How do you solve $y= x^2 + 5x + 2$ and $y= 3x + 5$ ?

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Answer 1 · 2016-04-14T02:31:46

Solve by substitution.

$y = 3x + 5 -> 3x + 5 = x^2 + 5x + 2$

$0 = x^2 + 2x - 3$

$0 = (x + 3)(x - 1)$

$x = -3 and 1$

$y = 3(-3) + 5 or y = 3(1) + 5$

$y = -4 or y = 8$

Your solution set is ${-3, -4}; {1, 8}$

Hopefully this helps!

How do you solve y= x^2 + 5x + 2y= x^2 + 5x + 2 and y= 3x + 5y= 3x + 5?