Solution 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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The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines
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{{Displayed math||<math>2x+y-1=0\qquad\text{and}\qquad y-2x-2=0\,\textrm{.}</math>}}
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<math>2x+y-1=0</math>
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If we make ''y'' the subject of the second equation <math>y=2x+2</math> and substitute it into the first equation, we obtain an equation which only contains ''x'',
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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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{{Displayed math||<math>2x+(2x+2)-1=0\quad\Leftrightarrow\quad 4x+1=0\,,</math>}}
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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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which gives that <math>x=-1/4\,</math>. Then, from the relation <math>y=2x+2</math>, we obtain <math>y = 2\cdot(-1/4)+2 = 3/2\,</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
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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<math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</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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<center>[[Image:2_2_6_e.gif|center]]</center>
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{{NAVCONTENT_START}}
 
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{{NAVCONTENT_STOP}}
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We check for safety's sake that <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math>
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[[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:
really satisfies both equations:
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:*2''x''&nbsp;+&nbsp;''y''&nbsp;-&nbsp;1&nbsp;=&nbsp;0: <math>\quad\textrm{LHS} = 2\cdot\bigl(-\tfrac{1}{4}\bigr) + \tfrac{3}{2} - 1 = -\tfrac{1}{2} + \tfrac{3}{2} - \tfrac{2}{2} = 0 = \textrm{RHS.}</math>
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<math>2x+y-1=0</math>:
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:*''y''&nbsp;-&nbsp;2''x''&nbsp;-&nbsp;2&nbsp;=&nbsp;0: <math>\quad\textrm{LHS} = \tfrac{3}{2}-2\cdot\bigl(-\tfrac{1}{4}\bigr)-2 = \tfrac{3}{2} + \tfrac{1}{2} - \tfrac{4}{2} = 0 = \textrm{RHS.}</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
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Current revision

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

\displaystyle 2x+y-1=0\qquad\text{and}\qquad y-2x-2=0\,\textrm{.}

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

\displaystyle 2x+(2x+2)-1=0\quad\Leftrightarrow\quad 4x+1=0\,,

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

The point of intersection is \displaystyle \bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr).



We check for safety's sake that \displaystyle \bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr) really satisfies both equations:

  • 2x + y - 1 = 0: \displaystyle \quad\textrm{LHS} = 2\cdot\bigl(-\tfrac{1}{4}\bigr) + \tfrac{3}{2} - 1 = -\tfrac{1}{2} + \tfrac{3}{2} - \tfrac{2}{2} = 0 = \textrm{RHS.}
  • y - 2x - 2 = 0: \displaystyle \quad\textrm{LHS} = \tfrac{3}{2}-2\cdot\bigl(-\tfrac{1}{4}\bigr)-2 = \tfrac{3}{2} + \tfrac{1}{2} - \tfrac{4}{2} = 0 = \textrm{RHS.}
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