Solve the system.

y= 3x^2−2x+3

y = x + 3

the solutions are and

(0,3) and (1,4) are solution sets of given system

Step-by-step explanation:

Given that:

y= 3x^2−2x+3   eqA

y = x + 3   eqB

Putting value of y from eqB to eqA

x + 3 = 3x^2−2x+3

Taking all variables to left side:

Signs pf transferred terms will be changed

3x^2−2x+3  - x - 3 = 0

Adding like terms we get:

3x^2−3x = 0

Dividing both sides by 3 we get:

x^2 - x = 0

Now take x common:

x(x-1)= 0

So

x = 0      and   x = 1

Now put value in eqB

y = 0 + 3   and    y= 1 + 3

we get:

y = 3 and y = 4

So solution sets are:

(0,3) and (1,4)

i hope it will help you!