Lösung 2.2:6e - Versionsgeschichte

Tek: Replaced figure with metapost figure

Tek — Tue, 18 Aug 2009 13:02:12 GMT

Replaced figure with metapost figure

← Nächstältere Version		Version vom 13:02, 18. Aug. 2009
Zeile 12:		Zeile 12:


-	<center>~~[[Image~~:~~2_2_6_e~~.~~gif\|center]]~~</center>	+	<center>{{:2.2.6e - Solution - The lines 2x + y -1 = 0 and y - 2x - 2 = 0 through the point (-1/4,3/2)}}</center>

	Wir kontrollieren zur Sicherheit, dass der Punkt <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math> auch wirklich beide Gleichungen erfüllt		Wir kontrollieren zur Sicherheit, dass der Punkt <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math> auch wirklich beide Gleichungen erfüllt

Markus.bez um 09:53, 9. Jun. 2009

Markus.bez — Tue, 09 Jun 2009 09:53:13 GMT

← Nächstältere Version		Version vom 09:53, 9. Jun. 2009
Zeile 1:		Zeile 1:
-	Der Schnittpunkt der Geraden muss die Gleichungen beider Geraden erfüllen	+	Der Schnittpunkt der Geraden muss die Gleichungen beider Geraden erfüllen, also

-	{{Abgesetzte Formel\|\|<math>2x+y-1=0\qquad\text{~~and~~}\qquad y-2x-2=0\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>2x+y-1=0\qquad\text{und}\qquad y-2x-2=0\,\textrm{.}</math>}}

	Wenn wir von der ersten Gleichung <math>y</math> mit <math>2x+2</math> ersetzen, bekommen wir eine Gleichung mit nur <math>x</math>		Wenn wir von der ersten Gleichung <math>y</math> mit <math>2x+2</math> ersetzen, bekommen wir eine Gleichung mit nur <math>x</math>
Zeile 14:		Zeile 14:
	<center>[[Image:2_2_6_e.gif\|center]]</center>		<center>[[Image:2_2_6_e.gif\|center]]</center>

-	Wir kontrollieren zur Sicherheit dass der Punkt <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math> auch wirklich beide Gleichungen erfüllt	+	Wir kontrollieren zur Sicherheit, dass der Punkt <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math> auch wirklich beide Gleichungen erfüllt

	:*2''x'' + ''y'' - 1 = 0: <math>\quad\textrm{Linke Seite} = 2\cdot\bigl(-\tfrac{1}{4}\bigr) + \tfrac{3}{2} - 1 = -\tfrac{1}{2} + \tfrac{3}{2} - \tfrac{2}{2} = 0 = \textrm{Rechte Seite.}</math>		:*2''x'' + ''y'' - 1 = 0: <math>\quad\textrm{Linke Seite} = 2\cdot\bigl(-\tfrac{1}{4}\bigr) + \tfrac{3}{2} - 1 = -\tfrac{1}{2} + \tfrac{3}{2} - \tfrac{2}{2} = 0 = \textrm{Rechte Seite.}</math>

	:*''y'' - 2''x'' - 2 = 0: <math>\quad\textrm{Linke Seite} = \tfrac{3}{2}-2\cdot\bigl(-\tfrac{1}{4}\bigr)-2 = \tfrac{3}{2} + \tfrac{1}{2} - \tfrac{4}{2} = 0 = \textrm{Rechte Seite.}</math>		:*''y'' - 2''x'' - 2 = 0: <math>\quad\textrm{Linke Seite} = \tfrac{3}{2}-2\cdot\bigl(-\tfrac{1}{4}\bigr)-2 = \tfrac{3}{2} + \tfrac{1}{2} - \tfrac{4}{2} = 0 = \textrm{Rechte Seite.}</math>

Martin um 12:02, 13. Mär. 2009

Martin — Fri, 13 Mar 2009 12:02:41 GMT

← Nächstältere Version		Version vom 12:02, 13. Mär. 2009
Zeile 1:		Zeile 1:
-	~~The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines~~	+	Der Schnittpunkt der Geraden muss die Gleichungen beider Geraden erfüllen

	{{Abgesetzte Formel\|\|<math>2x+y-1=0\qquad\text{and}\qquad y-2x-2=0\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>2x+y-1=0\qquad\text{and}\qquad y-2x-2=0\,\textrm{.}</math>}}

-	~~If we make ''y'' the subject of the second equation~~ <math>y=2x+2</math> ~~and substitute it into the first equation~~, ~~we obtain an equation which only contains ''~~x~~'',~~	+	Wenn wir von der ersten Gleichung <math>y</math> mit <math>2x+2</math> ersetzen, bekommen wir eine Gleichung mit nur <math>x</math>

	{{Abgesetzte Formel\|\|<math>2x+(2x+2)-1=0\quad\Leftrightarrow\quad 4x+1=0\,,</math>}}		{{Abgesetzte Formel\|\|<math>2x+(2x+2)-1=0\quad\Leftrightarrow\quad 4x+1=0\,,</math>}}

-	~~which gives that~~ <math>x=-1/4\,</math>~~. Then~~, ~~from the relation~~ <math>y=2x+2</math>, ~~we obtain~~ <math>y = 2\cdot(-1/4)+2 = 3/2\,</math>.	+	Wir erhalten <math>x=-1/4\,</math>, und die erste Gleichung, <math>y=2x+2</math>, gibt uns <math>y = 2\cdot(-1/4)+2 = 3/2\,</math>.

-	~~The point of intersection is~~	+	Also ist der Schnittpunkt <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math>.
-	<math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math>.	+


	<center>[[Image:2_2_6_e.gif\|center]]</center>		<center>[[Image:2_2_6_e.gif\|center]]</center>

		+	Wir kontrollieren zur Sicherheit dass der Punkt <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math> auch wirklich beide Gleichungen erfüllt

-	~~We check for safety~~'~~s sake that~~ <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\~~bigr)~~</math>	+	:*2''x'' + ''y'' - 1 = 0: <math>\quad\textrm{Linke Seite} = 2\cdot\bigl(-\tfrac{1}{4}\bigr) + \tfrac{3}{2} - 1 = -\tfrac{1}{2} + \tfrac{3}{2} - \tfrac{2}{2} = 0 = \textrm{Rechte Seite.}</math>
-	~~really satisfies both equations:~~	+

-	:*2''x'' + ''y'' - 1 = 0: <math>\quad\textrm{LHS} = 2\cdot\bigl(-\tfrac{1}{4}\bigr) + \tfrac{3}{2} - 1 = -\tfrac{1}{2} + \tfrac{3}{2} - \tfrac{2}{2} = 0 = \textrm{RHS.}</math>	+	:*''y'' - 2''x'' - 2 = 0: <math>\quad\textrm{Linke Seite} = \tfrac{3}{2}-2\cdot\bigl(-\tfrac{1}{4}\bigr)-2 = \tfrac{3}{2} + \tfrac{1}{2} - \tfrac{4}{2} = 0 = \textrm{Rechte Seite.}</math>
-		+
-	:*''y'' - 2''x'' - 2 = 0: <math>\quad\textrm{~~LHS~~} = \tfrac{3}{2}-2\cdot\bigl(-\tfrac{1}{4}\bigr)-2 = \tfrac{3}{2} + \tfrac{1}{2} - \tfrac{4}{2} = 0 = \textrm{~~RHS~~.}</math>	+

Tekbot: hat „Solution 2.2:6e“ nach „Lösung 2.2:6e“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 14:00:25 GMT

hat „Solution 2.2:6e“ nach „Lösung 2.2:6e“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:00, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:29:58 GMT

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

← Nächstältere Version		Version vom 08:29, 22. Okt. 2008
Zeile 1:		Zeile 1:
	The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines		The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines

-	{{~~Displayed math~~\|\|<math>2x+y-1=0\qquad\text{and}\qquad y-2x-2=0\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>2x+y-1=0\qquad\text{and}\qquad y-2x-2=0\,\textrm{.}</math>}}

	If we make ''y'' the subject of the second equation <math>y=2x+2</math> and substitute it into the first equation, we obtain an equation which only contains ''x'',		If we make ''y'' the subject of the second equation <math>y=2x+2</math> and substitute it into the first equation, we obtain an equation which only contains ''x'',

-	{{~~Displayed math~~\|\|<math>2x+(2x+2)-1=0\quad\Leftrightarrow\quad 4x+1=0\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>2x+(2x+2)-1=0\quad\Leftrightarrow\quad 4x+1=0\,,</math>}}

	which gives that <math>x=-1/4\,</math>. Then, from the relation <math>y=2x+2</math>, we obtain <math>y = 2\cdot(-1/4)+2 = 3/2\,</math>.		which gives that <math>x=-1/4\,</math>. Then, from the relation <math>y=2x+2</math>, we obtain <math>y = 2\cdot(-1/4)+2 = 3/2\,</math>.

Tek um 13:30, 24. Sep. 2008

Tek — Wed, 24 Sep 2008 13:30:54 GMT

← Nächstältere Version		Version vom 13:30, 24. Sep. 2008
Zeile 1:		Zeile 1:
-	The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines:	+	The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines

		+	{{Displayed math\|\|<math>2x+y-1=0\qquad\text{and}\qquad y-2x-2=0\,\textrm{.}</math>}}

-	<math>~~2x+~~y-1=~~0</math>~~	+	If we make ''y'' the subject of the second equation <math>y=2x+2</math> and substitute it into the first equation, we obtain an equation which only contains ''x'',
-	~~and~~	+
-	~~<math>y-~~2x-2=0</math>.	+

-	~~If we make~~	+	{{Displayed math\|\|<math>2x+(2x+2)-1=0\quad\Leftrightarrow\quad 4x+1=0\,,</math>}}
-	<math>y</math>	+
-	~~the subject of the second equation~~	+
-	~~<math>y-~~2x-2=0~~</math>~~	+
-	~~and substitute it into the first equation~~, ~~we obtain an equation which only contains~~	+
-	~~<math>x~~</math>,	+

-		+	which gives that <math>x=-1/4\,</math>. Then, from the relation <math>y=2x+2</math>, we obtain <math>y = 2\cdot(-1/4)+2 = 3/2\,</math>.
-	~~<math>2x+\left( 2x+2 \right)-1=0\ \Leftrightarrow \ 4x+1=0</math>~~	+
-		+
-		+
-	which gives that	+
-	<math>x=-{1}/{4}\;</math>. Then, from the relation	+
-	<math>y=2x+2</math>, we obtain	+
-	<math>y=2\~~left~~( -{1}/{4~~}\; \right~~)+2={3}/{2}\;</math>.	+

	The point of intersection is		The point of intersection is
-	<math>\~~left~~( -\~~frac~~{1}{4} ~~\right.~~,\~~left. \frac~~{3}{2} \~~right~~)</math>.	+	<math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math>.

-	We check for safety's sake that
-	<math>\left( -\frac{1}{4} \right.,\left. \frac{3}{2} \right)</math>
-	really satisfies both equations:

		+	<center>[[Image:2_2_6_e.gif\|center]]</center>

-	{{NAVCONTENT_START}}

-	~~{{NAVCONTENT_STOP}}~~	+	We check for safety's sake that <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math>
-	~~[[Image:2_2_6_e.gif\|center]]~~	+
-	We check for safety's sake that	+
-	<math>\~~left~~( -\~~frac~~{1}{4} ~~\right.~~,\~~left. \frac~~{3}{2} \~~right~~)</math>	+
	really satisfies both equations:		really satisfies both equations:

		+	:*2''x'' + ''y'' - 1 = 0: <math>\quad\textrm{LHS} = 2\cdot\bigl(-\tfrac{1}{4}\bigr) + \tfrac{3}{2} - 1 = -\tfrac{1}{2} + \tfrac{3}{2} - \tfrac{2}{2} = 0 = \textrm{RHS.}</math>

-	~~<math>2x+~~y-1=0~~</math>~~:	+	:*''y'' - 2''x'' - 2 = 0: <math>\quad\textrm{LHS} = \tfrac{3}{2}-2\cdot\bigl(-\tfrac{1}{4}\bigr)-2 = \tfrac{3}{2} + \tfrac{1}{2} - \tfrac{4}{2} = 0 = \textrm{RHS.}</math>
-	~~LHS =~~	+
-	<math>2\~~left( -~~\~~frac~~{1}~~{4} \right)+\frac{3}{2}-1~~=-\~~frac{1}{2}+\frac~~{3}{2}-~~\frac{~~2~~}{2}=0</math> =RHS~~	+
-		+
-	~~<math>y-2x-2=0</math>:~~	+
-	~~LHS =~~	+
-	~~<math>~~\~~frac{3}{2}-2~~\~~left~~( -\~~frac~~{1}{4} \~~right~~)-2=\~~frac~~{3}{2}+\~~frac~~{1}{2}-\~~frac~~{4}{2}=0</math> ~~=RHS~~	+

Ian um 12:11, 18. Sep. 2008

Ian — Thu, 18 Sep 2008 12:11:06 GMT

← Nächstältere Version		Version vom 12:11, 18. Sep. 2008
Zeile 1:		Zeile 1:
		+	The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines:
		+
		+
		+	<math>2x+y-1=0</math>
		+	and
		+	<math>y-2x-2=0</math>.
		+
		+	If we make
		+	<math>y</math>
		+	the subject of the second equation
		+	<math>y-2x-2=0</math>
		+	and substitute it into the first equation, we obtain an equation which only contains
		+	<math>x</math>,
		+
		+
		+	<math>2x+\left( 2x+2 \right)-1=0\ \Leftrightarrow \ 4x+1=0</math>
		+
		+
		+	which gives that
		+	<math>x=-{1}/{4}\;</math>. Then, from the relation
		+	<math>y=2x+2</math>, we obtain
		+	<math>y=2\left( -{1}/{4}\; \right)+2={3}/{2}\;</math>.
		+
		+	The point of intersection is
		+	<math>\left( -\frac{1}{4} \right.,\left. \frac{3}{2} \right)</math>.
		+
		+	We check for safety's sake that
		+	<math>\left( -\frac{1}{4} \right.,\left. \frac{3}{2} \right)</math>
		+	really satisfies both equations:
		+
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	~~<center> [[Image:2_2_6e-1(2).gif]] </center>~~	+
-	~~{{NAVCONTENT_STOP}}~~	+
-	~~{{NAVCONTENT_START}}~~	+
-	~~<center> [[Image:2_2_6e-2(2).gif]] </center>~~	+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	[[Image:2_2_6_e.gif\|center]]		[[Image:2_2_6_e.gif\|center]]
		+	We check for safety's sake that
		+	<math>\left( -\frac{1}{4} \right.,\left. \frac{3}{2} \right)</math>
		+	really satisfies both equations:
		+
		+
		+	<math>2x+y-1=0</math>:
		+	LHS =
		+	<math>2\left( -\frac{1}{4} \right)+\frac{3}{2}-1=-\frac{1}{2}+\frac{3}{2}-\frac{2}{2}=0</math> =RHS
		+
		+	<math>y-2x-2=0</math>:
		+	LHS =
		+	<math>\frac{3}{2}-2\left( -\frac{1}{4} \right)-2=\frac{3}{2}+\frac{1}{2}-\frac{4}{2}=0</math> =RHS

Tekbot: Lösning 2.2:6e moved to Solution 2.2:6e: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:50:03 GMT

Lösning 2.2:6e moved to Solution 2.2:6e: Robot: moved page

← Nächstältere Version

Version vom 08:50, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 06:42:09 GMT

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

← Nächstältere Version		Version vom 06:42, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_6e-1(2).gif]] </center>	+	<center> [[Image:2_2_6e-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_6e-2(2).gif]] </center>	+	<center> [[Image:2_2_6e-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
-	[[~~Bild~~:2_2_6_e.gif\|center]]	+	[[Image:2_2_6_e.gif\|center]]

Madde um 11:24, 15. Aug. 2008

Madde — Fri, 15 Aug 2008 11:24:43 GMT

← Nächstältere Version		Version vom 11:24, 15. Aug. 2008
Zeile 5:		Zeile 5:
	<center> [[Bild:2_2_6e-2(2).gif]] </center>		<center> [[Bild:2_2_6e-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+	[[Bild:2_2_6_e.gif\|center]]