Answer:
(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!
Answer:
the solutions are (1,4) and (0,3)
Step-by-step explanation:
step 1: Subtract the equations
step 2: refine
step 3:Plug the solutionsx=1,x=0 into y=3x^2-2x+3
step 4: verify
