Question

How to solve this........

Answer 1

hey buddy here is your answer

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hope it helps

Answer 2

$y = {x}^{2} + 1 \\ y = 3 - x \\ {x}^{2} + 1 = 3 - x \\ {x}^{2} + x - 2 = 0 \\ {x}^{2} - x + 2x - 2 = 0 \\ x(x - 1) + 2(x - 1) = 0 \\ (x + 2)(x - 1) = 0 \\ x = - 2 \: \: \\ x = 1$
now,
$for \: \: x = 1 \\ y = {x}^{2} + 1 = {1}^{2} + 1 = 2 \\ y = 3 - x = 3 - 1 = 2$
and,
$for \: \: x = - 2 \\ y = {x}^{2} + 1 = { (- 2)}^{2} + 1 = 4 + 1 = 5 \\ y = x - 3 = - 2 - 3 = - 5$
now x = -2 don't verify both relations so the answer is (x,y)= (1,2)