How do you solve $y² = x + 3$ and $x - 2y = 12$ using substitution?

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Answer 1 · 2016-05-04T00:26:39

See solution below.

$x = 12 + 2y -> y^2 = 12 + 2y + 3$

$y^2 - 2y - 15 = 0$

$(y - 5)(y + 3) = 0$

$y = 5 and -3$

$x - 2(5) = 12 or X - 2(-3) = 12$

$x = 22 or x = 6$

Your solution set is ${22, 5} and {6, -3}$

Hopefully this helps!

How do you solve y² = x + 3y² = x + 3 and x - 2y = 12x - 2y = 12 using substitution?