Solution 2.2:2c - Revision history

Tek at 14:17, 23 September 2008

2008-09-23T14:17:37Z

←Older revision		Revision as of 14:17, 23 September 2008
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-	We can simplify the left-hand side in the equation by expanding the squares using the squaring rule:	+	We can simplify the left-hand side in the equation by expanding the squares using the squaring rule
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \left( x+3 \right)^{2}-\left( x-5 \right)^{2}=\left( x^{2}+2\centerdot 3x+3^{2} \right)-\left( x^{2}-2\centerdot 5x+5^{2} \right) \\~~	+
-	~~& =x^{2}+6x+9-x^{2}+10x-25=16x-16 \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	(x+3)^{2}-(x-5)^{2}
		+	&= (x^{2}+2\cdot 3x+3^{2})-(x^{2}-2\cdot 5x+5^{2})\\[5pt]
		+	&= x^{2}+6x+9-x^{2}+10x-25\\[5pt]
		+	&=16x-16\,\textrm{.}
		+	\end{align}</math>}}

	Thus, the equation is		Thus, the equation is

		+	{{Displayed math\|\|<math>16x-16=6x+4\,\textrm{.}</math>}}

-	~~<math>16x-16=6x+4</math>~~	+	Now, move all ''x'''s to the left-hand side (subtract 6''x'' from both sides) and the constants to the right-hand side (add 16 to both sides)
-		+
-		+
-	Now, move all "	+
-	~~<math>~~x~~</math>~~	+
-	"s to the left-hand side (subtract	+
-	~~<math>6x</math>~~	+
-	from both sides) and the constants to the right-hand side (add	+
-	~~<math>~~16~~</math>~~	+
-	to both sides)	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& 16x-6x=4+16 \\~~	+
-	~~& 10x=20 \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	~~Divide both sides by~~	+
-	~~<math>10</math>~~	+
-	~~to get the answer~~	+
-		+
-		+
-	~~<math>x=\frac{20}{10}=2</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	16x-6x&=4+16\,,\\[5pt]
		+	10x&=20\,\textrm{.}
		+	\end{align}</math>}}

-	~~Finally, we check that~~	+	Divide both sides by 10 to get the answer
-	~~<math>x=2</math>~~	+
-	~~satisfies~~ the ~~equation in the exercise:~~	+

-	~~LHS =~~	+	{{Displayed math\|\|<math>x=\frac{20}{10}=2\,\textrm{.}</math>}}
-	<math>\~~left( 2+3 \right)^~~{2}~~-\left( 2-5 \right)^~~{2}=~~5^{~~2}-\~~left( -3~~ \~~right)^~~{2}~~=25-9=16~~</math>	+

		+	Finally, we check that <math>x=2</math> satisfies the equation in the exercise

-	~~RHS =~~	+	{{Displayed math\|\|<math>\begin{align}
-	<math>6\~~centerdot~~ 2+4=12+4=16</math>	+	\text{LHS} &= (2+3)^{2}-(2-5)^{2} = 5^{2}-(-3)^{2} = 25-9 = 16,\\[5pt]
		+	\text{RHS} &= 6\cdot 2+4 = 12+4 = 16\,\textrm{.}
		+	\end{align}</math>}}

Ian at 12:58, 17 September 2008

2008-09-17T12:58:49Z

←Older revision		Revision as of 12:58, 17 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	We can simplify the left-hand side in the equation by expanding the squares using the squaring rule:
-	<~~center~~> ~~[[Image:2_2_2c.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\begin{align}
		+	& \left( x+3 \right)^{2}-\left( x-5 \right)^{2}=\left( x^{2}+2\centerdot 3x+3^{2} \right)-\left( x^{2}-2\centerdot 5x+5^{2} \right) \\
		+	& =x^{2}+6x+9-x^{2}+10x-25=16x-16 \\
		+	\end{align}</math>
		+
		+
		+	Thus, the equation is
		+
		+
		+	<math>16x-16=6x+4</math>
		+
		+
		+	Now, move all "
		+	<math>x</math>
		+	"s to the left-hand side (subtract
		+	<math>6x</math>
		+	from both sides) and the constants to the right-hand side (add
		+	<math>16</math>
		+	to both sides)
		+
		+
		+	<math>\begin{align}
		+	& 16x-6x=4+16 \\
		+	& 10x=20 \\
		+	\end{align}</math>
		+
		+
		+	Divide both sides by
		+	<math>10</math>
		+	to get the answer
		+
		+
		+	<math>x=\frac{20}{10}=2</math>
		+
		+
		+	Finally, we check that
		+	<math>x=2</math>
		+	satisfies the equation in the exercise:
		+
		+	LHS =
		+	<math>\left( 2+3 \right)^{2}-\left( 2-5 \right)^{2}=5^{2}-\left( -3 \right)^{2}=25-9=16</math>
		+
		+
		+	RHS =
		+	<math>6\centerdot 2+4=12+4=16</math>

Tekbot: Lösning 2.2:2c moved to Solution 2.2:2c: Robot: moved page

2008-09-09T08:44:23Z

Lösning 2.2:2c moved to Solution 2.2:2c: Robot: moved page

←Older revision

Revision as of 08:44, 9 September 2008

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

2008-08-21T06:39:19Z

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

←Older revision		Revision as of 06:39, 21 August 2008
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2008-03-31T14:32:13Z

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