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The form of the three given equations shows that a, p, y are the three roots
of the equation
a+s b+s c+s
in which s is regarded as the unknown. On clearing of fractions, and arranging
in the form of a cubic equation in e, it is seen that the sum of the three roots is
-(a + 6+c) + (x+y+z).
Hence a+.p + y= — (a+b+ c) -\-(x-\-y+z) , and x-\-y + z— a-\-a-\-b-\-^-\-c-\-y.
Note. It may be of interest to state that if each letter be squared the re-
sult expresses the distance of any point from the origin in terms of ellipsoidal
1S6. Proposed by B. F. FIHKEL, A.M., M.Sc, Professor of Mathematics and Physics, Drury College, Spring-
(z-\-x)a— (z— x)b=2yz....(l); (x-\-y)b—(x—y)c=2zz....(2); (y + z)c—
(jy—z)a=2xy....(Z'). Find the values of x, y, and z by the method of linear si-
Solution by (J. B. M. ZEEE, A.M., Ph. D., Professor of Chemistry and Physics, The Temple College, Philadel-
Let x=^(b-\-c)u, y—i(a+C)v, z—$(a-\-~b)w.
.-. (a— b)w-\-(b+c)u=(a + c)vw....(l).
(c— a)v+(a + b)w=(b+ c)uv....(3).
We might eliminate v, w and get an equation of the fifth degree in u. We
will, however, proceed as follows: Add (1), (2), (3), then
aw(2— u— «)+&m(2— v— w)-\-cv(2— u— w)=0.
This is the case when u=v=w=0; or u=v—w=l; or m=0, w=v=2; or
d=0, m=w=2; w=0, u=v=2.
The first two sets of values satisfy the conditions.
.-. x=y=z=0; x=i(b+c), 2/=K a +c), z=i(a+b).
Notb. This is exercise 31, page 224, Systems of Linear Simultaneous Equations, of Fisher
and Schwatt's Higher Algebra , and has given teachers of algebra throughout the country considerable
trouble. Solving the equations for a, 6, and c, we readily find that
.-. x=h(b+c), y=i(a+c), z=J(a+6), as one set of values for a;, y, and z. Editor F.
Also solved by L. C. WALKER .