Question 27
If $$\frac{xy}{x+y}=1$$, $$\frac{yz}{y+z}=2$$ and $$\frac{zx}{z+x}=3$$, then the numerical value of $$15x-7y-z$$ is
Correct Answer: 36
Solution
Taking reciprocals gives $$\frac{1}{x}+\frac{1}{y}=1$$, $$\frac{1}{y}+\frac{1}{z}=\frac{1}{2}$$ and $$\frac{1}{z}+\frac{1}{x}=\frac{1}{3}$$. Solving these linear equations gives $$x=\frac{12}{5}$$, $$y=\frac{12}{7}$$ and $$z=-12$$. Therefore $$15x-7y-z=36-12+12=36$$.
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