SOLUTION: Find the points of intersection, if any, of the graphs in the system. Give the solution of: 4x^2 - 56x + 9y^2 + 160 = 0 and 4x^2 + y^2 - 64 = 0. Please and thank you. please please

Question 288664: Find the points of intersection, if any, of the graphs in the system. Give the solution of: 4x^2 - 56x + 9y^2 + 160 = 0 and 4x^2 + y^2 - 64 = 0. Please and thank you. please please help me find the right solution for x and y, I had some help and they told me that the equation to solve was 4x^2-56x+9-(-4x^2+64)+160=0, but I'm stuck on how to solve this, please help! thank you
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the second equation.

Solve for by getting every other term to the right side.

-------------------

Move back to the first equation.

Replace each term with (since the two are essentially the same or equivalent).

Distribute.

Combine like terms.

Now let's solve by use of the quadratic formula.

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=-32, b=-56, and c=736

Negate -56 to get 56

Square -56 to get 3136 (note: remember when you square -56, you must square the negative as well. This is because .)

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and -32 to get -64

So now the expression breaks down into two parts

or

Lets look at the first part:

Add the terms in the numerator

Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator

Divide

So another answer is

So our solutions are:

or

Since the solutions in terms of 'x' are or , we can use them to find the corresponding solutions in terms of 'y'.

So if , then...

Start with the given equation

Plug in

Square to get

Multiply

Reduce.

Combine like terms.

Take the square root of both sides.

Since the square root of a negative number is not a real number, this means that there are no real solutions of 'y' when . So we can ignore this value.

Now if , then...

Start with the given equation

Plug in

Square 4 to get 16.

Multiply

Combine like terms.

Take the square root of both sides.

Take the square root of 0 to get 0.

So when , giving us the ordered pair (4,0)

This means that the two graphs intersect at the only point of (4,0)
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