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7 September, 07:42

Determine how many solutions the system has.

y = x + 2

y = x^2 + 2x + 1

+1
Answers (1)
  1. 7 September, 08:06
    0
    The system can be presented just as a function of x, so for every x for which the following statement

    x+2 = (x+1) ^2 is true the system as described above should be true

    x^2+x-1=0, x (x+1) = 1 where x is a Real number has no real solution
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