To solve this set of two equations, we are going to use what is called substitution or elimination method.
1. First, we should recognize the system of equations which are given as follows:
2x^2 - y^2 = 7, (Equation 1)
3x^2 + 2y^2 = 14. (Equation 2)
2. Step two is to solve one of the equations for one variable. If we solve Equation 1 for y^2, we get:
y^2 = 2x^2 - 7.
3. Now we will substitute y^2 from Equation 1 into Equation 2. The new Equation 2 will be:
3x^2 + 2(2x^2 - 7) = 14.
This simplifies to 7x^2 - 14 = 14,
Further simplifying to 7x^2 = 28,
And simplifying to x^2 = 4.
4. The solutions for x can be found by square-rooting both sides, and we should remember that sqrt(4) can be either positive or negative, so:
x = 2 or x = -2.
5. Now that we've solved for x, we can plug it into the solved Equation 1 (y^2 = 2x^2 - 7) to find y. For x = 2:
y^2 = 2(2)^2 - 7,
y^2 = 8 - 7,
y^2 = 1.
Like before, the square root of 1 can be both positive and negative, so y = 1 or y = -1.
6. We follow the same steps for x = -2:
y^2 = 2(-2)^2 - 7,
y^2 = 8 - 7,
y^2 = 1.
Again, the square root of 1 is both positive and negative, so y = 1 or y = -1.
In conclusion, the solution to the system of equations are the points (-2, -1), (-2, 1), (2, -1), and (2, 1).