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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

2x+y1=0andy2x2=0.

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

2x+(2x+2)1=04x+1=0

which gives that x=14. Then, from the relation y=2x+2, we obtain y=2(14)+2=32.

The point of intersection is 4123 .



We check for safety's sake that 4123  really satisfies both equations:

  • 2x + y - 1 = 0: LHS=241+231=21+2322=0=RHS. 
  • y - 2x - 2 = 0: LHS=232412=23+2124=0=RHS.