Solution 2.2:6e - Revision history

Tek at 13:30, 24 September 2008

Tek — Wed, 24 Sep 2008 13:30:54 GMT

←Older revision		Revision as of 13:30, 24 September 2008
Line 1:		Line 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 at 12:11, 18 September 2008

Ian — Thu, 18 Sep 2008 12:11:06 GMT

←Older revision		Revision as of 12:11, 18 September 2008
Line 1:		Line 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

←Older revision

Revision as of 08:50, 9 September 2008

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

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

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

←Older revision		Revision as of 06:42, 21 August 2008
Line 1:		Line 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 at 11:24, 15 August 2008

Madde — Fri, 15 Aug 2008 11:24:43 GMT

←Older revision		Revision as of 11:24, 15 August 2008
Line 5:		Line 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]]

Lina: Ny sida: {{NAVCONTENT_START}} Bild:2_2_6e-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:2_2_6e-2(2).gif {{NAVCONTENT_STOP}}

Lina — Mon, 31 Mar 2008 14:55:50 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_2_6e-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:2_2_6e-2(2).gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:2_2_6e-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:2_2_6e-2(2).gif]] </center>
{{NAVCONTENT_STOP}}