Help Help Help!!! :) solve the system of equations y=2x^2-3 y=3x-1

1 answer:

4 0

Solution:
y = 2x^2 - 3 ---- equation 1
y = 3x - 1 ------equation 2
Then we get,
2x^2 - 3 = 3x - 1
Now, simplifying the equation, we get
2x^2 - 3x - 3 + 1 = 0
2x^2 - 3x - 2 = 0
After factoring the equation, we get
(2x+1)(x-2) = 0
So, this means
(2x+1) = 0 and (x-2) = 0
From this, we get
x = -1/2 and x = 2
Putting values of x = -1/2 in equation 2
y = 3(-1/2) - 1
y = -2.5
Now, putting value of x = 2 in equation 2
y = 3(2) -1
y = 5

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