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1-x-1-2-y-2-3-z-3-x-2y-3z-56-please-help-me-to-solve-it-in-R-3-




Question Number 73766 by mathocean1 last updated on 15/Nov/19
 { (((1/(x−1))=(2/(y−2))=(3/(z−3)))),((x+2y+3z=56)) :}    please help me to solve it in R^3
$$\begin{cases}{\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{y}−\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{z}−\mathrm{3}}}\\{\mathrm{x}+\mathrm{2y}+\mathrm{3z}=\mathrm{56}}\end{cases} \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\mathbb{R}^{\mathrm{3}} \\ $$
Answered by MJS last updated on 15/Nov/19
  (1/(x−1))=(2/(y−2)) ⇒ 2x−2=y−2 ⇒ y=2x  (2/(y−2))=(3/(z−3)) ⇒ 3y−6=2z−6 ⇒ z=((3y)/2)=3x    x+2×2x+3×3x=56  14x=56  x=4  y=8  z=12
$$ \\ $$$$\frac{\mathrm{1}}{{x}−\mathrm{1}}=\frac{\mathrm{2}}{{y}−\mathrm{2}}\:\Rightarrow\:\mathrm{2}{x}−\mathrm{2}={y}−\mathrm{2}\:\Rightarrow\:{y}=\mathrm{2}{x} \\ $$$$\frac{\mathrm{2}}{{y}−\mathrm{2}}=\frac{\mathrm{3}}{{z}−\mathrm{3}}\:\Rightarrow\:\mathrm{3}{y}−\mathrm{6}=\mathrm{2}{z}−\mathrm{6}\:\Rightarrow\:{z}=\frac{\mathrm{3}{y}}{\mathrm{2}}=\mathrm{3}{x} \\ $$$$ \\ $$$${x}+\mathrm{2}×\mathrm{2}{x}+\mathrm{3}×\mathrm{3}{x}=\mathrm{56} \\ $$$$\mathrm{14}{x}=\mathrm{56} \\ $$$${x}=\mathrm{4} \\ $$$${y}=\mathrm{8} \\ $$$${z}=\mathrm{12} \\ $$
Answered by ajfour last updated on 15/Nov/19
x+4x+9x=56  x=4 , y=8, z=12 .
$${x}+\mathrm{4}{x}+\mathrm{9}{x}=\mathrm{56} \\ $$$${x}=\mathrm{4}\:,\:{y}=\mathrm{8},\:{z}=\mathrm{12}\:. \\ $$
Answered by mr W last updated on 15/Nov/19
let (1/(x−1))=(2/(y−2))=(3/(z−3))=(1/k)  ⇒x=k+1  ⇒y=2(k+1)  ⇒z=3(k+1)    x+2y+3z=56  ⇒(k+1)(1+2×2+3×3)=56  ⇒k+1=4    ⇒x=4, y=8, z=12
$${let}\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{y}−\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{z}−\mathrm{3}}=\frac{\mathrm{1}}{{k}} \\ $$$$\Rightarrow{x}={k}+\mathrm{1} \\ $$$$\Rightarrow{y}=\mathrm{2}\left({k}+\mathrm{1}\right) \\ $$$$\Rightarrow{z}=\mathrm{3}\left({k}+\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{x}+\mathrm{2y}+\mathrm{3z}=\mathrm{56} \\ $$$$\Rightarrow\left({k}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{2}×\mathrm{2}+\mathrm{3}×\mathrm{3}\right)=\mathrm{56} \\ $$$$\Rightarrow{k}+\mathrm{1}=\mathrm{4} \\ $$$$ \\ $$$$\Rightarrow{x}=\mathrm{4},\:{y}=\mathrm{8},\:{z}=\mathrm{12} \\ $$

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