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Solution 2.2:3c

From Förberedande kurs i matematik 1

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Current revision (08:11, 24 September 2008) (edit) (undo)
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(The exercise is taken from an actual exam from Spring Term 1944!)
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Start by rewriting the terms on the left-hand side as one term having a common denominator
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Start by rewriting the terms on the left-hand side as one term having a common denominator:
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{{Displayed math||<math>\begin{align}
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\frac{1}{x-1}-\frac{1}{x+1} &= \frac{1}{x-1}\cdot\frac{x+1}{x+1} - \frac{1}{x+1}\cdot\frac{x-1}{x-1}\\[5pt]
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&= \frac{x+1}{(x-1)(x+1)} - \frac{x-1}{(x-1)(x+1)}\\[5pt]
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&= \frac{(x+1)-(x-1)}{(x-1)(x+1)}\\[5pt]
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&= \frac{2}{(x-1)(x+1)}\,\textrm{.}
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\end{align}</math>}}
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If we also write <math>3x-3=3(x-1)</math>, the equation can be rewritten as
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<math>\begin{align}
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{{Displayed math||<math>\frac{2}{(x-1)(x+1)}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3(x-1)}\,\textrm{.}</math>}}
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& \frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{x-1}\centerdot \frac{x+1}{x+1}-\frac{1}{x+1}\centerdot \frac{x-1}{x-1} \\
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& =\frac{x+1}{\left( x-1 \right)\left( x+1 \right)}-\frac{x-1}{\left( x-1 \right)\left( x+1 \right)}=\frac{\left( x+1 \right)-\left( x-1 \right)}{\left( x-1 \right)\left( x+1 \right)}=\frac{2}{\left( x-1 \right)\left( x+1 \right)} \\
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\end{align}</math>
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Because <math>x=1</math> cannot be a solution to the equation, the factor
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<math>x-1</math> can be removed from the denominator of both sides (i.e. actually, we multiply both sides by <math>x-1</math> and then eliminate it)
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If we also write
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{{Displayed math||<math>\frac{2}{x+1}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3}\,\textrm{.}</math>}}
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<math>3x-3=3\left( x-1 \right)</math>, the equation can be rewritten as
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Then, both sides are multiplied by 3 and <math>x+1</math>, so that we get an equation without any denominators
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<math>\frac{2}{\left( x-1 \right)\left( x+1 \right)}\left( x^{2}+\frac{1}{2} \right)=\frac{6x-1}{3\left( x-1 \right)}</math>
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{{Displayed math||<math>6\bigl(x^{2}+\tfrac{1}{2}\bigr) = (6x-1)(x+1)\,\textrm{.}</math>}}
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Because
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<math>x=1</math>
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cannot be a solution to the equation, the factor
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<math>x-1</math>
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can be removed from the denominator of both sides (i.e. actually, we multiply both sides by
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<math>x-1</math>
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and then eliminate it).
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<math>\frac{2}{x+1}\left( x^{2}+\frac{1}{2} \right)=\frac{6x-1}{3}</math>
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Then, both sides are multiplied by
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<math>3</math>
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and
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<math>x+1</math>, so that we get an equation without any denominators.
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<math>6\left( x^{2}+\frac{1}{2} \right)=\left( 6x-1 \right)\left( x+1 \right)</math>
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Expanding both sides
Expanding both sides
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{{Displayed math||<math>6x^{2}+3=6x^{2}+5x-1\,\textrm{.}</math>}}
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<math>6x^{2}+3=6x^{2}+5x-1</math>
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The ''x''² terms cancel each other out and we obtain a first-order equation,
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the x2 terms cancel each other out and we obtain a first-order equation,
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<math>3=5x-1</math>
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{{Displayed math||<math>3=5x-1\,,</math>}}
which has the solution
which has the solution
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{{Displayed math||<math>x=\frac{4}{5}\,\textrm{.}</math>}}
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<math>x=\frac{4}{5}</math>
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We check whether we have calculated correctly by substituting <math>x=4/5</math> into the original equation,
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We check whether we have calculated correctly by substituting x=4/5 into the original equation:
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<math>\begin{align}
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& \text{LHS}=\left( \frac{1}{\frac{4}{5}-1}-\frac{1}{\frac{4}{5}+1} \right)\left( \left( \frac{4}{5} \right)^{2}+\frac{1}{2} \right)=\left( \frac{1}{-\frac{1}{5}}-\frac{1}{\frac{9}{5}} \right)\left( \frac{16}{25}+\frac{1}{2} \right) \\
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& \\
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& =\left( -5-\frac{5}{9} \right)\frac{16\centerdot 2+25}{2\centerdot 25}=-\frac{50}{9}\centerdot \frac{57}{50}=-\frac{57}{9}=-\frac{19}{3}, \\
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\end{align}</math>
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<math>\text{RHS}=\frac{6\centerdot \frac{4}{5}-1}{3\centerdot \frac{4}{5}-3}=\frac{\frac{24}{5}-\frac{5}{5}}{\frac{12}{5}-\frac{15}{5}}=\frac{\frac{1}{5}\centerdot \left( 24-5 \right)}{\frac{1}{5}\centerdot \left( 12-15 \right)}=\frac{19}{-3}=-\frac{19}{3}</math>
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{{Displayed math||<math>\begin{align}
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\text{LHS} &= \biggl(\frac{1}{\frac{4}{5}-1}-\frac{1}{\frac{4}{5}+1}\biggr)\bigl( \bigl(\tfrac{4}{5}\bigr)^{2}+\tfrac{1}{2}\bigr)
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= \biggl(\frac{1}{-\frac{1}{5}}-\frac{1}{\frac{9}{5}}\biggr)\bigl(
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\tfrac{16}{25}+\tfrac{1}{2}\bigr)\\[5pt]
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&= \bigl(-5-\tfrac{5}{9}\bigr)\cdot\frac{16\cdot 2+25}{2\cdot 25}
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= -\frac{50}{9}\cdot\frac{57}{50}
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= -\frac{57}{9}
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= -\frac{19}{3}\,,\\[15pt]
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\text{RHS} &= \frac{6\cdot\frac{4}{5}-1}{3\cdot\frac{4}{5}-3}
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= \frac{\frac{24}{5}-\frac{5}{5}}{\frac{12}{5}-\frac{15}{5}}
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= \frac{\frac{1}{5}\cdot (24-5)}{\frac{1}{5}\cdot (12-15)}
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= \frac{19}{-3}
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= -\frac{19}{3}\,\textrm{.}
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\end{align}</math>}}

Current revision

Start by rewriting the terms on the left-hand side as one term having a common denominator

1x−1−1x+1=1x−1x+1x+1−1x+1x−1x−1=x+1(x−1)(x+1)−x−1(x−1)(x+1)=(x−1)(x+1)(x+1)−(x−1)=2(x−1)(x+1).

If we also write 3x−3=3(x−1), the equation can be rewritten as

2(x−1)(x+1)x2+21=6x−13(x−1). 

Because x=1 cannot be a solution to the equation, the factor x−1 can be removed from the denominator of both sides (i.e. actually, we multiply both sides by x−1 and then eliminate it)

2x+1x2+21=36x−1. 

Then, both sides are multiplied by 3 and x+1, so that we get an equation without any denominators

6x2+21=(6x−1)(x+1). 

Expanding both sides

6x2+3=6x2+5x−1.

The x² terms cancel each other out and we obtain a first-order equation,

3=5x−1

which has the solution

x=54.

We check whether we have calculated correctly by substituting x=45 into the original equation,

LHSRHS=154−1−154+1542+21=1−51−1592516+21=−5−95225162+25=−9505057=−957=−319=354−3654−1=524−55512−515=51(24−5)51(12−15)=19−3=−319.
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