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Question Number 46785 by Tawa1 last updated on 31/Oct/18

Solve the system:           x + y + z = 30         ..... equation (i)          (x/3) + (y/2) + 2z  =  30     ...... equation (ii)

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{30}\:\:\:\:\:\:\:\:\:.....\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{x}}{\mathrm{3}}\:+\:\frac{\mathrm{y}}{\mathrm{2}}\:+\:\mathrm{2z}\:\:=\:\:\mathrm{30}\:\:\:\:\:......\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Answered by MJS last updated on 31/Oct/18

these are 2 planes in R^3  ⇒ we get a line  (1) ⇒ z=−x−y+30  (2) ⇒ z=−(1/6)x−(1/4)y+15  −x−y+30=−(1/6)x−(1/4)y+15  y=−((10)/9)x+20 ⇒ z=(1/9)x+10  put x=9t  l:  ((x),(y),(z) ) = ((0),((20)),((10)) ) +t× ((9),((−10)),(1) )

$$\mathrm{these}\:\mathrm{are}\:\mathrm{2}\:\mathrm{planes}\:\mathrm{in}\:\mathbb{R}^{\mathrm{3}} \:\Rightarrow\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{line} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\:{z}=−{x}−{y}+\mathrm{30} \\ $$$$\left(\mathrm{2}\right)\:\Rightarrow\:{z}=−\frac{\mathrm{1}}{\mathrm{6}}{x}−\frac{\mathrm{1}}{\mathrm{4}}{y}+\mathrm{15} \\ $$$$−{x}−{y}+\mathrm{30}=−\frac{\mathrm{1}}{\mathrm{6}}{x}−\frac{\mathrm{1}}{\mathrm{4}}{y}+\mathrm{15} \\ $$$${y}=−\frac{\mathrm{10}}{\mathrm{9}}{x}+\mathrm{20}\:\Rightarrow\:{z}=\frac{\mathrm{1}}{\mathrm{9}}{x}+\mathrm{10} \\ $$$$\mathrm{put}\:{x}=\mathrm{9}{t} \\ $$$${l}:\:\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{20}}\\{\mathrm{10}}\end{pmatrix}\:+{t}×\begin{pmatrix}{\mathrm{9}}\\{−\mathrm{10}}\\{\mathrm{1}}\end{pmatrix} \\ $$

Commented by Tawa1 last updated on 31/Oct/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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