Question Number 88329 by TawaTawa1 last updated on 10/Apr/20
Commented by TawaTawa1 last updated on 10/Apr/20
$$\mathrm{Find}\:\:\:\mathrm{x},\:\:\mathrm{y}\:\:\mathrm{and}\:\:\:\mathrm{z} \\ $$
Commented by Tony Lin last updated on 10/Apr/20
$$\begin{cases}{{x}=\frac{\mathrm{2}{pq}}{{q}+{p}−{r}}}\\{{y}=\frac{\mathrm{2}{qr}}{{q}+{r}−{p}}\:\:\:\:{or}}\\{{z}=\frac{\mathrm{2}{pr}}{{p}+{r}−{q}}}\end{cases} \\ $$$$\begin{cases}{{x}=\mathrm{0}}\\{{y}=\mathrm{0}}\\{{z}=\mathrm{0}}\end{cases} \\ $$
Commented by Tony Lin last updated on 10/Apr/20
$${yz}−{qz}={py} \\ $$$${z}=\frac{{py}}{{y}−{q}} \\ $$$${xy}−{rx}={py} \\ $$$${x}=\frac{{py}}{{y}−{r}} \\ $$$$\frac{{p}^{\mathrm{2}} {y}^{\mathrm{2}} }{\left({y}−{r}\right)\left({y}−{q}\right)}=\frac{{pqy}}{{y}−{q}}+\frac{{pry}}{{y}−{r}} \\ $$$${p}^{\mathrm{2}} {y}^{\mathrm{2}} ={pqy}\left({y}−{r}\right)+{pry}\left({y}−{q}\right) \\ $$$${if}\:{p},{y}\neq\mathrm{0} \\ $$$${py}={qy}−{qr}+{ry}−{rq} \\ $$$${y}=\frac{\mathrm{2}{qr}}{{q}+{r}−{p}} \\ $$$${x}=\frac{{py}}{{y}−{r}}=\frac{\mathrm{2}{pq}}{{q}+{p}−{r}} \\ $$$${z}=\frac{{py}}{{y}−{q}}=\frac{\mathrm{2}{pr}}{{p}+{r}−{q}} \\ $$$${if}\:{y}=\mathrm{0} \\ $$$${then}\:{x}\:,\:{z}=\mathrm{0} \\ $$$${similarily} \\ $$$${if}\:{x}=\mathrm{0} \\ $$$${then}\:{y}\:,\:{z}=\mathrm{0} \\ $$$${if}\:{z}=\mathrm{0} \\ $$$${then}\:{x}\:,\:{y}=\mathrm{0} \\ $$$${if}\:{one}\:{of}\:{p}\:,\:{q}\:,\:{r}=\mathrm{0}\:{or}\:{more} \\ $$$${then}\:{the}\:{equation}\:{has}\:\infty\:{solution} \\ $$
Commented by TawaTawa1 last updated on 10/Apr/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$