Hey Gary

What you have is a system of equations. You have two equations and two unknowns, which makes it not very impossible to solve. I will put stars ( ) where it might not be clear what I am doing.

Equation A:

Equation B:

What you want to do is define one of the variables or in one of the equations and plug it into the other equation in order to solve. When you evaluate equation B, you will notice that you easily can define by manipulating a bit: . If you plug the found into equation A, you get the following:

This can obviously be a nicer looking expression:

Step one:

Step two:

As you might notice, we now have a quadratic equation of sorts, which we must solve for using the proper method. A quadratic equation has the form and therefore we must set our equation equal to 0. This we can do by adding 60 on both sides:

will become

To proceed, we must find the discriminant to plug into the solution formula used for quadratic equations. If the equation is to have a real solution, must be 0 or larger. If it's zero, one solution exists. If it's larger than zero, two solutions exist. The discriminant is generally defined

in this case we have:

There are two solutions to the equation. The solution formula is, generally

In this case:

Now that is known in numerical values, also can be defined numerically. Remember equation B? We simply plug the found values in instead of and get a very simple equation which will yield some numerical values:

,

thus we get

I am to tired to actually think about the other content of your problem, but I guess you can figure it out from here. Good night. I hope my reply was helpful and I didn't obscure too much in my drunkenness

Remember: