Solve the system.
y= 3x2−2x+3
y = x + 3

the solutions are and

(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!

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step-by-step explanation:

answer: 2 answers

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