Solution 2.2:6d

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Current revision (13:14, 24 September 2008) (edit) (undo)
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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:
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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,
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{{Displayed math||<math>x+y+1=0\qquad\text{and}\qquad x=12\,\textrm{.}</math>}}
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<math>x+y+1=0</math>
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We obtain the solution to this system of equations by substituting <math>x=12</math>
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and
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<math>x=12</math>.
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We obtain the solution to this system of equations by substituting
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<math>x=12</math>
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into the first equation
into the first equation
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{{Displayed math||<math>12+y+1=0\quad\Leftrightarrow\quad y=-13\,\textrm{,}</math>}}
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<math>12+y+1=0\ \Leftrightarrow \ y=-13</math>
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which gives us the point of intersection as (12,-13).
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which gives us the point of intersection as
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<math>\left( 12 \right.,\left. -13 \right)</math>.
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{{NAVCONTENT_START}}
 
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[[Image:2_2_6_d.gif|center]]
 
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{{NAVCONTENT_STOP}}
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<center>[[Image:2_2_6_d.gif|center]]</center>

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).


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