Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 08:45, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.2:3b moved to Solution 2.2:3b: Robot: moved page) ← Previous diff		Revision as of 14:45, 17 September 2008 (edit) (undo) Ian (Talk | contribs) Next diff →
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-	{{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>

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Revision as of 14:45, 17 September 2008

First, we move all the terms over to the left-hand side:

4x4x−7−12x−3−1=0

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,

4x4x−72x−32x−3−12x−34x−74x−7−2x−34x−72x−34x−7=0 

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

2x−34x−74x2x−3−4x−7−2x−34x−7=0 

We expand the numerator

2x−34x−78x2−12x−4x−7−8x2−14x−12x+21=0 

and simplify

10x−142x−34x−7=0 

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

10x−14=0

which gives x=75.

It can easily happen that we calculate incorrectly, so we must check that the answer x=75 satisfies the equation:

LHS    =    457457−7−1257−3 = multiply top and bottom by 5    = 457457−755−1257−355=4747−75−527−35=44−5−514−15=−4−−5=1    = RHS