Lösung 2.2:2a - Versionsgeschichte

Moni um 22:33, 8. Aug. 2009

2009-08-08T22:33:07Z

← Nächstältere Version		Version vom 22:33, 8. Aug. 2009
Zeile 6:		Zeile 6:
	\end{align}</math>}}		\end{align}</math>}}

-	Die linke Seite der Gleichung kann wie <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math> geschrieben werden, und wir bekommen	+	Die linke Seite der Gleichung kann wie <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math> geschrieben werden und wir bekommen

	{{Abgesetzte Formel\|\|<math>13x-4=9\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>13x-4=9\,\textrm{.}</math>}}

-	Jetzt besteht nur mehr eine lineare Gleichung, die wir ~~wir~~ vorher lösen	+	Jetzt besteht nur mehr eine lineare Gleichung, die wir wie vorher lösen

	<ol>		<ol>

Bayerer um 09:58, 22. Jul. 2009

2009-07-22T09:58:59Z

← Nächstältere Version		Version vom 09:58, 22. Jul. 2009
Zeile 13:		Zeile 13:

	<ol>		<ol>
-	<li>~~Add~~ 4 ~~to both sides~~, <math>\vphantom{x_2}13x-4+4=9+4\,,</math> ~~which gives~~ <math>\ 13x=13\,\textrm{.}</math></li>	+	<li> 4 auf beiden Seiten addiert, <math>\vphantom{x_2}13x-4+4=9+4\,,</math> liefert <math>\ 13x=13\,\textrm{.}</math></li>
-	<li>~~Divide both sides by~~ 13, <math>\frac{13x}{13}=\frac{13}{13}\,,</math> ~~which gives the answer~~ <math>\ x=1\,\textrm{.}</math></li>	+	<li>Teilen wir beide Seiten der Gleichung durch 13, <math>\frac{13x}{13}=\frac{13}{13}\,,</math> erhalten wir die Lösung <math>\ x=1\,\textrm{.}</math></li>
	</ol>		</ol>

Markus.bez um 09:18, 9. Jun. 2009

2009-06-09T09:18:20Z

← Nächstältere Version		Version vom 09:18, 9. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir teilen die Nenner in der Gleichung in ihre Primfaktoren auf, <math>6=2\cdot 3</math>, <math>9=3\cdot 3</math> und 2. Wir sehen dass der kleinste gemeinsamer Nenner <math>2\cdot 3\cdot 3=18</math> ist. Wir multiplizieren daher beide Seiten mit <math>2\cdot 3\cdot 3</math> um die Nenner zu eliminieren.	+	Wir teilen die Nenner in der Gleichung in ihre Primfaktoren auf, <math>6=2\cdot 3</math>, <math>9=3\cdot 3</math> und 2. Wir sehen, dass der kleinste gemeinsamer Nenner <math>2\cdot 3\cdot 3=18</math> ist. Wir multiplizieren daher beide Seiten mit <math>2\cdot 3\cdot 3</math>, um die Nenner zu eliminieren.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 6:		Zeile 6:
	\end{align}</math>}}		\end{align}</math>}}

-	Die linke Seite der Gleichung kann wie <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math> ~~geshrieben~~ werden, und wir bekommen	+	Die linke Seite der Gleichung kann wie <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math> geschrieben werden, und wir bekommen

	{{Abgesetzte Formel\|\|<math>13x-4=9\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>13x-4=9\,\textrm{.}</math>}}

-	Jetzt besteht nur mehr eine ~~Lineare~~ Gleichung, die wir wir vorher lösen	+	Jetzt besteht nur mehr eine lineare Gleichung, die wir wir vorher lösen

	<ol>		<ol>

Martin um 13:52, 12. Mär. 2009

2009-03-12T13:52:57Z

← Nächstältere Version		Version vom 13:52, 12. Mär. 2009
Zeile 22:		Zeile 22:

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\text{~~LHS~~}	+	\text{Linke Seite}
	&= \frac{5\cdot 1}{6}-\frac{1+2}{9} = \frac{5}{6}-\frac{3}{9} = \frac{5}{6}-\frac{1}{3}\\[5pt]		&= \frac{5\cdot 1}{6}-\frac{1+2}{9} = \frac{5}{6}-\frac{3}{9} = \frac{5}{6}-\frac{1}{3}\\[5pt]
	&= \frac{5}{6}-\frac{1\cdot 2}{3\cdot 2} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2}		&= \frac{5}{6}-\frac{1\cdot 2}{3\cdot 2} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2}
-	= \text{~~RHS~~}\,\textrm{.}	+	= \text{Rechte Seite}\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Martin um 13:51, 12. Mär. 2009

2009-03-12T13:51:47Z

← Nächstältere Version		Version vom 13:51, 12. Mär. 2009
Zeile 1:		Zeile 1:
-	~~If we divide up the denominators that appear~~ in ~~the equation into small integer factors~~ <math>6=2\cdot 3</math>, <math>9=3\cdot 3</math> ~~and~~ 2~~, we see that the lowest common denominator is~~ <math>2\cdot 3\cdot 3=18</math>. ~~Thus, we multiply both sides of the equation by~~ <math>2\cdot 3\cdot 3</math> ~~in order to avoid having denominators in the equation~~	+	Wir teilen die Nenner in der Gleichung in ihre Primfaktoren auf, <math>6=2\cdot 3</math>, <math>9=3\cdot 3</math> und 2. Wir sehen dass der kleinste gemeinsamer Nenner <math>2\cdot 3\cdot 3=18</math> ist. Wir multiplizieren daher beide Seiten mit <math>2\cdot 3\cdot 3</math> um die Nenner zu eliminieren.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 6:		Zeile 6:
	\end{align}</math>}}		\end{align}</math>}}

-	~~We can rewrite the left-hand side as~~ <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math>, ~~so that we get the equation~~	+	Die linke Seite der Gleichung kann wie <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math> geshrieben werden, und wir bekommen

	{{Abgesetzte Formel\|\|<math>13x-4=9\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>13x-4=9\,\textrm{.}</math>}}

-	~~We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get ''x'' by itself on one side:~~	+	Jetzt besteht nur mehr eine Lineare Gleichung, die wir wir vorher lösen

	<ol>		<ol>
Zeile 17:		Zeile 17:
	</ol>		</ol>

-	~~The equation has~~ <math>x=1</math> ~~as the solution~~.	+	Die Gleichung hat die Lösung <math>x=1</math>.

-	~~When we have obtained an answer~~, ~~it is important to go back to the original equation to check that <math>~~x~~=1</math> really is the correct answer (i.e. that we haven~~'~~t calculated incorrectly)~~	+	Wir kontrollieren die Lösung, indem wir ''x'' mit 1 in der ursprünglichen Gleichung substituieren

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Tekbot: hat „Solution 2.2:2a“ nach „Lösung 2.2:2a“ verschoben: Robot: moved page

2008-10-22T13:54:05Z

hat „Solution 2.2:2a“ nach „Lösung 2.2:2a“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 13:54, 22. Okt. 2008

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

2008-10-22T08:26:48Z

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 08:26, 22. Okt. 2008
Zeile 1:		Zeile 1:
	If we divide up the denominators that appear in the equation into small integer factors <math>6=2\cdot 3</math>, <math>9=3\cdot 3</math> and 2, we see that the lowest common denominator is <math>2\cdot 3\cdot 3=18</math>. Thus, we multiply both sides of the equation by <math>2\cdot 3\cdot 3</math> in order to avoid having denominators in the equation		If we divide up the denominators that appear in the equation into small integer factors <math>6=2\cdot 3</math>, <math>9=3\cdot 3</math> and 2, we see that the lowest common denominator is <math>2\cdot 3\cdot 3=18</math>. Thus, we multiply both sides of the equation by <math>2\cdot 3\cdot 3</math> in order to avoid having denominators in the equation

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	& \rlap{/}2\cdot{}\rlap{/}3\cdot 3\cdot\frac{5x}{\rlap{/}6} - 2\cdot{}\rlap{/}3\cdot{}\rlap{/}3\cdot\frac{x+2}{\rlap{/}9} = \rlap{/}2\cdot 3\cdot 3\cdot \frac{1}{\rlap{/}2} \\[5pt]		& \rlap{/}2\cdot{}\rlap{/}3\cdot 3\cdot\frac{5x}{\rlap{/}6} - 2\cdot{}\rlap{/}3\cdot{}\rlap{/}3\cdot\frac{x+2}{\rlap{/}9} = \rlap{/}2\cdot 3\cdot 3\cdot \frac{1}{\rlap{/}2} \\[5pt]
	&\qquad\Leftrightarrow\quad 3\cdot 5x-2\cdot (x+2) = 3\cdot 3\,\textrm{.}\\		&\qquad\Leftrightarrow\quad 3\cdot 5x-2\cdot (x+2) = 3\cdot 3\,\textrm{.}\\
Zeile 8:		Zeile 8:
	We can rewrite the left-hand side as <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math>, so that we get the equation		We can rewrite the left-hand side as <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math>, so that we get the equation

-	{{~~Displayed math~~\|\|<math>13x-4=9\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>13x-4=9\,\textrm{.}</math>}}

	We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get ''x'' by itself on one side:		We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get ''x'' by itself on one side:
Zeile 21:		Zeile 21:
	When we have obtained an answer, it is important to go back to the original equation to check that <math>x=1</math> really is the correct answer (i.e. that we haven't calculated incorrectly)		When we have obtained an answer, it is important to go back to the original equation to check that <math>x=1</math> really is the correct answer (i.e. that we haven't calculated incorrectly)

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\text{LHS}		\text{LHS}
	&= \frac{5\cdot 1}{6}-\frac{1+2}{9} = \frac{5}{6}-\frac{3}{9} = \frac{5}{6}-\frac{1}{3}\\[5pt]		&= \frac{5\cdot 1}{6}-\frac{1+2}{9} = \frac{5}{6}-\frac{3}{9} = \frac{5}{6}-\frac{1}{3}\\[5pt]

Tek um 13:44, 23. Sep. 2008

2008-09-23T13:44:59Z

← Nächstältere Version		Version vom 13:44, 23. Sep. 2008
Zeile 1:		Zeile 1:
-	If we divide up the denominators that appear in the equation into small integer factors	+	If we divide up the denominators that appear in the equation into small integer factors <math>6=2\cdot 3</math>, <math>9=3\cdot 3</math> and 2, we see that the lowest common denominator is <math>2\cdot 3\cdot 3=18</math>. Thus, we multiply both sides of the equation by <math>2\cdot 3\cdot 3</math> in order to avoid having denominators in the equation
-	<math>6=2\~~centerdot~~ 3</math>,	+
-	<math>9=3\~~centerdot~~ 3</math>	+
-	and 2, we see that the lowest common denominator is	+
-	<math>2\~~centerdot~~ 3\~~centerdot~~ 3=18</math>. Thus, we multiply both sides of the equation by	+
-	<math>2\~~centerdot~~ 3\~~centerdot~~ 3</math>	+
-	in order to avoid having denominators in the equation:	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	& \rlap{/}2\cdot{}\rlap{/}3\cdot 3\cdot\frac{5x}{\rlap{/}6} - 2\cdot{}\rlap{/}3\cdot{}\rlap{/}3\cdot\frac{x+2}{\rlap{/}9} = \rlap{/}2\cdot 3\cdot 3\cdot \frac{1}{\rlap{/}2} \\[5pt]
		+	&\qquad\Leftrightarrow\quad 3\cdot 5x-2\cdot (x+2) = 3\cdot 3\,\textrm{.}\\
		+	\end{align}</math>}}

-	<math>~~\begin{align}~~	+	We can rewrite the left-hand side as <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math>, so that we get the equation
-	~~& 2\centerdot~~ 3\~~centerdot 3\centerdot \frac{~~5x~~}{6}~~-2\~~centerdot 3\centerdot 3\centerdot \frac{x+2}{9}=2\centerdot 3\centerdot 3\centerdot \frac{1}{2} \\~~	+
-	~~& \Leftrightarrow 3\centerdot 5x-2\centerdot \left~~( x+2 ~~\right~~)=~~3\centerdot 3 \\~~	+
-	~~\end{align}~~</math>	+

		+	{{Displayed math\|\|<math>13x-4=9\,\textrm{.}</math>}}

-	We can ~~rewrite the left~~-~~hand side~~ as	+	We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get ''x'' by itself on one side:
-	~~<math>3\centerdot 5x-2\centerdot \left(~~ x~~+2 \right)=15x-2x-4=13x-4</math>, so that we get the equation~~	+

		+	<ol>
		+	<li>Add 4 to both sides, <math>\vphantom{x_2}13x-4+4=9+4\,,</math> which gives <math>\ 13x=13\,\textrm{.}</math></li>
		+	<li>Divide both sides by 13, <math>\frac{13x}{13}=\frac{13}{13}\,,</math> which gives the answer <math>\ x=1\,\textrm{.}</math></li>
		+	</ol>

-	<math>~~13x-4~~=9</math>	+	The equation has <math>x=1</math> as the solution.

		+	When we have obtained an answer, it is important to go back to the original equation to check that <math>x=1</math> really is the correct answer (i.e. that we haven't calculated incorrectly)

-	~~We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get~~	+	{{Displayed math\|\|<math>\begin{align}
-	<math~~>x</math>~~	+	\text{LHS}
-	~~by itself on one side:~~	+	&= \frac{5\cdot 1}{6}-\frac{1+2}{9} = \frac{5}{6}-\frac{3}{9} = \frac{5}{6}-\frac{1}{3}\\[5pt]
-		+	&= \frac{5}{6}-\frac{1\cdot 2}{3\cdot 2} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2}
-	~~1. Add 4 to both sides:~~	+	= \text{RHS}\,\textrm{.}
-		+	\end{align}</math>}}
-	~~<math>13x-+4=9+4</math>~~	+
-		+
-	~~which gives~~	+
-		+
-	~~<math>13x=13</math>~~	+
-	.	+
-		+
-	~~2. Divide both sides by 13:~~	+
-		+
-	<math>\~~frac~~{~~13x~~}~~{13}=~~\~~frac~~{13}~~{13}</math>~~	+
-		+
-	~~which gives the answer~~	+
-	~~<math>x~~=~~1</math>.~~	+
-		+
-	~~The equation has~~	+
-	~~<math>1</math>~~	+
-	~~as the solution.~~	+
-		+
-	~~When we have obtained an answer, it is important to go back to the original equation to check that~~	+
-	~~<math>x=1</math>~~	+
-	~~really is the correct answer( i.e. that we haven't calculated incorrectly):~~	+
-		+
-	~~LHS =~~	+
-	~~<math>~~\frac{5\~~centerdot~~ 1}{6}-\frac{1~~\centerdot~~ 2}{9}=\frac{5}{6}-\frac{3}{9}=\frac{5}{6}-\frac{1}{3}=\frac{5}{6}-\frac{1\~~centerdot~~ 2}{3\~~centerdot~~ 2}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2}</math>	+
-	~~= RHS~~	+

Ian um 12:25, 17. Sep. 2008

2008-09-17T12:25:29Z

← Nächstältere Version		Version vom 12:25, 17. Sep. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	If we divide up the denominators that appear in the equation into small integer factors
-	<~~center~~> ~~[[Image:2_2_2a-1(~~2).~~gif]]~~ </~~center~~>	+	<math>6=2\centerdot 3</math>,
-	{{~~NAVCONTENT_STOP~~}}	+	<math>9=3\centerdot 3</math>
-	{{~~NAVCONTENT_START~~}}	+	and 2, we see that the lowest common denominator is
-	<~~center~~> ~~[[Image:2_2_2a~~-2(2).~~gif]]~~ </~~center~~>	+	<math>2\centerdot 3\centerdot 3=18</math>. Thus, we multiply both sides of the equation by
-	{{~~NAVCONTENT_STOP~~}}	+	<math>2\centerdot 3\centerdot 3</math>
		+	in order to avoid having denominators in the equation:
		+
		+
		+	<math>\begin{align}
		+	& 2\centerdot 3\centerdot 3\centerdot \frac{5x}{6}-2\centerdot 3\centerdot 3\centerdot \frac{x+2}{9}=2\centerdot 3\centerdot 3\centerdot \frac{1}{2} \\
		+	& \Leftrightarrow 3\centerdot 5x-2\centerdot \left( x+2 \right)=3\centerdot 3 \\
		+	\end{align}</math>
		+
		+
		+	We can rewrite the left-hand side as
		+	<math>3\centerdot 5x-2\centerdot \left( x+2 \right)=15x-2x-4=13x-4</math>, so that we get the equation
		+
		+
		+	<math>13x-4=9</math>
		+
		+
		+	We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get
		+	<math>x</math>
		+	by itself on one side:
		+
		+	1. Add 4 to both sides:
		+
		+	<math>13x-+4=9+4</math>
		+
		+	which gives
		+
		+	<math>13x=13</math>
		+	.
		+
		+	2. Divide both sides by 13:
		+
		+	<math>\frac{13x}{13}=\frac{13}{13}</math>
		+
		+	which gives the answer
		+	<math>x=1</math>.
		+
		+	The equation has
		+	<math>1</math>
		+	as the solution.
		+
		+	When we have obtained an answer, it is important to go back to the original equation to check that
		+	<math>x=1</math>
		+	really is the correct answer( i.e. that we haven't calculated incorrectly):
		+
		+	LHS =
		+	<math>\frac{5\centerdot 1}{6}-\frac{1\centerdot 2}{9}=\frac{5}{6}-\frac{3}{9}=\frac{5}{6}-\frac{1}{3}=\frac{5}{6}-\frac{1\centerdot 2}{3\centerdot 2}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2}</math>
		+	= RHS

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

2008-09-09T08:43:43Z

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

← Nächstältere Version

Version vom 08:43, 9. Sep. 2008