Solution 2.2:6d
From Förberedande kurs i matematik 1
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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> | ||
+ | and | ||
+ | <math>x=12</math>. | ||
+ | |||
+ | We obtain the solution to this system of equations by substituting | ||
+ | <math>x=12</math> | ||
+ | into the first equation | ||
+ | |||
+ | |||
+ | <math>12+y+1=0\ \Leftrightarrow \ y=-13</math> | ||
+ | |||
+ | which gives us the point of intersection as | ||
+ | <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).