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314 Answers to Exercises and Solutions to Problems

whence

vy(y - x) = c - bx

vy2(y - x) = d - cx

(c - bx)2 = (b - ux)(d - cx).
The solutions for which y = x can occur only if x = b/u = c/b = d/c. In
this case, u and v are restricted only by u + v = a. Henceforth, we exclude
this possibility.
Consider the possibility that y = 0. This is feasible only if bd = c2, in
which case x = c/b = d/c, u = b/x and v = a - u.
Now (ac - b2)x2 - (ad - bc)x + (bd - c2) = 0. Suppose that UC - b2 #

0. Then we can solve for x and determine the remaining variables from
   y(b-ux)=c-bx,v(y-x)=b-ux,u=u-v.Ifac-b2=O,ud-bc#O,
   then the quadratic equation collapses to a linear equation with a single
   root x = b/u = c/b. But then 0 # d - cx = y2(b - ax) = 0, a contradiction.
   The remaining possibility is that all coefficients of the quadratic vanish. In
   this case, t/u = c/b = d/c. From the three equations at the head of the
   solution, we find that

(i) v = 0, x = b/u so that u = a and y is arbitrary;

(ii) y = b/u = c/b = d/ c so that either u = 0, v = a and x is arbitrary or

else u # 0 and x = y = b/a.

A method for dealing with general equations of this type is given in S.
Ramanujan, Note on a set of simultaneous equations, J. Ind. Math. Sot. 4
(1912), 94-96 = Collected papers (#3), 18-19 (Chelsea, 1927, 1962).

8.18. From the given system, we can derive

(xy + yz + ZX) + 3u + 1Ov = 0 or 8 + 3u + 1Ov = 0

(v + Y)(Z - x) = 0
(v + z>(y - x) = 0
(v + x)(z - y) = 0.
The first two equations of the given system show that x, y, z cannot all
be equal. Also, x, y, z cannot all be distinct, since then v + y, v + z, v + x
could not vanish simultaneously. Hence, exactly two of x, y, z are equal,
sayx=y#z.Then2x+z=5,2x2+r2=9,v+x=Oand3u+-lOv=-8.
Hence (x, y, z, u, v) = (2,2,1,4, -2) or (4/3, 4/3, 7/3, 16/9, -4/3).

8.19. If any of x, y, z vanish, then (x, y, z) is one of (O,O,O), (a, O,O),
(0, b,O), (O,O, c). Suppose, if possible, that xyz # 0. Adding these three
equations yields