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Solution 2.2:6b

From Förberedande kurs i matematik 1

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Current revision (13:04, 24 September 2008) (edit) (undo)
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Because the point of intersection lies on both lines, it must satisfy the equations of both lines
Because the point of intersection lies on both lines, it must satisfy the equations of both lines
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{{Displayed math||<math>y=-x+5\qquad\text{and}\qquad x=0\,,</math>}}
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<math>y=-x+5</math>
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where <math>x=0</math> is the equation of the ''y''-axis. Substituting the second equation, <math>x=0</math>, into the first equation gives <math>y=-0+5=5</math>. This means that the point of intersection is (0,5).
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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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<center>[[Image:2_2_6_b.gif]]</center>
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{{NAVCONTENT_START}}
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[[Image:2_2_6_b.gif]]
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{{NAVCONTENT_STOP}}
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Current revision

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

y=−x+5andx=0

where x=0 is the equation of the y-axis. Substituting the second equation, x=0, into the first equation gives y=−0+5=5. This means that the point of intersection is (0,5).


Image:2_2_6_b.gif
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