Alg. II - Solving Linear-Quadratic Systems Algebraically (Simple)

In this example problem, I work through a solution of a linear-quadratic system of equations using substitution.

Generally speaking, substitution is the EASIEST way to solve these, and I encourage you to always try and work these out when I ask you to solve linear-quadratic systems algebraically.

[DerBubboTea]

There's only one solution.. it's (-0.5, -9.5)
When you factored and switched over to the new page, you put (x-3) instead of (x+3). Also.. If you plugged -0.5 into the y=15x-2 equation, you'd get y= -9.5 and not -11/2