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Question Number 71282 by mr W last updated on 13/Oct/19

$$solve$$$$x(y+z)=27$$$$y(z+x)=32$$$$z(x+y)=35$$

Answered by ajfour last updated on 13/Oct/19

$$letxy=p,yz=q,zx=r$$$$p+r=27$$$$q+p=32$$$$r+q=35$$$$\Rightarrow p+q+r=47$$$$\Rightarrow p=12,q=20,r=15$$$$xyz=\sqrt{pqr}=60$$$$x=\frac{xyz}{q}=\frac{60}{20}=3$$$$y=\frac{xyz}{r}=\frac{60}{15}=4$$$$z=\frac{xyz}{p}=\frac{60}{12}=5.$$

Commented by MJS last updated on 13/Oct/19

$$xyz=\pm \sqrt{pqr}$$

Commented by ajfour last updated on 13/Oct/19

$$thankssir,{wasn}^{\prime }tcareful!$$

Commented by Rasheed.Sindhi last updated on 13/Oct/19

$$\vee \cap i\subset \in sir!$$

Answered by MJS last updated on 13/Oct/19

$$\mathrm{if}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{an}‘‘{\mathrm{easy}}^{″}\mathrm{solution}$$$$x\mid 27\wedge y\mid 32\wedge z\mid 35$$$$\mathrm{trying}\mathrm{I}\mathrm{found}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\pm \left(\begin{array}{c}3\\ 4\\ 5\end{array}\right)$$

Answered by behi83417@gmail.com last updated on 13/Oct/19

$$\mathrm{xy}+\mathrm{xz}+\mathrm{yz}=\frac{27+32+35}{2}=47$$$$\Rightarrow \{\begin{array}{l}\mathrm{yz}=20\\ \mathrm{zx}=15\\ \mathrm{xy}=12\end{array}\Rightarrow {\left(\mathrm{xyz}\right)}^2=3600$$$$\Rightarrow \mathrm{xyz}=\pm 60\Rightarrow \{\begin{array}{l}\mathrm{x}=\frac{\pm 60}{-20}=\pm 3\\ \mathrm{y}=\frac{\pm 60}{-15}=\pm 4\\ \mathrm{z}=\frac{\pm 60}{-12}=\pm 5\end{array}$$

Commented by Rasheed.Sindhi last updated on 13/Oct/19

$${\mathbf{e}}^{\mathbf{x}}\mathbf{cellent}\mathbf{sir}!$$

Answered by MJS last updated on 13/Oct/19

$$y=p-q\wedge z=p+q$$$$\left(1\right)2px=27$$$$\left(2\right)(p-q)(x+p+q)=32$$$$\left(3\right)(p+q)(x+p-q)=35$$$$$$$$\left(1\right)x=\frac{27}{2p}$$$$\left(2\right)2p^3-2{pq}^2-37p-27q=0$$$$\left(3\right)2p^3-2{pq}^2-43p+27q=0$$$$$$$$\left(2\right)-\left(3\right)6p-54q=0\Rightarrow p=9q$$$$$$$$\left(2\right)=\left(3\right)1440q^3-360q=0$$$$\Rightarrow q=\pm \frac{1}{2}[q=0\mathrm{not}\mathrm{valid}]$$$$\Rightarrow p=\pm \frac{9}{2}$$$$\Rightarrow x=\pm 3;y=\pm 4;z=\pm 5$$$$$$$$========================$$$$$$$$\mathrm{or}$$$$y=px\wedge z=qx$$$$\left(1\right)(p+q)x^2=27$$$$\left(2\right)p(q+1)x^2=32$$$$\left(3\right)(p+1){qx}^2=35$$$$$$$$\left(1\right)x^2=\frac{27}{p+q}$$$$\left(2\right)\Rightarrow 27pq-5p-32q=0$$$$\left(3\right)\Rightarrow 27pa-35p-8q=0$$$$$$$$\left(2\right)-\left(3\right)30p-24q=0\Rightarrow p=\frac{4}{5}q$$$$$$$$\left(2\right)=\left(3\right)\frac{108}{5}{q}^2-36q=0$$$$\Rightarrow q=\frac{5}{3}[q=0\mathrm{not}\mathrm{valid}]$$$$\Rightarrow p=\frac{4}{3}$$$$\Rightarrow x^2=9$$$$\Rightarrow x=\pm 3;y=\pm 4;z=\pm 5$$

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