Answer:
To solve the system of equations:
1. x^2 + y^2 = 58
2. x^2 - y^2 = 40
We can use the method of elimination. Add the two equations together to eliminate the y^2 terms:
(x^2 + y^2) + (x^2 - y^2) = 58 + 40
Simplify:
2x^2 = 98
Now, divide both sides by 2 to solve for x^2:
x^2 = 98 / 2
x^2 = 49
Take the square root of both sides:
x = ±√49
x = ±7
Now that we have the values of x, we can substitute them into either equation (1) or (2) to find the corresponding y values. Let's use equation (1):
1. x^2 + y^2 = 58
For x = 7:
7^2 + y^2 = 58
49 + y^2 = 58
Subtract 49 from both sides:
y^2 = 58 - 49
y^2 = 9
Take the square root of both sides:
y = ±√9
y = ±3
So, we have two sets of solutions:
1. When x = 7, y = 3 or y = -3
2. When x = -7, y = 3 or y = -3
Therefore, the system of equations has four solutions:
1. (x, y) = (7, 3)
2. (x, y) = (7, -3)
3. (x, y) = (-7, 3)
4. (x, y) = (-7, -3)