Show that if (a, b) satisfies a2 - 3ab + 2b2 + a - b = a2 - 2ab + b2 - 5a + 7b = 0, then it also satisfies ab - 12a + 15b = 0.
Solution
One approach is simply to solve the equations. The trick is to notice that we can factorise the first as (a-b)(a-2b+1) = 0. If a = b, then the second equation gives a = 0, so the solution is (a,b) = (0,0). If a-2b+1 = 0, then the second equation gives (4b2-4b+1) - 2b(2b-1) + b2 - 5(2b-1) + 7b = b2 - 5b + 6 = 0, so b = 2 or 3, giving solutions (a,b) = (3,2) or (5,3). It is easy to check that all of (0,0), (3,2), (5,3) satisfy ab - 12a + 15b = 0.
Another approach is to find a suitable combination of the two equations. We obviously cannot take m(1) + n(2) with m and n reals, but suppose we take m, n to have the form pa + qb + r. So suppose we take (pa + qb + r)(a2-3ab+2b2+a-b) + (p'a+q'b+r')(a2-2ab+b2-5a+7b). To avoid an a3 term we need p' = -p. To avoid a b3 term we need q' = -2q. Then to avoid an a2b term we need -3p+q+2p-2q = 0, so q = -p. So we try (a-b+r)(a2-3ab+2b2+a-b) + (-a+2b+r')(a2-2ab+b2-5a+7b). After some more algebra we find this works with r=-9, r'=3.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03