Medale For Help -
Medal for help :)
Solve the system of equations-
y=2x2 -3
y=3x-1

A. no solution
b.(-1/2,5), (2,-5/2)
c.(-1/2,-5/2),(2,5)
d(1/2,5/2),(2,5)

Question

Medale For Help -
Medal for help :)
Solve the system of equations-
y=2x2 -3
y=3x-1

A. no solution
b.(-1/2,5), (2,-5/2)
c.(-1/2,-5/2),(2,5)
d(1/2,5/2),(2,5)

anonymous · Answer

A.?

whpalmer4 · Answer

$$y = 2x^2-3$$$$y = 3x-1$$Are those the equations?

anonymous · Answer

Yes

whpalmer4 · Answer

Then no, A is not correct.

Here's how you can solve it.  The solution to the system of equations will be a pair of numbers (x,y) that satisfy both equations.  Your equations are both written $y = \text{<stuff>}$ and the value of $y$ is the same, so we can just set the other two sides of the equation equal to each other:

$$2x^2-3 = 3x-1$$

Does that make sense?

anonymous · Answer

Its c ( -1/2, -5/2 )  ( 2 , 5 )

whpalmer4 · Answer

That's a quadratic equation which you can solve in a number of different ways.

whpalmer4 · Answer

start by moving all the terms to one side:
$$2x^2-3 - 3x = 3x - 3x - 1$$$$2x^2-3-3x+1 = 3x-3x-1+1$$$$2x^2-3x-2 = 0$$Probably the easiest thing here is to use the quadratic formula for the solution of $ax^2+bx+c=0$:$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$