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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

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

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

$$$$$$thesystemofthisequationshaveone$$$$moreanswers:x=y=z=0$$

Commented by mrW1 last updated on 19/Mar/17

$$youareright.thiscaseshouldbeconsidered$$$$inthestep2.$$

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