# Solve-x-y-and-z-in-terms-of-p-q-and-r-yz-py-qz-i-zx-qz-rx-ii-xy-rx-py-iii-

Question Number 11268 by tawa last updated on 18/Mar/17
$$\mathrm{Solve}\:\mathrm{x},\:\mathrm{y}\:\mathrm{and}\:\mathrm{z}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{p},\:\mathrm{q}\:\mathrm{and}\:\mathrm{r} \\$$$$\mathrm{yz}\:=\:\mathrm{py}\:+\:\mathrm{qz}\:\:\:\:\:……..\:\left(\mathrm{i}\right) \\$$$$\mathrm{zx}\:=\:\mathrm{qz}\:+\:\mathrm{rx}\:\:\:\:\:\:…….\:\left(\mathrm{ii}\right) \\$$$$\mathrm{xy}\:=\:\mathrm{rx}\:+\:\mathrm{py}\:\:\:\:\:…….\:\left(\mathrm{iii}\right) \\$$
Answered by mrW1 last updated on 18/Mar/17
$$\left({iii}\right)\Rightarrow{y}=\frac{{rx}}{{x}−{p}} \\$$$$\left({i}\right)\Rightarrow{z}=\frac{{py}}{{y}−{q}}=\frac{{p}}{\mathrm{1}−\frac{{q}}{{y}}}=\frac{{p}}{\mathrm{1}−\frac{{q}\left({x}−{p}\right)}{{rx}}} \\$$$$\left({ii}\right)\Rightarrow{z}=\frac{{rx}}{{x}−{q}} \\$$$$\Rightarrow\frac{{rx}}{{x}−{q}}=\frac{{p}}{\mathrm{1}−\frac{{q}\left({x}−{p}\right)}{{rx}}} \\$$$${rx}\left[\mathrm{1}−\frac{{q}\left({x}−{p}\right)}{{rx}}\right]={p}\left({x}−{q}\right) \\$$$${rx}−{q}\left({x}−{p}\right)={p}\left({x}−{q}\right) \\$$$$\left({p}+{q}−{r}\right){x}=\mathrm{2}{pq} \\$$$$\Rightarrow{x}=\frac{\mathrm{2}{pq}}{{p}+{q}−{r}} \\$$$${similarly} \\$$$$\Rightarrow{y}=\frac{\mathrm{2}{qr}}{{q}+{r}−{p}} \\$$$$\Rightarrow{z}=\frac{\mathrm{2}{pr}}{{p}+{r}−{q}} \\$$
Commented by tawa last updated on 18/Mar/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\$$
Commented by tawa last updated on 18/Mar/17
$$\mathrm{please}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{need}\:\mathrm{the}\:\mathrm{cubic}\:\mathrm{equation}\:\mathrm{formulars}\:\mathrm{you}\:\mathrm{posted}\:\mathrm{sometimes}\:\mathrm{ago}. \\$$
Commented by mrW1 last updated on 18/Mar/17
Commented by tawa last updated on 18/Mar/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\$$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/Mar/17
$$\\$$$${the}\:{system}\:{of}\:{this}\:{equations}\:{have}\:{one} \\$$$${more}\:{answers}:\:\:{x}={y}={z}=\mathrm{0} \\$$
Commented by mrW1 last updated on 19/Mar/17
$${you}\:{are}\:{right}.\:{this}\:{case}\:{should}\:{be}\:{considered} \\$$$${in}\:{the}\:{step}\:\mathrm{2}. \\$$