Solution 2.2:6d

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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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<math>x+y+1=0</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
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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
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<math>\left( 12 \right.,\left. -13 \right)</math>.
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Revision as of 11:39, 18 September 2008

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 and \displaystyle x=12.

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


\displaystyle 12+y+1=0\ \Leftrightarrow \ y=-13

which gives us the point of intersection as \displaystyle \left( 12 \right.,\left. -13 \right).


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