Lösung 2.2:6e

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The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines:
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<math>2x+y-1=0</math>
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and
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<math>y-2x-2=0</math>.
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If we make
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<math>y</math>
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the subject of the second equation
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<math>y-2x-2=0</math>
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and substitute it into the first equation, we obtain an equation which only contains
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<math>x</math>,
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<math>2x+\left( 2x+2 \right)-1=0\ \Leftrightarrow \ 4x+1=0</math>
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which gives that
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<math>x=-{1}/{4}\;</math>. Then, from the relation
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<math>y=2x+2</math>, we obtain
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<math>y=2\left( -{1}/{4}\; \right)+2={3}/{2}\;</math>.
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The point of intersection is
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<math>\left( -\frac{1}{4} \right.,\left. \frac{3}{2} \right)</math>.
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We check for safety's sake that
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<math>\left( -\frac{1}{4} \right.,\left. \frac{3}{2} \right)</math>
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really satisfies both equations:
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[[Image:2_2_6_e.gif|center]]
[[Image:2_2_6_e.gif|center]]
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We check for safety's sake that
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<math>\left( -\frac{1}{4} \right.,\left. \frac{3}{2} \right)</math>
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really satisfies both equations:
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<math>2x+y-1=0</math>:
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LHS =
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<math>2\left( -\frac{1}{4} \right)+\frac{3}{2}-1=-\frac{1}{2}+\frac{3}{2}-\frac{2}{2}=0</math> =RHS
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<math>y-2x-2=0</math>:
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LHS =
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<math>\frac{3}{2}-2\left( -\frac{1}{4} \right)-2=\frac{3}{2}+\frac{1}{2}-\frac{4}{2}=0</math> =RHS

Version vom 12:11, 18. Sep. 2008

The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines:


\displaystyle 2x+y-1=0 and \displaystyle y-2x-2=0.

If we make \displaystyle y the subject of the second equation \displaystyle y-2x-2=0 and substitute it into the first equation, we obtain an equation which only contains \displaystyle x,


\displaystyle 2x+\left( 2x+2 \right)-1=0\ \Leftrightarrow \ 4x+1=0


which gives that \displaystyle x=-{1}/{4}\;. Then, from the relation \displaystyle y=2x+2, we obtain \displaystyle y=2\left( -{1}/{4}\; \right)+2={3}/{2}\;.

The point of intersection is \displaystyle \left( -\frac{1}{4} \right.,\left. \frac{3}{2} \right).

We check for safety's sake that \displaystyle \left( -\frac{1}{4} \right.,\left. \frac{3}{2} \right) really satisfies both equations:


We check for safety's sake that \displaystyle \left( -\frac{1}{4} \right.,\left. \frac{3}{2} \right) really satisfies both equations:


\displaystyle 2x+y-1=0: LHS = \displaystyle 2\left( -\frac{1}{4} \right)+\frac{3}{2}-1=-\frac{1}{2}+\frac{3}{2}-\frac{2}{2}=0 =RHS

\displaystyle y-2x-2=0: LHS = \displaystyle \frac{3}{2}-2\left( -\frac{1}{4} \right)-2=\frac{3}{2}+\frac{1}{2}-\frac{4}{2}=0 =RHS

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