SOLUTION: the product of two numbers is six times their sum. the sum of their squares is 325 what are the two numbers?

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Question 912068: the product of two numbers is six times their sum. the sum of their squares is 325 what are the two numbers?
Found 2 solutions by CubeyThePenguin, greenestamps:
Answer by CubeyThePenguin(3113) (Show Source):
You can put this solution on YOUR website!
xy = 6(x + y)
x^2 + y^2 = 325

xy - 6x - 6y = 0
(x - 6)(y - 6) = 36

x - 6 = 1, y - 6 = 36 ---> (x, y) = (7, 42)
x - 6 = 2, y - 6 = 18 ---> (x, y) = (8, 24)
x - 6 = 3, y - 6 = 12 ---> (x, y) = (9, 18)
x - 6 = 4, y - 6 = 9 ---> (x, y) = (10, 15)
x - 6 = 6, y - 6 = 6 ---> (x, y) = (12, 12)
x - 6 = 9, y - 6 = 4 ---> (x, y) = (15, 10)
x - 6 = 12, y - 6 = 3 ---> (x, y) = (18, 9)
x - 6 = 18, y - 6 = 2 ---> (x, y) = (24, 8)
x - 6 = 36, y - 6 = 1 ---> (x, y) = (42, 7)

The pairs that work are (x, y) = (10, 15) and (15, 10).

Answer by greenestamps(13203) (Show Source):
You can put this solution on YOUR website!

The response from the other tutor shows a solution using a bit of formal algebra and then trial and error.

You can find the solution quickly and easily by looking for pairs of numbers for which the sum of the squares is 325 and the product of the two numbers is 6 times their sum.

On first reading the problem, it took only a few seconds for me to see 325 = 225+100 = 15^2+10^2, and 10*15 = 150 = 6(10+15).

So an informal solution can take a very short amount of time, if you are good with mental arithmetic.

However, there are actually two very different solutions to the problem. The other tutor found one of them (which is undoubtedly the answer the author of the problem intended); but the other is missing from his response.

Following is a path to finding both solutions.

The given information tells us

Here is the key to the path I used to the solution: Whenever the information given in a problem involves and , see what can be done with

We will do that here.

View this as a quadratic equation in which the "variable" is x+y and factor:

OR

Solve each of these together with to find the two solutions to the problem.

(1) if x+y=25, xy = 6(x+y) = 150; then...

The two numbers are 15 and 10.

CHECK:
10^2+15^2 = 100+225 = 325
10(15) = 150; 6(10+15) = 150

(2) if x+y = -13, xy = 6(x+y) = -78; then...

This one doesn't factor; we need to use the quadratic formula.

This leads to the two numbers being

and

CHECK: