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!