Me and my friend are revising for the [UK] STEP exams, and came across this problem:

Given:

p² + 4q² + 9r² = 729 and 8p + 8q + 3r = 243

Solve p, q, and r.

Now, this question is actually the 3rd part of one question; so I was convinced that we had to make use of one of our previous answers for that question, in order to solve this one (as is usually the case on STEP papers).

However, my friend used the fact that 243 was a factor of 729, and solved it as so:

3 * (8p + 8q + 3r) = 3 * 243

24p + 24q + 9r = 729

Seeing as p² + 4q² + 9r² also equals 729, then:

p² + 4q² + 9r² = 24p + 24q + 9r

We then move everything from the RHS to the LHS:

p² - 24p + 4q² - 24q + 9r² - 9r = 0

Now factor out p, q, and r:

p(p - 24) + q(4q - 24) + r(9r - 9) = 0

Now this is where I had a problem. My mate seemed to think that in this case, each term must each equal 0 -- despite my teacher and I both saying that you can't assume that.

However, we had a bit of a surprise when he done it anyway...

p(p - 24) = 0 => p=0, p=24

q(4q - 24) = 0 => q=0, q=6

r(9r - 9) = 0 => r=0, r=1

The p=q=r=0 solution does not work, but p=24, q=6, and r=1 is exactly what me and my teacher achieved when we used a [longer] way which *did* refer back to one of our previous questions.

My teacher (who is actually very good at maths) could not see if there was any sort of logic as to how and why this worked; his first thoughts were that it was coincidence, but perhaps it was intentional?

Can anyone see whether or not there's some logic behind why it worked?

Thanks

(sorry for the long post)

it wasn't a coincedence, because if they were any 2 eqn's with 3 unknows this method will work.

since this is true

p^2 + 4q^2 + 9r^2 = 24p + 24q + 9r

and u have infinite solutions, equating each part of the equation is a solution

that is p^2 = 24p

divide by p

p=24

and so on.

this is similar to what ur friend did, and the reason it worked is coz there are infinite solutions and u only need to find one.