system of equation

Question

x^2-xy-y^2=7

x^2+y^2=10

find x and y

Answer 1

If you're sure that you wrote down the equations correctly, then I can point the way, but I'm not going to work out the whole solution.

 

x² - xy - y² = 7
x² + y² = 10

 

From the second equation, we know that  y² = 10 - x².  That also means that y = ±√(10 - x²)

 

So we have two values for y that we can use in the first equation.  That will give us two different equations we can solve.

 

x² - xy - y² = 7

 

Equation 1:  x² - x√(10 - x²) - (10 - x²) = 7

Equation 2:  x² + x√(10 - x²) - (10 - x²) = 7

 

Each one of these can be solved.  I'll show you how to do the first, but not all the way.

 

First simplify

 

x² - x√(10 - x²) - (10 - x²) = 7

2x² - x√(10 - x²) - 10 = 7

2x² - x√(10 - x²) - 17 = 0

2x² - 17 = x√(10 - x²)

 

Square both sides:

4x⁴ - 68x² + 289 = x²(10 - x²)

4x⁴ - 68x² + 289 = 10x² - x⁴

5x⁴ - 78x² + 289 = 0

 

This is really a quadratic equation, because there are no odd powers of x.  You can let y = x²:

5x⁴ - 78x² + 289 = 0

5y² - 78y + 289 = 0

 

Then solve for y, using the quadratic formula.  You'll get some expression for y, but since y = x², you then take the positive and negative square roots of y to get an expression for x.  So you'll have two solutions for x.

 

Then do the same for Equation 2 and get two more solutions for x.

 

In the end, you'll have 4 solutions for x.  You'd need to check all of them, because when you squared the equations, there's the possibility that you introduced invalid solutions. 

 

However, it's very possible that there are 4 solutions.  I don't know if you've seen graphs of these types of equations yet, but the first one is a hyperbola and the second one is a circle.  A circle and a hyperbola can intersect at 4 points so it does make sense.

 

Anyway, I know this is very messy and there might be some trick that I'm missing.  If someone else finds a better way to solve this, I'd be very interested!