How do you solve the system #x^2 - x - y = 2#, #4x - 3y = 0#?
1 Answer
Dec 21, 2015
Explanation:
First modify the equation
Now, take this and plug it into
Thus, we have two possible values for
Plug these both into
When
Solution point:
When
Solution point:
graph{(x^2-x-y-2)(4x-3y)=0 [-6.92, 13.08, -3.04, 6.96]}