How to solve for $x,y$ this system ${(x = y-2),(4y+x^2=16):}$ ?

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Answer 1 · 2017-05-26T16:55:33

See below.

Calling $x, y$ the two numbers such that $x < y$ we have:

${(x = y-2),(4y+x^2=16):}$

Substituting the first equation into the second

$4(x+2)+x^2=16$ now solving for $x$ we obtain

$x = -2(1pm sqrt3)$ so

$y = 2-2(1pm sqrt3)$

If we discard negative solutions, then

$x = 2(sqrt3-1)$ and

$y = 2 sqrt3$

How to solve for x,yx,y this system {(x = y-2),(4y+x^2=16):}{(x = y-2),(4y+x^2=16):} ?