Question-88329

Question Number 88329 by TawaTawa1 last updated on 10/Apr/20

Commented by TawaTawa1 last updated on 10/Apr/20

Find x, y and z

Commented by Tony Lin last updated on 10/Apr/20

{\begin{cases} x = \frac{2 p q}{q + p - r} \\ y = \frac{2 q r}{q + r - p} o r \\ z = \frac{2 p r}{p + r - q} \end{cases}

{\begin{cases} x = 0 \\ y = 0 \\ z = 0 \end{cases}

Commented by Tony Lin last updated on 10/Apr/20

y z - q z = p y

z = \frac{p y}{y - q}

x y - r x = p y

x = \frac{p y}{y - r}

\frac{p^{2} y^{2}}{(y - r) (y - q)} = \frac{p q y}{y - q} + \frac{p r y}{y - r}

p^{2} y^{2} = p q y (y - r) + p r y (y - q)

i f p, y \neq 0

p y = q y - q r + r y - r q

y = \frac{2 q r}{q + r - p}

x = \frac{p y}{y - r} = \frac{2 p q}{q + p - r}

z = \frac{p y}{y - q} = \frac{2 p r}{p + r - q}

i f y = 0

t h e n x, z = 0

s i m i l a r i l y

i f x = 0

t h e n y, z = 0

i f z = 0

t h e n x, y = 0

i f o n e o f p, q, r = 0 o r m o r e

t h e n t h e e q u a t i o n h a s \infty s o l u t i o n

Commented by TawaTawa1 last updated on 10/Apr/20

God bless you sir .