Answer:
(0,3) and (1,4)
Step-by-step explanation:
Given that:
y= 3x2−2x+3 ------------- eq1
y = x + 3 ------------------ eq2
Putting value from eq2 to eq1
x + 3 = 3x2−2x+3
3x2−2x+3 - x - 3 = 0
3x2−3x = 0
Taking 3 common
x2 - x = 0
x(x-1)= 0
So
x = 0 and x = 1
Now put value in eq2
y = 0 + 3 and y= 1 + 3
y = 3 and y = 4
So solution sets are:
(0,3) and (1,4)
i hope it will help you!
