Not Three Egyptian Fractions

$\frac{1}{x}+\frac{1}{y} - (\frac{1}{x}+\frac{1}{z}) = \frac{1}{3} - \frac{1}{5}$

$\frac{z-y}{zy} = \frac{2}{15}$

similarly $\frac{z+y}{zy} = \frac{1}{7}$

by dividing $\frac{z-y}{z+y} = \frac{14}{15}$

now by Componendo and Dividendo ,

$\frac{z-y+(z+y)}{z+y-(z-y)} = \frac{14+15}{15-14}$

$\frac{2z}{2y} = \frac{29}{1}$

hence $\frac{z}{y} = 29$

It is given that:

$\begin{cases} \dfrac {1}{x} + \dfrac {1}{y} = \dfrac {1}{3} & ...(1) \\ \dfrac {1}{x} + \dfrac {1}{z} = \dfrac {1}{5} & ...(2) \\ \dfrac {1}{y} + \dfrac {1}{z} = \dfrac {1}{7} & ...(3) \end{cases}$

$\begin{array} {llll} Eq.1+Eq.2-Eq.3: & \dfrac {2}{x} = \dfrac {1}{3} + \dfrac {1}{5} - \dfrac {1}{7} = \dfrac {41} {105} & \Rightarrow \dfrac {1}{x} = \dfrac {41} {210} & ...(3a) \\ Eq.3a \rightarrow Eq.1: & \dfrac {41}{210} + \dfrac {1}{y} = \dfrac {1}{3} & \Rightarrow \dfrac {1}{y} = \dfrac {29}{210} & ...(1a) \\ Eq.3a \rightarrow Eq.2: & \dfrac {41}{210} + \dfrac {1}{z} = \dfrac {1}{5} & \Rightarrow \dfrac {1}{z} = \dfrac {1}{210} & ...(2a) \\ \dfrac {Eq.1a}{Eq.2a}: & \dfrac {z}{y} = \boxed{29} && \end{array}$

$\begin{cases} \dfrac {1}{x} + \dfrac {1}{y} = \dfrac {1}{3} & ...(1) \\ \dfrac {1}{x} + \dfrac {1}{z} = \dfrac {1}{5} & ...(2) \\ \dfrac {1}{y} + \dfrac {1}{z} = \dfrac {1}{7} & ...(3) \end{cases}\\~~\\$ $\\~~\\\begin{cases} (2)~-~(1)~~\dfrac {1}{y} -\dfrac {1}{z} = \dfrac {2}{15} ....(4)\\Subtract~(4)~from~(3)~~\dfrac {2}{z}=\dfrac {1}{105}..(5)\\Add~(3)~and~(4)~~\dfrac {2}{y}=\dfrac {29}{105}...(6)\\ \end{cases} \\\dfrac {z}{y}=\dfrac {(6)}{(5)}= \color{#D61F06} { \Large 29}$

Add the three equations together to get $2( \frac{1}{x} + \frac{1}{y} + \frac{1}{z})=(1/3)+(1/5) + (1/7)$ (equation $1$ ).Then subtract the $2( \frac{1}{x} + \frac{1}{y}) = (2/3)$ from both sides to get $\frac{2}{z} = (71/105) - (70/105) = (1/105)$ , so $z = 210$ . Subtract $2( \frac{1}{x} + \frac{1}{z} )= (2/5)$ from both side of equation $1$ yields $\frac{2}{y} = (71/105) - (42/105) = (29/105)$ , so $y=(210/29)$ . It follows that $\frac{z}{y} = \frac{210}{210/29} = 29$ .

Being (1), (2) and (3) the equations, making (1)-(2)+(3) yields

$\dfrac{2}{y}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}=\dfrac{29}{105}$

and making (1)+(2)-(3) we have

$\dfrac{2}{z}=\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{1}{105}$

Divinding these expressions by one another, we have

$\dfrac{2/y}{2/z}=\dfrac{z}{y}=\dfrac{29/105}{1/105}=29$

Equation (1) + (3) - (2) Gives you:

2/y = 29/105

Therefore y = 210/29

Substitute it Equation (3) to give z = 210

z/y = 210/(210/29) = 29

let 1/x = a, 1/y = b, 1/z = c

a + b = 1/3

a + c = 1/5

b + c = 1/7

from equation 1,

b = 1/3 - a

from equation 2,

c = 1/5 - a

substitute these in equation 3

1/3 - a + 1/5 - a = 1/7

-2a = -41/105

a = 41/210

b = 1/3 - 41/210 = 29/210

c = 1/5 - 41/210 = 1/210

Now, solve for x,y and z

1/x = 41/210

x = 210/41

1/y = 29/210

y = 210/29

1/z = 1/210

z = 210

z/y = 210/(210/29)

z/y = (210/1) x (29/210) = 29

Even we can solve by putting 1/x=a 1/y=b 1/z=c Then solve simultaneously for a,b,c And then find b/c!!!!