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Question Number 177037 by mr W last updated on 30/Sep/22

Answered by Rasheed.Sindhi last updated on 30/Sep/22

$$\{\begin{array}{l}\frac{x+y+z}{x(y+z)}=\frac{1}{2}\Rightarrow x+y+z=\frac{x(y+z)}{2}\\ \frac{x+y+z}{y(z+x)}=\frac{1}{3}\Rightarrow x+y+z=\frac{y(z+x)}{3}\\ \frac{x+y+z}{z(x+y)}=\frac{1}{4}\Rightarrow x+y+z=\frac{z(x+y)}{4}\end{array}$$$$\{\begin{array}{l}\frac{x(y+z)}{2}=\frac{y(z+x)}{3}\Rightarrow 3xy+3zx=2yz+2xy\\ \frac{y(z+x)}{3}=\frac{z(x+y)}{4}\Rightarrow 4yz+4xy=3zx+3yz\\ \frac{z(x+y)}{4}=\frac{x(y+z)}{2}\Rightarrow 4xy+4zx=2zx+2yz\end{array}$$$$\Rightarrow \{\begin{array}{l}3xy+3zx=2yz+2xy\\ 4yz+4xy=3zx+3yz\\ 4xy+4zx=2zx+2yz\end{array}$$$$\Rightarrow \{\begin{array}{l}xy-2yz+3zx=0\\ 4xy+yz-3zx=0\\ 4xy-2yz+2zx=0\end{array}$$$$Letxy=a,yz=b,zx=c$$$$\Rightarrow \{\begin{array}{l}a-2b+3c=0...\left(i\right)\\ 4a+b-3c=0...\left(ii\right)\\ 4a-2b+2c=0...\left(iii\right)\end{array}$$$$\left(i\right)+\left(ii\right):5a-b=0\Rightarrow b=5a$$$$\left(ii\right)-\left(iii\right):3b-5c=0\Rightarrow 3\left(5a\right)=5c\Rightarrow c=3a$$$$5a=b\Rightarrow 5xy=yz\Rightarrow z=5x;y\ne 0$$$$3a=c\Rightarrow 3xy=zx\Rightarrow z=3y;x\ne 0$$$$5x=3y\Rightarrow y=\frac{5}{3}x,z=5x$$$$putthesevaluesofy\mathrm{\&}zinfirst$$$$givenequationtoget:$$$$x=\frac{23}{10},y=\frac{23}{6},z=\frac{23}{2}$$

Commented by mr W last updated on 30/Sep/22

$$thanksalotsir!$$

Commented by Rasheed.Sindhi last updated on 30/Sep/22

$$\mathit{S}\mathit{i}\mathit{r}\mathit{m}\mathit{r}\mathit{W},whatshouldweunderstand$$$$bythefollowingsituation?$$$$\Rightarrow \{\begin{array}{l}a-2b+3c=0\\ 4a+b-3c=0\\ 4a-2b+2c=0\end{array}$$$$\left|\begin{array}{ccc}1& -2& 3\\ 4& 1& -3\\ 4& -2& 2\end{array}\right|=0$$$$\mathcal{T}hematrixofcoefficientsissingular$$$$\mathcal{D}{oesn}^{\prime }tthismeanthattheequation$$$$hasnouniquesolution?Butwecan$$$$seethatabovehasuniquesolution!$$

Commented by Frix last updated on 30/Sep/22

$$a,b,c\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{variables}$$

Commented by mr W last updated on 30/Sep/22

$$\left|\begin{array}{ccc}1& -2& 3\\ 4& 1& -3\\ 4& -2& 2\end{array}\right|=0$$$$meansonlythatthereisnounique$$$$solutionfora,b,c.butthis{doesn}^{\prime }t$$$$meanthatthereisalsonounique$$$$solutionforx,y,z.$$

Commented by Rasheed.Sindhi last updated on 30/Sep/22

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathrm{you}\mathbf{sirs}!$$

Commented by Rasheed.Sindhi last updated on 30/Sep/22

$$\mathcal{S}\mathit{i}\mathit{r},doesa=xy=\frac{{23}^2}{60},b=yz=\frac{{23}^2}{12},$$$$c=zx=\frac{{23}^2}{20}isnotauniquesolutionof$$$$\{\begin{array}{l}a-2b+3c=0...\left(i\right)\\ 4a+b-3c=0...\left(ii\right)\\ 4a-2b+2c=0...\left(iii\right)\end{array}?$$$$$$

Commented by Frix last updated on 30/Sep/22

$$\mathrm{this}\mathrm{solution}\mathrm{is}\mathrm{not}\mathrm{unique}\mathrm{for}a,b,c\mathrm{but}$$$$\mathrm{only}\mathrm{if}a=xy\wedge b=yz\wedge c=zx.\mathrm{if}\mathrm{we}\mathrm{let}$$$$a=3xy\wedge b=4yz\wedge c=5zx\mathrm{we}\mathrm{get}\mathrm{a}\mathrm{different}$$$$\mathrm{solution}\left(\mathrm{if}\mathrm{we}\mathrm{get}\mathrm{one}\mathrm{at}\mathrm{all}\right)$$

Commented by Tawa11 last updated on 02/Oct/22

$$\mathrm{Great}\mathrm{sirs}$$

Answered by mr W last updated on 30/Sep/22

$$lety=px,z=qx$$$$\frac{p+q+1}{p+q}=\frac{x}{2}$$$$\frac{p+q+1}{p(q+1)}=\frac{x}{3}$$$$\frac{p+q+1}{q(p+1)}=\frac{x}{4}$$$$$$$$\frac{p(q+1)}{p+q}=\frac{3}{2}\Rightarrow pq+p=\frac{3(p+q)}{2}$$$$\frac{q(p+1)}{p+q}=\frac{4}{2}\Rightarrow pq+q=2(p+q)$$$$\Rightarrow pq=\frac{5(p+q)}{4}$$$$p=\frac{3(p+q)}{2}-\frac{5(p+q)}{4}=\frac{p+q}{4}\Rightarrow 3p=q$$$$\frac{p(3p+1)}{p+3p}=\frac{3}{2}\Rightarrow p=\frac{5}{3}\Rightarrow q=5$$$$1+\frac{1}{\frac{5}{3}+5}=\frac{x}{2}$$$$\Rightarrow x=\frac{23}{10}$$$$\Rightarrow y=\frac{23}{10}\times \frac{5}{3}=\frac{23}{6}$$$$\Rightarrow z=\frac{23}{10}\times 5=\frac{23}{2}$$

Answered by mr W last updated on 30/Sep/22

$$anotherway:$$$$\frac{x+y+z}{xy+zx}=\frac{1}{2}$$$$\Rightarrow xy+zx=2(x+y+z)...\left(i\right)$$$$similarly$$$$\Rightarrow yz+xy=3(x+y+z)...\left(ii\right)$$$$\Rightarrow zx+yz=4(x+y+z)...\left(iii\right)$$$$[\left(i\right)+\left(ii\right)+\left(iii\right)]/2:$$$$xy+yz+zx=\frac{9(x+y+z)}{2}...\left(iv\right)$$$$\left(iv\right)-\left(i\right):$$$$yz=\frac{5(x+y+z)}{2}=5\lambda with\lambda =\frac{x+y+z}{2}$$$$similarly$$$$zx=\frac{3(x+y+z)}{2}=3\lambda$$$$xy=\frac{x+y+z}{2}=\lambda$$$${\left(xyz\right)}^2=15{\lambda }^3$$$$\Rightarrow xyz=\sqrt{15{\lambda }^3}$$$$\Rightarrow x=\frac{\sqrt{15\lambda }}{5}$$$$\Rightarrow y=\frac{\sqrt{15\lambda }}{3}$$$$\Rightarrow z=\frac{\sqrt{15\lambda }}{1}$$$$x+y+z=2\lambda =(\frac{1}{5}+\frac{1}{3}+1)\sqrt{15\lambda }$$$$\Rightarrow \sqrt{15\lambda }=\frac{23}{2}$$$$\Rightarrow x=\frac{23}{2\times 5}=\frac{23}{10}$$$$\Rightarrow y=\frac{23}{2\times 3}=\frac{23}{6}$$$$\Rightarrow z=\frac{23}{2}$$

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