The three examples below give a feel for the range of solutions we received. They are interesting in their range of approaches and detail. Well done to all of you.

Firstly, Rachael and Katy of Ardingly College Junior School who sent a solution obtained through trial and improvement. Some people undervalue this valid mathematical method and it can often lead us to a very complete answer or, at the very least, throw light on what to be looking for when finding a fuller solution to the problem.

For this problem we have found more than 1 solution.
To work this out we used trial and error.
We started with $0.5 giving 0.5 x 0.5 = 0.25 + 0.5 = 0.75$
The answer was too low so we tried $1 giving 1 x 1 = 1 + 1 = 2$
This is a correct solution. The numbers are the same and so equal the same whichever way round they go. Then I remembered that if you square a negative number the answer is positive so I tried:
$- 2 giving - 2 x - 2 = 4 + (- 2) = 2$
This is another possible solution.

James sent us this solution

From the question we can form 3 simultaneous equations.

\begin{matrix} xy + z & = & 2 \\ xz + y & = & 2 \\ yz + x & = & 2 \end{matrix}

So,

xy + z = xz + y

Take y and z from each side;

x (y - 1) = x (z - 1)

y = z (if x is not equal to zero)

Since any two of the three equations could have been chosen, by symmetry we must have

x = y = z

This gives us the quadratic,

\begin{matrix} x^{2} + x & = & 2 \\ x^{2} + x - 2 & = & 0 \\ (x - 1) (x + 2) & = & 0 \end{matrix}

So the only 2 solutions are,

x = y = z = 1 and - 2

A very complete solution, which explains the equality of the three unknowns more fully was supplied by Derek of Tin College.

Let

x, y, z

denote the set of all numbers that satisfy the given problem, namely:

\begin{matrix} xy + z & = & 2 & (I) \\ yz + x & = & 2 & (II) \\ zx + y & = & 2 & (III) \end{matrix}

Take (I), (II) and (III) and multiply through by z, x and y respectively to obtain:

\begin{matrix} xyz + z^{2} & = & 2 z & (IV) \\ xyz + x^{2} & = & 2 x & (V) \\ xyz + y^{2} & = & 2 y & (VI) \end{matrix}

Eliminate xyz from the trio of equations:

\begin{matrix} (IV) - (V) : \\ z^{2} - x^{2} & = & 2 z - 2 x \\ (z - x) (z + x) & = & 2 (z - x) \\ z + x & = & 2 & (VII) \\ (V) - (VI) : \\ x^{2} - y^{2} & = & 2 x - 2 y \\ (x - y) (x + y) & = & 2 (x - y) \\ x + y & = & 2 & (VIII) \\ (VI) - (IV) : \\ y^{2} - z^{2} & = & 2 y - 2 z \\ (y - z) (y + z) & = & 2 (y - z) \\ y + z & = & 2 & (IX) \end{matrix}

Note that by dividing through by

(z - x), (x - y) and (y - z)

we have assumed

z - x \neq 0, x - y \neq 0 and y - z \neq 0

.
If the assumption was not true, that is,

z - x = 0, x - y = 0 and y - z = 0,

then it follows that

z = x, x = y and y = z,

which as you shall see, are results we will still reach.
Elimate 2 from the trio of equations by equating:

\begin{matrix} (VII) and (VIII) : \\ z + x & = & x + y \\ z & = & y & (X) \\ (VIII) and & (IX) : \\ x + y & = & y + z \\ x & = & z & (XI) \\ (IX) and & (VII) : \\ y + z & = & z + x \\ y & = & x & (XII) \end{matrix}

Equating the trio of equations:
(X)=(XI)=(XII):

z = x = y

Therefore, the three variables are equal, which transforms (I) (or (II) or (III)) into:

\begin{matrix} x^{2} + x & = & 2 \\ x^{2} + x - 2 & = & = 0 \\ x & = & 1 or - 2 \end{matrix}

Therefore the solution sets are
1,1,1 and -2,-2,-2.