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Solution 2.2:3a

From Förberedande kurs i matematik 1

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m (Lösning 2.2:3a moved to Solution 2.2:3a: Robot: moved page)
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We multiply the top and bottom of the terms on the left-hand side by appropriate factors so that they have the same common denominator, in the following way:
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<center> [[Image:2_2_3a-1(2).gif]] </center>
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<math>\frac{x+3}{x-3}\centerdot \frac{x-2}{x-2}-\frac{x+5}{x-2}\centerdot \frac{x-3}{x-3}=0</math>
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<center> [[Image:2_2_3a-2(2).gif]] </center>
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Now, the numerators can be subtracted:
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<math>\frac{\left( x+3 \right)\left( x-2 \right)-\left( x+5 \right)\left( x-3 \right)}{\left( x-2 \right)\left( x-3 \right)}=0</math>
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Expand the brackets in the numerator
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<math>\frac{x^{2}-2x+3x-6-\left( x^{2}-3x+5x-15 \right)}{\left( x-2 \right)\left( x-3 \right)}=0</math>
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and simplify
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<math>\frac{-x+9}{\left( x-2 \right)\left( x-3 \right)}=0</math>
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The left-hand side will be zero only when its numerator is zero (provided the denominator is not also zero), which gives us that the equation's solutions are given by the solutions to
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<math>-x+9=0</math>
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i.e.
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<math>x=9</math>.
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Substituting
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<math>x=9</math>
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into the original equation shows that we have calculated correctly:
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<math>\text{LHS}=\frac{9+3}{9-3}-\frac{9+5}{9-2}=\frac{12}{6}-\frac{14}{7}=2-2=0=\text{RHS}</math>

Revision as of 14:06, 17 September 2008

We multiply the top and bottom of the terms on the left-hand side by appropriate factors so that they have the same common denominator, in the following way:


\displaystyle \frac{x+3}{x-3}\centerdot \frac{x-2}{x-2}-\frac{x+5}{x-2}\centerdot \frac{x-3}{x-3}=0


Now, the numerators can be subtracted:


\displaystyle \frac{\left( x+3 \right)\left( x-2 \right)-\left( x+5 \right)\left( x-3 \right)}{\left( x-2 \right)\left( x-3 \right)}=0


Expand the brackets in the numerator


\displaystyle \frac{x^{2}-2x+3x-6-\left( x^{2}-3x+5x-15 \right)}{\left( x-2 \right)\left( x-3 \right)}=0


and simplify


\displaystyle \frac{-x+9}{\left( x-2 \right)\left( x-3 \right)}=0


The left-hand side will be zero only when its numerator is zero (provided the denominator is not also zero), which gives us that the equation's solutions are given by the solutions to


\displaystyle -x+9=0

i.e. \displaystyle x=9.

Substituting \displaystyle x=9 into the original equation shows that we have calculated correctly:


\displaystyle \text{LHS}=\frac{9+3}{9-3}-\frac{9+5}{9-2}=\frac{12}{6}-\frac{14}{7}=2-2=0=\text{RHS}

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