jmc

Question

Let a and b be nonzero real constants such that |a| |b|. Find the number of distinct values of x satisfying x - ab + x - ba = bx - a + ax - b.

Accepted Answer

Combining the fractions on each side, we get

\frac{a x - a ^{2} + b x - b ^{2}}{ab} = \frac{a x - a ^{2} + b x - b ^{2}}{( x - a ) ( x - b )} .

Note that the numerators are equal. The solution to

a x - a^{2} + b x - b^{2} = 0

is

x = \frac{a ^{2} + b ^{2}}{a + b} .

Otherwise,

\frac{1}{ab} = \frac{1}{( x - a ) ( x - b )},

so

(x - a) (x - b) = ab .

Then

x^{2} - (a + b) x + ab = ab,

so

x^{2} - (a + b) x = 0.

Hence,

x = 0

or

x = a + b .

Thus, there are

3

solutions, namely

x = 0,

x = a + b,

and

x = \frac{a ^{2} + b ^{2}}{a + b} .

(If

\frac{a ^{2} + b ^{2}}{a + b} = a + b,

then

a^{2} + b^{2} = a^{2} + 2 ab + b^{2},

so

2 ab = 0.

This is impossible, since

a

and

b

are non-zero, so all three solutions are distinct.)