If $$\frac{{xy}}{{x + y}} = a,$$ $$\frac{{xz}}{{x + z}} = b$$ and $$\frac{{yz}}{{y + z}} = c{\text{,}}$$ where a, b, c are all non - zero numbers, then x equals to?
A. $$\frac{{2abc}}{{ab + bc - ac}}$$
B. $$\frac{{2abc}}{{ab - bc + ac}}$$
C. $$\frac{{2abc}}{{bc + ac - ab}}$$
D. $$\frac{{2abc}}{{ab + bc + ac}}$$
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \frac{{xy}}{{x + y}} = a,\,\frac{{xz}}{{x + z}} = b,\,\frac{{yz}}{{y + z}} = c{\text{ }} \cr & {\text{Now,}} \cr & \frac{{x + y}}{{xy}} = \frac{1}{a} \cr & \frac{{x + z}}{{xz}} = \frac{1}{b} \cr & \frac{{y + z}}{{yz}} = \frac{1}{c} \cr & \Rightarrow \frac{1}{y} + \frac{1}{x} = \frac{1}{a},\frac{1}{z} + \frac{1}{x} = \frac{1}{b},\frac{1}{x} + \frac{1}{y} = \frac{1}{c} \cr & {\text{Now we have to find the value of }}x \cr & \therefore \frac{1}{a} + \frac{1}{b} - \frac{1}{c} = \frac{1}{y} + \frac{1}{x} + \frac{1}{z} + \frac{1}{x} - \frac{1}{y} - \frac{1}{z} \cr & \therefore \frac{1}{a} + \frac{1}{b} - \frac{1}{c} = \frac{2}{x} \cr & \Rightarrow \frac{{bc + ac - ab}}{{abc}} = \frac{2}{x} \cr & \Rightarrow x = \frac{{2abc}}{{bc + ac - ab}} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
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B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
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D. $$\frac{8}{6}$$
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