Solution 2.1:7c
From Förberedande kurs i matematik 1
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- | We multiply the top and bottom of the first term by | + | We multiply the top and bottom of the first term by <math>a+1</math>, so that both terms then have the same denominator, |
- | <math>a+1</math>, so that both terms then have the same | + | |
- | denominator, | + | |
+ | {{Displayed math||<math>\frac{ax}{a+1}\cdot\frac{a+1}{a+1} - \frac{ax^{2}}{(a+1)^{2}} = \frac{ax(a+1)-ax^{2}}{(a+1)^{2}}\,\textrm{.}</math>}} | ||
- | <math> | + | Because both terms in the numerator contain the factor <math>ax</math>, we take out that factor, obtaining |
- | + | {{Displayed math||<math>\frac{ax(a+1-x)}{(a+1)^{2}}</math>}} | |
- | + | ||
- | + | ||
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- | <math>\frac{ax | + | |
and see that the answer cannot be simplified any further. | and see that the answer cannot be simplified any further. | ||
- | + | Note: It is only factors in the numerator and denominator that can cancel each other out, not individual terms. Hence, the following "cancellation" is wrong | |
- | individual terms. Hence, the following "cancellation" is wrong | + | |
- | + | ||
- | <math>\frac{ax | + | {{Displayed math||<math>\frac{ax(\rlap{/\,/\,/\,/}a+1)-ax^2}{(a+1)^{2\llap{/}}} = \frac{ax-ax^{2}}{a+1}\,\textrm{.}</math>}} |
Current revision
We multiply the top and bottom of the first term by \displaystyle a+1, so that both terms then have the same denominator,
\displaystyle \frac{ax}{a+1}\cdot\frac{a+1}{a+1} - \frac{ax^{2}}{(a+1)^{2}} = \frac{ax(a+1)-ax^{2}}{(a+1)^{2}}\,\textrm{.} |
Because both terms in the numerator contain the factor \displaystyle ax, we take out that factor, obtaining
\displaystyle \frac{ax(a+1-x)}{(a+1)^{2}} |
and see that the answer cannot be simplified any further.
Note: It is only factors in the numerator and denominator that can cancel each other out, not individual terms. Hence, the following "cancellation" is wrong
\displaystyle \frac{ax(\rlap{/\,/\,/\,/}a+1)-ax^2}{(a+1)^{2\llap{/}}} = \frac{ax-ax^{2}}{a+1}\,\textrm{.} |