SOLUTION: solve the system using your choice of method: x^2-y^2-4x+6y-4=0 x^2+y^2-4x-6y+12=0

Algebra ->  Systems-of-equations -> SOLUTION: solve the system using your choice of method: x^2-y^2-4x+6y-4=0 x^2+y^2-4x-6y+12=0      Log On


   

Question 152174: solve the system using your choice of method:
x^2-y^2-4x+6y-4=0
x^2+y^2-4x-6y+12=0
Answer by kev82(151) About Me  (Show Source):
You can put this solution on YOUR website!
Adding the equations together looks like it may be a good place to start, where exactly are you stuck? Can you show some working?
-------------------------------------------------
I added them and came up with 2x^2-8x+8=0, this is where I was stuck

This is a quadratic equation. If you are given a problem like the one above, you should really be quite comfortable with this type of equation already, and the techniques of how to solve it. You have no idea?
I don't really like the site solver, but try and follow this. However I suggest you go and see your teacher and cover solving quadratics.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-8x%2B-8+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-8%29%5E2-4%2A2%2A-8=128.

Discriminant d=128 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--8%2B-sqrt%28+128+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-8%29%2Bsqrt%28+128+%29%29%2F2%5C2+=+4.82842712474619
x%5B2%5D+=+%28-%28-8%29-sqrt%28+128+%29%29%2F2%5C2+=+-0.82842712474619

Quadratic expression 2x%5E2%2B-8x%2B-8 can be factored:
2x%5E2%2B-8x%2B-8+=+2%28x-4.82842712474619%29%2A%28x--0.82842712474619%29
Again, the answer is: 4.82842712474619, -0.82842712474619. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-8%2Ax%2B-8+%29


When you have solved for the possible values of x, you must then find y. Substituting for x should leave you some more quadratic equations to solve to give y, you can use the same technique again.