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Solution 2.2:3b

From Förberedande kurs i matematik 1

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m (Lösning 2.2:3b moved to Solution 2.2:3b: Robot: moved page)
Current revision (07:40, 24 September 2008) (edit) (undo)
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First, we move all the terms over to the left-hand side,
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<center> [[Image:2_2_3b-1(2).gif]] </center>
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{{NAVCONTENT_STOP}}
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{{Displayed math||<math>\frac{4x}{4x-7}-\frac{1}{2x-3}-1=0\,\textrm{.}</math>}}
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{{NAVCONTENT_START}}
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<center> [[Image:2_2_3b-2(2).gif]] </center>
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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,
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{{NAVCONTENT_STOP}}
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{{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>}}
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and so that we can rewrite the left-hand side giving
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{{Displayed math||<math>\frac{4x(2x-3) - (4x-7) - (2x-3)(4x-7)}{(2x-3)(4x-7)}=0\,\textrm{.}</math>}}
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We expand the numerator
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{{Displayed math||<math>\frac{8x^{2}-12x-(4x-7)-(8x^{2}-14x-12x+21)}{(2x-3)(4x-7)} = 0</math>}}
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and simplify
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{{Displayed math||<math>\frac{10x-14}{(2x-3)(4x-7)}=0\,\textrm{.}</math>}}
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This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when
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{{Displayed math||<math>10x-14=0\,,</math>}}
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which gives <math>x=7/5\,</math>.
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It can easily happen that we calculate incorrectly, so we check that the answer <math>x=7/5</math> satisfies the equation,
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{{Displayed math||<math>\begin{align}
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\text{LHS } &= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7} - \frac{1}{2\cdot\frac{7}{5}-3}
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= \{\,\text{multiply top and bottom by 5}\,\}\\[5pt]
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&= \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}
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= \frac{4\cdot 7}{4\cdot 7-7\cdot 5}-\frac{5}{2\cdot 7-3\cdot 5}\\[5pt]
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&= \frac{4}{4-5}-\frac{5}{14-15}
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= -4-(-5) = 1 = \text{RHS.}
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\end{align}</math>}}

Current revision

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−3)(4x−7)(2x−3)(4x−7)=0

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

(2x−3)(4x−7)4x(2x−3)−(4x−7)−(2x−3)(4x−7)=0.

We expand the numerator

(2x−3)(4x−7)8x2−12x−(4x−7)−(8x2−14x−12x+21)=0

and simplify

10x−14(2x−3)(4x−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 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.
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