Processing Math: Done
Solution 2.1:6b
From Förberedande kurs i matematik 1
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| - | {{ | + | The lowest common denominator for the three terms is <math>(x-2)(x+3)</math> and we expand each term so that all terms have the same denominator |
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| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | \frac{x}{x-2}+\frac{x}{x+3}-2 | ||
| + | &= \frac{x}{x-2}\cdot\frac{x+3}{x+3} + \frac{x}{x+3}\cdot\frac{x-2}{x-2} - 2\cdot\frac{(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] | ||
| + | &= \frac{x(x+3)+x(x-2)-2(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] | ||
| + | &= \frac{x^{2}+3x+x^{2}-2x-2(x^{2}+3x-2x-6)}{(x-2)(x+3)}\\[5pt] | ||
| + | &= \frac{x^{2}+3x+x^{2}-2x-2x^{2}-6x+4x+12}{(x-2)(x+3)}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
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| + | Now, collect the terms in the numerator | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{x}{x-2}+\frac{x}{x+3}-2 &= \frac{(x^{2}+x^{2}-2x^{2})+(3x-2x-6x+4x)+12}{(x-2)(x+3)}\\[5pt] | ||
| + | &= \frac{-x+12}{(x-2)(x+3)}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | Note: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further. | ||
Current revision
The lowest common denominator for the three terms is
 x+3x+3+xx+3x−2x−2−2(x−2)(x+3)(x−2)(x+3)=(x−2)(x+3)x(x+3)+x(x−2)−2(x−2)(x+3)=(x−2)(x+3)x2+3x+x2−2x−2(x2+3x−2x−6)=(x−2)(x+3)x2+3x+x2−2x−2x2−6x+4x+12. |
Now, collect the terms in the numerator
Note: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.
