From Förberedande kurs i matematik 1

Revision as of 11:34, 18 September 2008 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (13:10, 24 September 2008) (edit) (undo) Tek (Talk | contribs) m
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	The point of intersection is that point which satisfies the equations of both lines		The point of intersection is that point which satisfies the equations of both lines
		+	{{Displayed math||<math>4x+5y+6=0\qquad\text{and}\qquad x=0\,\textrm{.}</math>}}
-	<math>4x+5y+6=0</math>	+	Substituting <math>x=0</math> into <math>4x+5y+6=0</math> gives
-	and	+
-	<math>x=0</math>.	+
-	Substituting	+	{{Displayed math||<math>4\cdot 0+5y+6=0\quad\Leftrightarrow\quad y=-\frac{6}{5}\,\textrm{.}</math>}}
-	<math>4x+5y+6=0</math>	+
-	into	+
-	<math>x=0</math>	+
-	gives	+
		+	This gives the point of intersection as <math>\bigl(0,-\tfrac{6}{5}\bigr)</math>.
-	<math>4\centerdot 0+5y+6=0\ \Leftrightarrow \ y=-\frac{6}{5}</math>
-		+	<center>[[Image:2_2_6_c.gif|center]]</center>
-	This gives the point of intersection as	+
-	<math>\left( 0 \right.,\left. -\frac{6}{5} \right)</math>.	+
-		+
-		+
-		+
-	{{NAVCONTENT_START}}	+
-		+
-	[[Image:2_2_6_c.gif|center]]	+
-	{{NAVCONTENT_STOP}}	+

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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).