From Förberedande kurs i matematik 1

Revision as of 08:49, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.2:6c moved to Solution 2.2:6c: Robot: moved page) ← Previous diff		Current revision (13:10, 24 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	The point of intersection is that point which satisfies the equations of both lines
-	<center> [[Image:2_2_6c.gif]] </center>	+
-	[[Image:2_2_6_c.gif|center]]	+	{{Displayed math||<math>4x+5y+6=0\qquad\text{and}\qquad x=0\,\textrm{.}</math>}}
-	{{NAVCONTENT_STOP}}	+
		+	Substituting <math>x=0</math> into <math>4x+5y+6=0</math> gives
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		+	{{Displayed math||<math>4\cdot 0+5y+6=0\quad\Leftrightarrow\quad y=-\frac{6}{5}\,\textrm{.}</math>}}
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		+	This gives the point of intersection as <math>\bigl(0,-\tfrac{6}{5}\bigr)</math>.
		+
		+
		+	<center>[[Image:2_2_6_c.gif|center]]</center>

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Current revision

The point of intersection is that point which satisfies the equations of both lines

\displaystyle 4x+5y+6=0\qquad\text{and}\qquad x=0\,\textrm{.}

Substituting \displaystyle x=0 into \displaystyle 4x+5y+6=0 gives

\displaystyle 4\cdot 0+5y+6=0\quad\Leftrightarrow\quad y=-\frac{6}{5}\,\textrm{.}

This gives the point of intersection as \displaystyle \bigl(0,-\tfrac{6}{5}\bigr).