x-y-z-R-If-1-x-1-y-z-1-2-1-y-1-x-z-1-3-1-z-1-x-y-1-4-x-y-z-

Question Number 183668 by mnjuly1970 last updated on 28/Dec/22

x, y, z \in R :

If {\begin{cases} \frac{1}{x} + \frac{1}{y + z} = \frac{1}{2} \\ \frac{1}{y} + \frac{1}{x + z} = \frac{1}{3} \\ \frac{1}{z_{}} + \frac{1}{x + y} = \frac{1}{4} \end{cases}

\Rightarrow x, y, z = ?

Commented by Frix last updated on 28/Dec/22

This is easy to solve with

y = a x \land z = b x \land a, b \neq 0 \land a + b + 1 \neq 0

W e get

a = \frac{5}{3} \land b = 5 \Rightarrow x = \frac{23}{10} \land y = \frac{23}{6} \land z = \frac{23}{2}

Answered by Rasheed.Sindhi last updated on 29/Dec/22

{\begin{cases} \frac{1}{x} + \frac{1}{y + z} = \frac{1}{2} \\ \frac{1}{y} + \frac{1}{x + z} = \frac{1}{3} \\ \frac{1}{z_{}} + \frac{1}{x + y} = \frac{1}{4} \end{cases}

{\begin{cases} \frac{x + y + z}{x (y + z)} = \frac{1}{2} \Rightarrow x + y + z = \frac{x (y + z)}{2} \\ \frac{x + y + z}{y (x + z)} = \frac{1}{3} \Rightarrow x + y + z = \frac{y (x + z)}{3} \\ \frac{x + y + z}{z (x + y)} = \frac{1}{4} \Rightarrow x + y + z = \frac{z (x + y)}{4} \end{cases}

\Rightarrow \frac{x y + x z}{2} = \frac{x y + y z}{3} = \frac{x z + y z}{4}

= \frac{x y + x z + x y + y z + x z + y z}{2 + 3 + 4}

= \frac{2 (x y + y z + x z)}{9}

{\begin{cases} 9 (x y + x z) = 4 (x y + y z + x z) \\ 9 (x y + y z) = 6 (x y + y z + x z) \\ 9 (x z + y z) = 8 (x y + y z + x z) \end{cases}

{\begin{cases} 5 x y - 4 y z + 5 x z = 0 \\ 3 x y + 3 y z - 6 x z = 0 \\ - 8 x y + y z + x z = 0 \end{cases}

D i v i d i n g b y x y z t o b o t h s i d e s :

{\begin{cases} \frac{5}{z} - \frac{4}{x} + \frac{5}{y} = 0 \dots . (i) \\ \frac{3}{z} + \frac{3}{x} - \frac{6}{y} = 0 \dots . (i i) \\ \frac{- 8}{z} + \frac{1}{x} + \frac{1}{y} = 0 \dots . (i i i) \end{cases}

(i) \times 3 : \frac{15}{z} - \frac{12}{x} + \frac{15}{y} = 0 \dots (i v)

(i i) \times 5 : \frac{15}{z} + \frac{15}{x} - \frac{30}{y} = 0 \dots (v)

(i v) - (v) : \frac{- 27}{x} + \frac{45}{y} = 0

\frac{27}{x} = \frac{45}{y}

\frac{3}{x} = \frac{5}{y} \dots . (v i i i)

(i i) \times 1 : \frac{3}{z} + \frac{3}{x} - \frac{6}{y} = 0 \dots (v i)

(i i i) \times 3 : \frac{- 24}{z} + \frac{3}{x} + \frac{3}{y} = 0 \dots (v i i)

(v i) - (v i i) : \frac{27}{z} - \frac{9}{y} = 0

\frac{3}{z} = \frac{1}{y}

\frac{15}{z} = \frac{5}{y}

\frac{3}{x} = \frac{5}{y} \dots \dots . (i x)

(v i i i) & (i x) : \frac{3}{x} = \frac{5}{y} = \frac{15}{z}

x = 3 k, y = 5 k, z = 15 k

\frac{1}{x} + \frac{1}{y + z} = \frac{1}{2} (f i r s t g i v e n e q n)

\Rightarrow \frac{1}{3 k} + \frac{1}{5 k + 15 k} = \frac{1}{2}

60 k \cdot \frac{1}{3 k} + 60 k \cdot \frac{1}{20 k} = 60 k \cdot \frac{1}{2}

20 + 3 = 30 k \Rightarrow k = \frac{23}{30}

x = 3 k = 3 (\frac{23}{30}) = \frac{23}{10}

y = 5 k = 5 (\frac{23}{30}) = \frac{23}{6}

z = 15 k = 15 (\frac{23}{30}) = \frac{23}{2}

Commented by mnjuly1970 last updated on 29/Dec/22

t a n k s a l o t a l i \dots