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Question Number 73766 by mathocean1 last updated on 15/Nov/19

$$\{\begin{array}{l}\frac{1}{\mathrm{x}-1}=\frac{2}{\mathrm{y}-2}=\frac{3}{\mathrm{z}-3}\\ \mathrm{x}+2\mathrm{y}+3\mathrm{z}=56\end{array}$$$$$$$$\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{to}\mathrm{solve}\mathrm{it}\mathrm{in}{\mathbb{R}}^3$$

Answered by MJS last updated on 15/Nov/19

$$$$$$\frac{1}{x-1}=\frac{2}{y-2}\Rightarrow 2x-2=y-2\Rightarrow y=2x$$$$\frac{2}{y-2}=\frac{3}{z-3}\Rightarrow 3y-6=2z-6\Rightarrow z=\frac{3y}{2}=3x$$$$$$$$x+2\times 2x+3\times 3x=56$$$$14x=56$$$$x=4$$$$y=8$$$$z=12$$

Answered by ajfour last updated on 15/Nov/19

$$x+4x+9x=56$$$$x=4,y=8,z=12.$$

Answered by mr W last updated on 15/Nov/19

$$let\frac{1}{\mathrm{x}-1}=\frac{2}{\mathrm{y}-2}=\frac{3}{\mathrm{z}-3}=\frac{1}{k}$$$$\Rightarrow x=k+1$$$$\Rightarrow y=2(k+1)$$$$\Rightarrow z=3(k+1)$$$$$$$$\mathrm{x}+2\mathrm{y}+3\mathrm{z}=56$$$$\Rightarrow (k+1)(1+2\times 2+3\times 3)=56$$$$\Rightarrow k+1=4$$$$$$$$\Rightarrow x=4,y=8,z=12$$

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