Solution 2.1:6a - Revision history

Tek at 11:24, 23 September 2008

Tek — Tue, 23 Sep 2008 11:24:08 GMT

←Older revision		Revision as of 11:24, 23 September 2008
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-	Before we try dealing with the whole expression, we focus on simplifying the two factors individually by rewriting them using a common denominator:	+	Before we try dealing with the whole expression, we focus on simplifying the two factors individually by rewriting them using a common denominator
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& x-y+\frac{x^{2}}{y-x}=\frac{\left( x-y \right)\left( y-x \right)}{y-x}+\frac{x^{2}}{y-x} \\~~	+
-	~~& \left\{ y-x=-\left( x-y \right) \right\} \\~~	+
-	~~& =\frac{-\left( x-y \right)^{2}}{y-x}+\frac{x^{2}}{y-x}=\frac{-\left( x-y \right)^{2}+x^{2}}{y-x}=\frac{-\left( x^{2}-2xy+y^{2} \right)+x^{2}}{y-x} \\~~	+
-	~~& =\frac{-x^{2}+2xy-y^{2}+x^{2}}{y-x}=\frac{2xy-y^{2}}{y-x}=\frac{y\left( 2x-y \right)}{y-x}, \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \frac{y}{2x-y}-1=\frac{y}{2x-y}-\frac{2x-y}{2x-y}=\frac{y-\left( 2x-y \right)}{2x-y}=\frac{y-2x+y}{2x-y} \\~~	+
-	~~& =\frac{2y-2x}{2x-y}=\frac{2\left( y-x \right)}{2x-y} \\~~	+
-	\\	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	x-y+\frac{x^{2}}{y-x} &= \frac{\left( x-y \right)\left( y-x \right)}{y-x}+\frac{x^{2}}{y-x} = \{\ y-x=-(x-y)\ \}\\[5pt]
		+	&= \frac{-(x-y)^{2}}{y-x}+\frac{x^{2}}{y-x} = \frac{-(x-y)^{2}+x^{2}}{y-x}\\[5pt]
		+	&= \frac{-(x^{2}-2xy+y^{2})+x^{2}}{y-x} = \frac{-x^{2}+2xy-y^{2}+x^{2}}{y-x}\\[5pt]
		+	&= \frac{2xy-y^{2}}{y-x} = \frac{y(2x-y)}{y-x},\\[15pt]
		+	\frac{y}{2x-y}-1
		+	&= \frac{y}{2x-y}-\frac{2x-y}{2x-y} = \frac{y-(2x-y)}{2x-y} = \frac{y-2x+y}{2x-y}\\[5pt]
		+	& =\frac{2y-2x}{2x-y} = \frac{2(y-x)}{2x-y}\,\textrm{.}
		+	\end{align}</math>}}

	Then, we multiply the factors together and simplify by elimination:		Then, we multiply the factors together and simplify by elimination:

-		+	{{Displayed math\|\|<math>\biggl(x-y+\frac{x^{2}}{y-x}\biggr) \biggl(\frac{y}{2x-y}-1\biggr) = \frac{y(2x-y)}{y-x}\cdot\frac{2(y-x)}{2x-y}=2y\,\textrm{.}</math>}}
-	<math>\~~left~~( x-y+\frac{x^{2}}{y-x} \~~right~~)\~~left~~( \frac{y}{2x-y}-1 \~~right~~)=\frac{y~~\left~~( 2x-y ~~\right~~)}{y-x}\~~centerdot~~ \frac{2~~\left~~( y-x ~~\right~~)}{2x-y}=2y.</math>	+

Ian at 10:37, 16 September 2008

Ian — Tue, 16 Sep 2008 10:37:29 GMT

←Older revision		Revision as of 10:37, 16 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Before we try dealing with the whole expression, we focus on simplifying the two factors individually by rewriting them using a common denominator:
-	<~~center~~> ~~[[Image:2_1_6a.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\begin{align}
		+	& x-y+\frac{x^{2}}{y-x}=\frac{\left( x-y \right)\left( y-x \right)}{y-x}+\frac{x^{2}}{y-x} \\
		+	& \left\{ y-x=-\left( x-y \right) \right\} \\
		+	& =\frac{-\left( x-y \right)^{2}}{y-x}+\frac{x^{2}}{y-x}=\frac{-\left( x-y \right)^{2}+x^{2}}{y-x}=\frac{-\left( x^{2}-2xy+y^{2} \right)+x^{2}}{y-x} \\
		+	& =\frac{-x^{2}+2xy-y^{2}+x^{2}}{y-x}=\frac{2xy-y^{2}}{y-x}=\frac{y\left( 2x-y \right)}{y-x}, \\
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& \frac{y}{2x-y}-1=\frac{y}{2x-y}-\frac{2x-y}{2x-y}=\frac{y-\left( 2x-y \right)}{2x-y}=\frac{y-2x+y}{2x-y} \\
		+	& =\frac{2y-2x}{2x-y}=\frac{2\left( y-x \right)}{2x-y} \\
		+	\\
		+	\end{align}</math>
		+
		+
		+	Then, we multiply the factors together and simplify by elimination:
		+
		+
		+	<math>\left( x-y+\frac{x^{2}}{y-x} \right)\left( \frac{y}{2x-y}-1 \right)=\frac{y\left( 2x-y \right)}{y-x}\centerdot \frac{2\left( y-x \right)}{2x-y}=2y.</math>

Tekbot: Lösning 2.1:6a moved to Solution 2.1:6a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:39:03 GMT

Lösning 2.1:6a moved to Solution 2.1:6a: Robot: moved page

←Older revision

Revision as of 08:39, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 06:36:39 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Lina — Mon, 31 Mar 2008 11:15:33 GMT

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