Solution 2.3:3d - Revision history

Tek at 08:39, 29 September 2008

Tek — Mon, 29 Sep 2008 08:39:34 GMT

←Older revision		Revision as of 08:39, 29 September 2008
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-	Because both terms,	+	Because both terms, <math>x(x+3)</math> and <math>x(2x-9)</math>, contain the factor <math>x</math>, we can take out <math>x</math> from the left-hand side and collect together the remaining expression,
-	<math>x~~\left~~( x+3 ~~\right~~)</math>	+
-	and	+
-	<math>x~~\left~~( 2x-9 ~~\right~~)</math>	+
-	contain the factor	+
-	<math>x</math>, we can take out	+
-	<math>x</math> from the left-hand side and collect together the remaining expression:	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& x\left( x+3 \right)-x\left( 2x-9 \right)=x\left( \left( x+3 \right)-\left( 2x-9 \right) \right) \\~~	+
-	~~& =x\left( x+3-2x+9 \right)=x\left( -x+12 \right) \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	x(x+3)-x(2x-9)
		+	&= x\bigl((x+3)-(2x-9)\bigr)\\[5pt]
		+	&= x(x+3-2x+9)\\[5pt]
		+	&= x(-x+12)\,\textrm{.}
		+	\end{align}</math>}}

	The equation is thus		The equation is thus

		+	{{Displayed math\|\|<math>x(-x+12) = 0</math>}}

-	~~<math>x\left( -x+12 \right)=0</math>~~	+	and we obtain directly that the equation is satisfied if either <math>x</math> or <math>-x+12</math> is zero. The solutions to the equation are therefore <math>x=0</math> and <math>x=12</math>.
-		+
-	and we obtain directly that the equation is satisfied if either	+
-	<math>x</math>	+
-	or	+
-	<math>-x+~~\text{~~12}</math>	+
-	is zero. The solutions to the equation are therefore	+
-	<math>x=0~~\text{ }~~</math>	+
-	and	+
-	<math>x=~~\text{~~12}</math>.	+

-	Here, it can be worth checking that	+	Here, it can be worth checking that <math>x=12</math> is a solution (the case
-	<math>x=~~\text{~~12 }</math>	+	<math>x=0</math> is obvious)
-	is a solution (the case	+
-	<math>x=0</math>	+
-	is obvious):	+

-	~~LHS~~	+	{{Displayed math\|\|<math>\text{LHS} = 12\cdot (12+3) - 12\cdot (2\cdot 12-9) = 2\cdot 15 - 12\cdot 15 = 0 = \text{RHS.}</math>}}
-	<math>=12\~~centerdot \left~~( 12+3 ~~\right~~)-12\~~centerdot \left~~( 2\~~centerdot~~ 12-9 ~~\right~~)=2\~~centerdot~~ 15-12\~~centerdot~~ 15=0=</math>	+
-	~~RHS~~	+

Ian at 15:05, 20 September 2008

Ian — Sat, 20 Sep 2008 15:05:55 GMT

←Older revision		Revision as of 15:05, 20 September 2008
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-	{{~~NAVCONTENT_START}~~}	+	Because both terms,
-	<~~center~~> ~~[[Image:2_3_3d.gif]]~~ </~~center~~>	+	<math>x\left( x+3 \right)</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>x\left( 2x-9 \right)</math>
		+	contain the factor
		+	<math>x</math>, we can take out
		+	<math>x</math> from the left-hand side and collect together the remaining expression:
		+
		+
		+	<math>\begin{align}
		+	& x\left( x+3 \right)-x\left( 2x-9 \right)=x\left( \left( x+3 \right)-\left( 2x-9 \right) \right) \\
		+	& =x\left( x+3-2x+9 \right)=x\left( -x+12 \right) \\
		+	\end{align}</math>
		+
		+
		+	The equation is thus
		+
		+
		+	<math>x\left( -x+12 \right)=0</math>
		+
		+	and we obtain directly that the equation is satisfied if either
		+	<math>x</math>
		+	or
		+	<math>-x+\text{12}</math>
		+	is zero. The solutions to the equation are therefore
		+	<math>x=0\text{ }</math>
		+	and
		+	<math>x=\text{12}</math>.
		+
		+	Here, it can be worth checking that
		+	<math>x=\text{12 }</math>
		+	is a solution (the case
		+	<math>x=0</math>
		+	is obvious):
		+
		+	LHS
		+	<math>=12\centerdot \left( 12+3 \right)-12\centerdot \left( 2\centerdot 12-9 \right)=2\centerdot 15-12\centerdot 15=0=</math>
		+	RHS

Tekbot: Lösning 2.3:3d moved to Solution 2.3:3d: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:59:03 GMT

Lösning 2.3:3d moved to Solution 2.3:3d: Robot: moved page

←Older revision

Revision as of 08:59, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:46:39 GMT

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

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Oskar — Tue, 01 Apr 2008 10:03:10 GMT

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