Solution 2.2:3c
From Förberedande kurs i matematik 1
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| - | {{ | + | Start by rewriting the terms on the left-hand side as one term having a common denominator |
| - | + | ||
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| - | {{ | + | \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] |
| - | < | + | &= \frac{x+1}{(x-1)(x+1)} - \frac{x-1}{(x-1)(x+1)}\\[5pt] |
| - | {{ | + | &= \frac{(x+1)-(x-1)}{(x-1)(x+1)}\\[5pt] |
| - | {{ | + | &= \frac{2}{(x-1)(x+1)}\,\textrm{.} |
| - | < | + | \end{align}</math>}} |
| - | + | ||
| + | If we also write <math>3x-3=3(x-1)</math>, the equation can be rewritten as | ||
| + | |||
| + | {{Displayed math||<math>\frac{2}{(x-1)(x+1)}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3(x-1)}\,\textrm{.}</math>}} | ||
| + | |||
| + | Because <math>x=1</math> cannot be a solution to the equation, the factor | ||
| + | <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) | ||
| + | |||
| + | {{Displayed math||<math>\frac{2}{x+1}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3}\,\textrm{.}</math>}} | ||
| + | |||
| + | Then, both sides are multiplied by 3 and <math>x+1</math>, so that we get an equation without any denominators | ||
| + | |||
| + | {{Displayed math||<math>6\bigl(x^{2}+\tfrac{1}{2}\bigr) = (6x-1)(x+1)\,\textrm{.}</math>}} | ||
| + | |||
| + | Expanding both sides | ||
| + | |||
| + | {{Displayed math||<math>6x^{2}+3=6x^{2}+5x-1\,\textrm{.}</math>}} | ||
| + | |||
| + | The ''x''² terms cancel each other out and we obtain a first-order equation, | ||
| + | |||
| + | {{Displayed math||<math>3=5x-1\,,</math>}} | ||
| + | |||
| + | which has the solution | ||
| + | |||
| + | {{Displayed math||<math>x=\frac{4}{5}\,\textrm{.}</math>}} | ||
| + | |||
| + | We check whether we have calculated correctly by substituting <math>x=4/5</math> into the original equation, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \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) | ||
| + | = \biggl(\frac{1}{-\frac{1}{5}}-\frac{1}{\frac{9}{5}}\biggr)\bigl( | ||
| + | \tfrac{16}{25}+\tfrac{1}{2}\bigr)\\[5pt] | ||
| + | &= \bigl(-5-\tfrac{5}{9}\bigr)\cdot\frac{16\cdot 2+25}{2\cdot 25} | ||
| + | = -\frac{50}{9}\cdot\frac{57}{50} | ||
| + | = -\frac{57}{9} | ||
| + | = -\frac{19}{3}\,,\\[15pt] | ||
| + | \text{RHS} &= \frac{6\cdot\frac{4}{5}-1}{3\cdot\frac{4}{5}-3} | ||
| + | = \frac{\frac{24}{5}-\frac{5}{5}}{\frac{12}{5}-\frac{15}{5}} | ||
| + | = \frac{\frac{1}{5}\cdot (24-5)}{\frac{1}{5}\cdot (12-15)} | ||
| + | = \frac{19}{-3} | ||
| + | = -\frac{19}{3}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
Start by rewriting the terms on the left-hand side as one term having a common denominator
| \displaystyle \begin{align}
\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] &= \frac{x+1}{(x-1)(x+1)} - \frac{x-1}{(x-1)(x+1)}\\[5pt] &= \frac{(x+1)-(x-1)}{(x-1)(x+1)}\\[5pt] &= \frac{2}{(x-1)(x+1)}\,\textrm{.} \end{align} |
If we also write \displaystyle 3x-3=3(x-1), the equation can be rewritten as
| \displaystyle \frac{2}{(x-1)(x+1)}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3(x-1)}\,\textrm{.} |
Because \displaystyle x=1 cannot be a solution to the equation, the factor \displaystyle x-1 can be removed from the denominator of both sides (i.e. actually, we multiply both sides by \displaystyle x-1 and then eliminate it)
| \displaystyle \frac{2}{x+1}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3}\,\textrm{.} |
Then, both sides are multiplied by 3 and \displaystyle x+1, so that we get an equation without any denominators
| \displaystyle 6\bigl(x^{2}+\tfrac{1}{2}\bigr) = (6x-1)(x+1)\,\textrm{.} |
Expanding both sides
| \displaystyle 6x^{2}+3=6x^{2}+5x-1\,\textrm{.} |
The x² terms cancel each other out and we obtain a first-order equation,
| \displaystyle 3=5x-1\,, |
which has the solution
| \displaystyle x=\frac{4}{5}\,\textrm{.} |
We check whether we have calculated correctly by substituting \displaystyle x=4/5 into the original equation,
| \displaystyle \begin{align}
\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) = \biggl(\frac{1}{-\frac{1}{5}}-\frac{1}{\frac{9}{5}}\biggr)\bigl( \tfrac{16}{25}+\tfrac{1}{2}\bigr)\\[5pt] &= \bigl(-5-\tfrac{5}{9}\bigr)\cdot\frac{16\cdot 2+25}{2\cdot 25} = -\frac{50}{9}\cdot\frac{57}{50} = -\frac{57}{9} = -\frac{19}{3}\,,\\[15pt] \text{RHS} &= \frac{6\cdot\frac{4}{5}-1}{3\cdot\frac{4}{5}-3} = \frac{\frac{24}{5}-\frac{5}{5}}{\frac{12}{5}-\frac{15}{5}} = \frac{\frac{1}{5}\cdot (24-5)}{\frac{1}{5}\cdot (12-15)} = \frac{19}{-3} = -\frac{19}{3}\,\textrm{.} \end{align} |
