What are the solutions of the system?
y = x^2 - 3x + 2
y = 4x - 4

a.(1, 0) and (–6, –28)
b.(1, 20) and (6, 0)
c. (1, 0) and (6, 20)
d. no solution

Question

What are the solutions of the system?
y = x^2 - 3x + 2 
y = 4x - 4 

a.(1, 0) and (–6, –28)
b.(1, 20) and (6, 0)
c. (1, 0) and (6, 20)
d. no solution

campbell_st · Answer

equate the functions

$$4x - 4 = x^2 - 3x + 2$$

simplifying gives 

$$x^2 - 7x + 6 = 0$$

solve the qadratic for x by factorising 

then substitute into one of the original equations to get y

anonymous · Answer

both are equal to y, so why not set them equal to each other :)?
you can proceed from there, isolating x!

anonymous · Answer

actually, i was wrong. Quadratic or graph for sure, didn't notice the x^2 part.