Question

Solve the system.

y= 3x2−2x+3​
y = x + 3

The solutions are (__,__) and (__,__).

Answer 1

(0,3) and (1,4)

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!