The value of x satisfying the equations \(\frac{1}{x} + \frac{1}{y} = 8\),…

Concept.

When the unknowns of a system appear only through their reciprocals, put \(a=\frac{1}{x}\), \(b=\frac{1}{y}\), \(c=\frac{1}{z}\); the system then becomes linear in \(a\), \(b\), \(c\). If the three equations give the three pairwise sums of \(a\), \(b\), \(c\), then adding all of them counts every unknown twice, so the grand total \(a+b+c\) is half the sum of the constants — and any single unknown equals that grand total minus the pairwise sum that leaves it out.

Application.

1. Write the system in the reciprocal variables: \(a+b=8\), \(b+c=12\), \(c+a=10\).

2. Add the three equations: \((a+b)+(b+c)+(c+a)=8+12+10=30\), that is \(2(a+b+c)=30\).

3. Halve both sides to get the grand total: \(a+b+c=15\).

4. Subtract the pairwise sum that omits \(a\): \(a=(a+b+c)-(b+c)=15-12=3\).

5. Convert back to the original unknown: \(\frac{1}{x}=3\), hence \(x=\frac{1}{3}\).

Cross-check.

The same subtraction gives \(b=15-10=5\) and \(c=15-8=7\), so \(y=\frac{1}{5}\) and \(z=\frac{1}{7}\). Substituting the reciprocals back: \(3+5=8\), \(5+7=12\) and \(7+3=10\) — all three equations hold, confirming \(x=\frac{1}{3}\).

Contrast.

• \(3\) is the value of the reciprocal \(\frac{1}{x}\) reached at the subtraction step; the question asks for \(x\) itself, so that value still has to be inverted.

• \(3\frac{1}{3}=\frac{10}{3}\) blends the reciprocal \(3\) with \(\frac{1}{3}\); no step of this elimination produces such a sum.

• The reciprocal system is linear with a single solution, so exactly one value of \(x\) can work; a claim that several of the listed values work, or that none of them does, is therefore ruled out.