From Förberedande kurs i matematik 1

Revision as of 08:49, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.2:6d moved to Solution 2.2:6d: Robot: moved page) ← Previous diff		Current revision (13:14, 24 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	At the point where the lines cut each other, we have a point that lies on both lines and which must therefore satisfy the equations of both lines,
-	[[Image:2_2_6_d.gif|center]]	+
-	<center> [[Image:2_2_6d.gif]] </center>	+	{{Displayed math||<math>x+y+1=0\qquad\text{and}\qquad x=12\,\textrm{.}</math>}}
-	{{NAVCONTENT_STOP}}	+
		+	We obtain the solution to this system of equations by substituting <math>x=12</math>
		+	into the first equation
		+
		+	{{Displayed math||<math>12+y+1=0\quad\Leftrightarrow\quad y=-13\,\textrm{,}</math>}}
		+
		+	which gives us the point of intersection as (12,-13).
		+
		+
		+	<center>[[Image:2_2_6_d.gif|center]]</center>

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Current revision

At the point where the lines cut each other, we have a point that lies on both lines and which must therefore satisfy the equations of both lines,

\displaystyle x+y+1=0\qquad\text{and}\qquad x=12\,\textrm{.}

We obtain the solution to this system of equations by substituting \displaystyle x=12 into the first equation

\displaystyle 12+y+1=0\quad\Leftrightarrow\quad y=-13\,\textrm{,}

which gives us the point of intersection as (12,-13).