If \(x + \dfrac{y}{2} = \dfrac{1}{4}\), \(y + \dfrac{z}{2} = \dfrac{1}{4}\) and \(z + \dfrac{x}{2} = \dfrac{1}{4}\), then what is the value of \(x + y + z\)?
\(\dfrac{1}{2}\)
The three equations are cyclic: each variable appears once with coefficient 1 and once with coefficient one-half. Adding all three equations collects every variable with the same total coefficient, so the sum can be found without solving for each variable separately.
Step 1 — Add the three equations: \(\left(x + y + z\right) + \dfrac{1}{2}\left(x + y + z\right) = \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4}\).
Step 2 — \(\dfrac{3}{2}\left(x + y + z\right) = \dfrac{3}{4}\).
Step 3 — \(x + y + z = \dfrac{3}{4} \times \dfrac{2}{3} = \dfrac{1}{2}\).
Check — by symmetry \(x = y = z = \dfrac{1}{6}\) satisfies each equation, since \(\dfrac{1}{6} + \dfrac{1}{12} = \dfrac{1}{4}\), and their sum is \(\dfrac{1}{2}\), not 1, one-third or one-fourth.
\(\dfrac{1}{2}\) option is correct.
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