Question

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

Answer 1

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