Processing Math: Done
Solution 2.2:6e
From Förberedande kurs i matematik 1
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- | {{ | + | The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines |
- | < | + | |
- | + | {{Displayed math||<math>2x+y-1=0\qquad\text{and}\qquad y-2x-2=0\,\textrm{.}</math>}} | |
- | {{ | + | |
- | < | + | 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'', |
- | {{ | + | |
- | [[Image:2_2_6_e.gif|center]] | + | {{Displayed math||<math>2x+(2x+2)-1=0\quad\Leftrightarrow\quad 4x+1=0\,,</math>}} |
+ | |||
+ | 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>. | ||
+ | |||
+ | The point of intersection is | ||
+ | <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math>. | ||
+ | |||
+ | |||
+ | <center>[[Image:2_2_6_e.gif|center]]</center> | ||
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+ | |||
+ | We check for safety's sake that <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math> | ||
+ | really satisfies both equations: | ||
+ | |||
+ | :*2''x'' + ''y'' - 1 = 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> | ||
+ | |||
+ | :*''y'' - 2''x'' - 2 = 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> |
Current revision
The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines
If we make y the subject of the second equation
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which gives that 4
(−1
4)+2=3
2
The point of intersection is
−41
23
We check for safety's sake that −41
23
- 2x + y - 1 = 0:
LHS=2 −41
+23−1=−21+23−22=0=RHS.
- 2x + y - 1 = 0:
- y - 2x - 2 = 0:
LHS=23−2 −41
−2=23+21−24=0=RHS.
- y - 2x - 2 = 0: