Processing Math: Done

From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	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
-	<center> [[Bild:2_1_6b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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.

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Current revision

The lowest common denominator for the three terms is (x−2)(x+3) and we expand each term so that all terms have the same denominator

xx−2+xx+3−2=xx−2x+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

xx−2+xx+3−2=(x−2)(x+3)(x2+x2−2x2)+(3x−2x−6x+4x)+12=−x+12(x−2)(x+3).

Note: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.