Use the elimination method to find all solutions of the system of equations. (Order your answers from smallest to largest x, then from smallest to largest y.)


x − y2 + = 0

2x2+ y2 −9 = 0

add the equations.

x+2x^2-9=0

2x^2+x-9=0

x=(-1+-sqrt(1+72))/4

To solve the system of equations using the elimination method, we need to eliminate one variable by manipulating the equations and adding or subtracting them.

First, let's rewrite the equations in standard form:

Equation 1: x - y^2 = 0
Equation 2: 2x^2 + y^2 - 9 = 0

To eliminate one of the variables, we can multiply Equation 1 by 2 and Equation 2 by -1 to make the coefficients of the squared terms the same:

2(x - y^2) = 0 --> 2x - 2y^2 = 0
-1(2x^2 + y^2 - 9) = 0 --> -2x^2 - y^2 + 9 = 0

Now, add both equations together:

(2x - 2y^2) + (-2x^2 - y^2 + 9) = 0

Simplifying, we get:

-2x^2 - 3y^2 + 9 = 0

Now, rearrange the equation to solve for y^2:

-3y^2 = 2x^2 - 9

Divide both sides by -3:

y^2 = (2/3)x^2 - 3

Now, let's substitute this expression for y^2 into Equation 1:

x - (2/3)x^2 + 3 = 0

To find the solutions, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -2/3, b = 1, and c = 3. Substituting these values into the quadratic formula, we have:

x = (-(1) ± √((1)^2 - 4(-2/3)(3))) / (2(-2/3))
x = (-1 ± √(1 + (8/3)(3))) / (-4/3)
x = (-1 ± √(1 + 8)) / (-4/3)
x = (-1 ± √9) / (-4/3)
x = (-1 ± 3) / (-4/3)

Now, let's find the values of x:

x1 = (-1 + 3) / (-4/3) = 2 / (-4/3) = -3/2
x2 = (-1 - 3) / (-4/3) = -4 / (-4/3) = 3

To find the corresponding values of y, we substitute the values of x into Equation 1:

For x = -3/2:
-3/2 - y^2 = 0
y^2 = -3/2
y = ±√(-3/2)
The square root of a negative number is not a real number, so there are no real solutions for y when x = -3/2.

For x = 3:
3 - y^2 = 0
y^2 = 3
y = ±√3

Therefore, the solutions to the system of equations are:
(-3/2, √3) and (-3/2, -√3)
(3, √3) and (3, -√3)