Quadratic Systems

y=3x^2-5
x-y=3

Could you tell me the answer.I want to make sure I'm right and could you show work

It's better if you show your work first, but here goes.

Since y = x-3,
x-3 = 3x^2-5
3x^2 - x - 2 = 0
(3x+2)(x-1) = 0
x = 1 or -2/3
and you can now evaluate y for each x.

as shown at

http://www.wolframalpha.com/input/?i=solve+y%3D3x^2-5%2C+x-y%3D3

Thank you so much it help a lot

To solve the given quadratic system of equations, you can use the method of substitution or elimination. Let's use the substitution method.

Step 1: Solve one of the equations for one variable in terms of the other. We'll solve the second equation for x.

Given: x - y = 3
Rearranging the equation, we get:
x = y + 3

Step 2: Substitute the value of x from the second equation into the first equation. Substitute y + 3 for x in the equation y = 3x^2 - 5.

y = 3(y + 3)^2 - 5

Step 3: Simplify the equation and solve for y.

y = 3(y^2 + 6y + 9) - 5
y = 3y^2 + 18y + 27 - 5
y = 3y^2 + 18y + 22

Step 4: Rearrange the equation to make it quadratic.

3y^2 + 18y + 22 - y = 0
3y^2 + 17y + 22 = 0

The resulting equation is a quadratic equation in terms of y.

Step 5: Solve the quadratic equation. You can either use factoring, completing the square, or the quadratic formula. In this case, the equation does not factor easily, so we'll use the quadratic formula.

Using the quadratic formula, where a = 3, b = 17, and c = 22, we have:

y = (-b ± √(b^2 - 4ac)) / (2a)
y = (-17 ± √(17^2 - 4(3)(22))) / (2(3))
y = (-17 ± √(289 - 264)) / 6
y = (-17 ± √25) / 6

Now we have two potential solutions for y:

y1 = (-17 + 5) / 6
y1 = -2/3

y2 = (-17 - 5) / 6
y2 = -22/3

Step 6: Substitute the y-values back into the first equation (x = y + 3) to solve for x.

For y = -2/3:
x = (-2/3) + 3
x = 7/3

For y = -22/3:
x = (-22/3) + 3
x = 7/3

So the solutions to the quadratic system of equations y = 3x^2 - 5 and x - y = 3 are x = 7/3 and y = -2/3.