Simple System of Equations Part 2

It is given that: $\quad \begin{cases} xy + z = 26 &...(1) \\ xz+y = 29 &...(2) \\ yz+ x = 34 &... (3) \end{cases}$

$(2)-(1): \quad x(z-y)-(z-y) = 3 \implies x = \dfrac{3}{z-y} + 1\quad ...(4)$

Since $x$ , $y$ and $z$ are integers, for equation $(4)$ to be true $\implies \dfrac 3{z-y}$ must be an integer.

$\implies \begin{cases} z-y = -3 & \implies z = y-3 & \implies x = 0 \\ z-y = -1 & \implies z = y-1 & \implies x = -2 \\ z-y = 1 & \implies z = y+1 & \implies x = 4 \\ z-y = 3 & \implies z = y+3 & \implies x = 2 \end{cases}$

From equation $(3): \ \ yz+x = 34$ , we have:

$\implies \begin{cases} x = 0 & \implies y(y-3) + 0 = 34 & \implies y^2-3y-34 = 0 & \color{#D61F06}{\implies y \text{ is not integral}} \\ x = -2 & \implies y(y-1) + -2 = 34 & \implies y^2-y-36 = 0 & \color{#D61F06}{\implies y \text{ is not integral}} \\ x = 4 & \implies y(y+1) + 4 = 34 & \implies y^2+y-30 = 0 & \implies \begin{cases} y = 5 & \implies z = 6 & \\ y = -6 & \implies z = -5 \end{cases} \\ x = 2 & \implies y(y+3) + 2 = 34 & \implies y^2+3y-32 = 0 & \color{#D61F06}{\implies y \text{ is not integral}} \end{cases}$

Of the two integral cases, we find that $(4,-6,-5)$ is not a solution of the system of equations. Therefore, $x+y+z = 4+5+6 = \boxed{15}$

What a dumb problem.
29 - 26 = 3.
So right there you know that one of the numbers is 3 or 3 multiplied by some number.
34 -29 = 5. So right there you know that one of the numbers is 5 or 5 multiplied by some number. A few simple substitutions gives you the answer.

Simple guess the numbers as 4,5,6. All of them satisfy the 3 equations. Hence x + y + z = 15

Subtracting these equations taking two at a time, we get (x-1)(z-y) = 3 ............. {1} (y-1)(z-x) = 8 ............. {2} (z-1)(y-x) = 5 ............. {3} We know that x,y,z are integers and that 3 and 5 can be factorised only as 3×1 and 5×1 respectively. So from {1} and {3}, we can conclude that (x-1) = ±3 OR ±1 which implies that x = 4,-2,2 or 0. Also (z-1) = ±5 OR ±1 which implies that z = 6,-4,2 or 0 Now eliminating y from the first two equations (which are given in the question), we get (x²-1)z = 29x-26 Now we can put values of x in this equation and check the values of z that come from this equation. Then we find that only x=4 and z=6 is a valid combination. Plugging these in any of the original equations we get y=5 ∴ x+y+z = 4+5+6 = 15

$x=4$ $y=5$ $z=6$

$4 \times 5+ 6 =26$

$4 \times 6+ 5 = 29$

$5 \times 6 + 4 = 34$

$4+5+6 = \boxed{15}$