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

(iii)⇒y=((rx)/(x−p)) (i)⇒z=((py)/(y−q))=(p/(1−(q/y)))=(p/(1−((q(x−p))/(rx)))) (ii)⇒z=((rx)/(x−q)) ⇒((rx)/(x−q))=(p/(1−((q(x−p))/(rx)))) rx[1−((q(x−p))/(rx))]=p(x−q) rx−q(x−p)=p(x−q) (p+q−r)x=2pq ⇒x=((2pq)/(p+q−r)) similarly ⇒y=((2qr)/(q+r−p)) ⇒z=((2pr)/(p+r−q))

$$\left({iii}\right)\Rightarrow{y}=\frac{{rx}}{{x}−{p}} \\ $$$$\left({i}\right)\Rightarrow{z}=\frac{{py}}{{y}−{q}}=\frac{{p}}{\mathrm{1}−\frac{{q}}{{y}}}=\frac{{p}}{\mathrm{1}−\frac{{q}\left({x}−{p}\right)}{{rx}}} \\ $$$$\left({ii}\right)\Rightarrow{z}=\frac{{rx}}{{x}−{q}} \\ $$$$\Rightarrow\frac{{rx}}{{x}−{q}}=\frac{{p}}{\mathrm{1}−\frac{{q}\left({x}−{p}\right)}{{rx}}} \\ $$$${rx}\left[\mathrm{1}−\frac{{q}\left({x}−{p}\right)}{{rx}}\right]={p}\left({x}−{q}\right) \\ $$$${rx}−{q}\left({x}−{p}\right)={p}\left({x}−{q}\right) \\ $$$$\left({p}+{q}−{r}\right){x}=\mathrm{2}{pq} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}{pq}}{{p}+{q}−{r}} \\ $$$${similarly} \\ $$$$\Rightarrow{y}=\frac{\mathrm{2}{qr}}{{q}+{r}−{p}} \\ $$$$\Rightarrow{z}=\frac{\mathrm{2}{pr}}{{p}+{r}−{q}} \\ $$

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

please sir. i need the cubic equation formulars you posted sometimes ago.

$$\mathrm{please}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{need}\:\mathrm{the}\:\mathrm{cubic}\:\mathrm{equation}\:\mathrm{formulars}\:\mathrm{you}\:\mathrm{posted}\:\mathrm{sometimes}\:\mathrm{ago}. \\ $$

Commented by mrW1 last updated on 18/Mar/17