Lösung 2.2:6b

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Because the point of intersection lies on both lines, it must satisfy the equations of both lines
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<math>y=-x+5</math>
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and
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<math>x=0</math>,
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where
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<math>x=0</math>
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is the equation of the
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<math>y</math>
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-axis. Substituting the other equation,
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<math>x=0</math>, into the first equation gives
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<math>y=-0+5=5</math>. This means that the point of intersection is
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<math>\left( 0 \right.,\left. 5 \right)</math>.
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Version vom 10:24, 18. Sep. 2008

Because the point of intersection lies on both lines, it must satisfy the equations of both lines


\displaystyle y=-x+5 and \displaystyle x=0,

where \displaystyle x=0 is the equation of the \displaystyle y -axis. Substituting the other equation, \displaystyle x=0, into the first equation gives \displaystyle y=-0+5=5. This means that the point of intersection is \displaystyle \left( 0 \right.,\left. 5 \right).


Image:2_2_6_b.gif

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:6b“