Question Number 11268 by tawa last updated on 18/Mar/17
![Solve x, y and z in terms of p, q and r yz = py + qz ........ (i) zx = qz + rx ....... (ii) xy = rx + py ....... (iii)](https://www.tinkutara.com/question/Q11268.png)
$$\mathrm{Solve}\:\mathrm{x},\:\mathrm{y}\:\mathrm{and}\:\mathrm{z}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{p},\:\mathrm{q}\:\mathrm{and}\:\mathrm{r} \\ $$$$\mathrm{yz}\:=\:\mathrm{py}\:+\:\mathrm{qz}\:\:\:\:\:……..\:\left(\mathrm{i}\right) \\ $$$$\mathrm{zx}\:=\:\mathrm{qz}\:+\:\mathrm{rx}\:\:\:\:\:\:…….\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{xy}\:=\:\mathrm{rx}\:+\:\mathrm{py}\:\:\:\:\:…….\:\left(\mathrm{iii}\right) \\ $$
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))](https://www.tinkutara.com/question/Q11269.png)
$$\left({iii}\right)\Rightarrow{y}=\frac{{rx}}{{x}−{p}} \\ $$$$\left({i}\right)\Rightarrow{z}=\frac{{py}}{{y}−{q}}=\frac{{p}}{\mathrm{1}−\frac{{q}}{{y}}}=\frac{{p}}{\mathrm{1}−\frac{{q}\left({x}−{p}\right)}{{rx}}} \\ $$$$\left({ii}\right)\Rightarrow{z}=\frac{{rx}}{{x}−{q}} \\ $$$$\Rightarrow\frac{{rx}}{{x}−{q}}=\frac{{p}}{\mathrm{1}−\frac{{q}\left({x}−{p}\right)}{{rx}}} \\ $$$${rx}\left[\mathrm{1}−\frac{{q}\left({x}−{p}\right)}{{rx}}\right]={p}\left({x}−{q}\right) \\ $$$${rx}−{q}\left({x}−{p}\right)={p}\left({x}−{q}\right) \\ $$$$\left({p}+{q}−{r}\right){x}=\mathrm{2}{pq} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}{pq}}{{p}+{q}−{r}} \\ $$$${similarly} \\ $$$$\Rightarrow{y}=\frac{\mathrm{2}{qr}}{{q}+{r}−{p}} \\ $$$$\Rightarrow{z}=\frac{\mathrm{2}{pr}}{{p}+{r}−{q}} \\ $$
Commented by tawa last updated on 18/Mar/17
![God bless you sir. i really appreciate.](https://www.tinkutara.com/question/Q11270.png)
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Commented by tawa last updated on 18/Mar/17
![please sir. i need the cubic equation formulars you posted sometimes ago.](https://www.tinkutara.com/question/Q11271.png)
$$\mathrm{please}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{need}\:\mathrm{the}\:\mathrm{cubic}\:\mathrm{equation}\:\mathrm{formulars}\:\mathrm{you}\:\mathrm{posted}\:\mathrm{sometimes}\:\mathrm{ago}. \\ $$
Commented by mrW1 last updated on 18/Mar/17
![](https://www.tinkutara.com/question/928.png)
Commented by tawa last updated on 18/Mar/17
![God bless you sir.](https://www.tinkutara.com/question/Q11276.png)
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
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=0](https://www.tinkutara.com/question/Q11290.png)
$$ \\ $$$${the}\:{system}\:{of}\:{this}\:{equations}\:{have}\:{one} \\ $$$${more}\:{answers}:\:\:{x}={y}={z}=\mathrm{0} \\ $$
Commented by mrW1 last updated on 19/Mar/17
![you are right. this case should be considered in the step 2.](https://www.tinkutara.com/question/Q11292.png)
$${you}\:{are}\:{right}.\:{this}\:{case}\:{should}\:{be}\:{considered} \\ $$$${in}\:{the}\:{step}\:\mathrm{2}. \\ $$