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 |
- | + | ||
- | {{ | + | {{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>}} | ||
+ | |||
+ | 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 \displaystyle (x-2)(x+3) and we expand each term so that all terms have the same denominator
\displaystyle \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} |
Now, collect the terms in the numerator
\displaystyle \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} |
Note: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.