Question

if xy/x+y =a

xz/x+z=b

yz/y+z=c

where a,b,c are not zero

the x equals:

a)abc/ab+bc+ca

b)2abc/ab+bc+ca

c)2abc/ab+ac-bc

d)2abc/bc+ca-ab

Answer 1

xy/(x+y) is the same as 1/(1/x+1/y)

so we can rewrite xy/(x+y) = a

as

1/x+1/y = 1/a (1)

also rewrite xz/(x+z) = b

as

1/x+1/z = 1/b (2)

add (1) and (2)

==> 2/x+1/y+1/z = 1/a+1/b

but we also know that 1/z+1/y = 1/c

so 2/x+1/c = 1/a+1/b

so 2/x = 1/a+1/b-1/c = (bc+ac-ab)/abc

so x = 2abc / (bc+ac-ab)

QED

Answer 2

Answer:

The required answer is (d) $\frac{2abc}{bc + ac - ab}$

Step-by-step explanation:

Given that,

 $\frac{xy}{x+y} = a$

$or, \frac{1}{\frac{x}{xy} +\frac{y}{xy} } = a$

$or, \frac{1}{\frac{1}{x} + \frac{1}{y} } = a$

$or, \frac{1}{x} + \frac{1}{y} = \frac{1}{a}\rightarrow(1)$

$Similarly, \frac{xz}{x+z} = b$

$or, \frac{1}{\frac{x}{xz}+ \frac{z}{xz} } = b$

$or, \frac{1}{\frac{1}{z} + \frac{1}{x} } = b$

$or, \frac{1}{z} + \frac{1}{x} = \frac{1}{b} \rightarrow (2)$

$and, \frac{yz}{y+z} = c$

$or, \frac{1}{\frac{y}{yz} + \frac{z}{yz} } = c$

$or, \frac{1}{\frac{1}{z} + \frac{1}{y} } = c$

$or, \frac{1}{z} + \frac{1}{y} = c \rightarrow (3)$

Subtracting equation (3) from (1) we get,

$\frac{1}{x} - \frac{1}{z} = \frac{1}{a} - \frac{1}{c} \rightarrow (4)$

Adding equation (2) and (4) we get,

$\frac{2}{x} = \frac{1}{a} + \frac{1}{b} - \frac{1}{c}$

$\Rightarrow \frac{2 abc}{x} = bc + ac - ab$

$\Rightarrow x = \frac{2abc}{bc +ac -ab}$