{2x-3}-1=0</math>~~	+

		+	{{Displayed math\|\|<math>\frac{4x}{4x-7}-\frac{1}{2x-3}-1=0\,\textrm{.}</math>}}

	Then, we multiply the top and bottom of all three terms by appropriate factors so that they have the same common denominator, in the following way,		Then, we multiply the top and bottom of all three terms by appropriate factors so that they have the same common denominator, in the following way,

-		+	{{Displayed math\|\|<math>\frac{4x}{4x-7}\cdot\frac{2x-3}{2x-3} - \frac{1}{2x-3}\cdot\frac{4x-7}{4x-7} - \frac{(2x-3)(4x-7)}{(2x-3)(4x-7)} = 0</math>}}
-	<math>\frac{4x}{4x-7}\~~centerdot~~ \frac{2x-3}{2x-3}-\frac{1}{2x-3}\~~centerdot~~ \frac{4x-7}{4x-7}-\frac{~~\left~~( 2x-3 ~~\right~~)~~\left~~( 4x-7 ~~\right~~)}{~~\left~~( 2x-3 ~~\right~~)~~\left~~( 4x-7 ~~\right~~)}=0</math>	+
-		+

	and so that we can rewrite the left-hand side giving		and so that we can rewrite the left-hand side giving

-		+	{{Displayed math\|\|<math>\frac{4x(2x-3) - (4x-7) - (2x-3)(4x-7)}{(2x-3)(4x-7)}=0\,\textrm{.}</math>}}
-	<math>\frac{4x~~\left~~( 2x-3 ~~\right~~)-~~\left~~( 4x-7 ~~\right~~)-~~\left~~( 2x-3 ~~\right~~)~~\left~~( 4x-7 ~~\right~~)}{~~\left~~( 2x-3 ~~\right~~)~~\left~~( 4x-7 ~~\right~~)}=0</math>	+
-		+

	We expand the numerator		We expand the numerator

-		+	{{Displayed math\|\|<math>\frac{8x^{2}-12x-(4x-7)-(8x^{2}-14x-12x+21)}{(2x-3)(4x-7)} = 0</math>}}
-	<math>\frac{8x^{2}-12x-~~\left~~( 4x-7 ~~\right~~)-~~\left~~( 8x^{2}-14x-12x+21 ~~\right~~)}{~~\left~~( 2x-3 ~~\right~~)~~\left~~( 4x-7 ~~\right~~)}=0</math>	+
-		+

	and simplify		and simplify

-		+	{{Displayed math\|\|<math>\frac{10x-14}{(2x-3)(4x-7)}=0\,\textrm{.}</math>}}
-	<math>\frac{10x-14}{~~\left~~( 2x-3 ~~\right~~)~~\left~~( 4x-7 ~~\right~~)}=0</math>	+
-		+

	This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when		This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when

		+	{{Displayed math\|\|<math>10x-14=0\,,</math>}}

-	~~<math>10x-14=0</math>~~	+	which gives <math>x=7/5\,</math>.
-		+
-		+
-	which gives	+
-	<math>x={7}/{5}\;</math>.	+
-		+
-	~~It can easily happen that we calculate incorrectly, so we must check that the answer~~	+
-	~~<math>x={7}/{5}\;</math>~~	+
-	~~satisfies the equation:~~	+

		+	It can easily happen that we calculate incorrectly, so we check that the answer <math>x=7/5</math> satisfies the equation,

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \text{LHS }~~~~~~~=~~\text{ }~~~~~\frac{4\~~centerdot~~ \frac{7}{5}}{4\~~centerdot~~ \frac{7}{5}-7}-\frac{1}{2\~~centerdot~~ \frac{7}{5}-3}~~\text{ }~~=\~~text~~{ }\~~left\{~~ \text{ multiply top and bottom by 5} \~~right~~\} \\	+	\text{LHS } &= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7} - \frac{1}{2\cdot\frac{7}{5}-3}
-	~~& \text{ }~~~ \\~~	+	= \{\,\text{multiply top and bottom by 5}\,\}\\[5pt]
-	& =~\frac{4\~~centerdot~~ \frac{7}{5}}{4\~~centerdot~~ \frac{7}{5}-7}\~~centerdot~~ \frac{5}{5}-\frac{1}{2\~~centerdot~~ \frac{7}{5}-3}\~~centerdot~~ \frac{5}{5}=\frac{4\~~centerdot~~ 7}{4\~~centerdot~~ 7-7\~~centerdot~~ 5}-\frac{5}{2\~~centerdot~~ 7-3\~~centerdot~~ 5} \\	+	&= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7}\cdot\frac{5}{5} - \frac{1}{2\cdot \frac{7}{5}-3}\cdot\frac{5}{5}
-	~~& \\~~	+	= \frac{4\cdot 7}{4\cdot 7-7\cdot 5}-\frac{5}{2\cdot 7-3\cdot 5}\\[5pt]
-	& =\frac{4}{4-5}-\frac{5}{14-15}=-4-~~\left~~( -5 ~~\right~~)=1~~\text{ }~~~~~=\text{ RHS}~~~~~ \\~~	+	&= \frac{4}{4-5}-\frac{5}{14-15}
-	\end{align}</math>	+	= -4-(-5) = 1 = \text{RHS.}
		+	\end{align}</math>}}

Ian at 14:45, 17 September 2008

2008-09-17T14:45:41Z

←Older revision		Revision as of 14:45, 17 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	First, we move all the terms over to the left-hand side:
-	<~~center~~> ~~[[Image:2_2_3b~~-1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP}~~}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\frac{4x}{4x-7}-\frac{1}{2x-3}-1=0</math>
-	<~~center~~> ~~[[Image:2_2_3b~~-2(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	Then, we multiply the top and bottom of all three terms by appropriate factors so that they have the same common denominator, in the following way,
		+
		+
		+	<math>\frac{4x}{4x-7}\centerdot \frac{2x-3}{2x-3}-\frac{1}{2x-3}\centerdot \frac{4x-7}{4x-7}-\frac{\left( 2x-3 \right)\left( 4x-7 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0</math>
		+
		+
		+	and so that we can rewrite the left-hand side giving
		+
		+
		+	<math>\frac{4x\left( 2x-3 \right)-\left( 4x-7 \right)-\left( 2x-3 \right)\left( 4x-7 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0</math>
		+
		+
		+	We expand the numerator
		+
		+
		+	<math>\frac{8x^{2}-12x-\left( 4x-7 \right)-\left( 8x^{2}-14x-12x+21 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0</math>
		+
		+
		+	and simplify
		+
		+
		+	<math>\frac{10x-14}{\left( 2x-3 \right)\left( 4x-7 \right)}=0</math>
		+
		+
		+	This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when
		+
		+
		+	<math>10x-14=0</math>
		+
		+
		+	which gives
		+	<math>x={7}/{5}\;</math>.
		+
		+	It can easily happen that we calculate incorrectly, so we must check that the answer
		+	<math>x={7}/{5}\;</math>
		+	satisfies the equation:
		+
		+
		+	<math>\begin{align}
		+	& \text{LHS }~~~=\text{ }~~~\frac{4\centerdot \frac{7}{5}}{4\centerdot \frac{7}{5}-7}-\frac{1}{2\centerdot \frac{7}{5}-3}\text{ }=\text{ }\left\{ \text{ multiply top and bottom by 5} \right\} \\
		+	& \text{ }~~~ \\
		+	& =~\frac{4\centerdot \frac{7}{5}}{4\centerdot \frac{7}{5}-7}\centerdot \frac{5}{5}-\frac{1}{2\centerdot \frac{7}{5}-3}\centerdot \frac{5}{5}=\frac{4\centerdot 7}{4\centerdot 7-7\centerdot 5}-\frac{5}{2\centerdot 7-3\centerdot 5} \\
		+	& \\
		+	& =\frac{4}{4-5}-\frac{5}{14-15}=-4-\left( -5 \right)=1\text{ }~~~=\text{ RHS}~~~ \\
		+	\end{align}</math>

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

2008-09-09T08:45:23Z

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

←Older revision

Revision as of 08:45, 9 September 2008

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

2008-08-21T06:39:49Z

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

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2008-03-31T14:37:47Z

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