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Question Number 9438 by tawakalitu last updated on 08/Dec/16

Solve for x, y and z 2xy = x + y ...... (i) 6xz = 6z − 2x ....... (ii) 3yz = 3y + 4z ......... (iii) That is the correct question sir.

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x},\:\mathrm{y}\:\mathrm{and}\:\mathrm{z} \\ $$$$\mathrm{2xy}\:=\:\mathrm{x}\:+\:\mathrm{y}\:\:\:\:\:\:......\:\left(\mathrm{i}\right) \\ $$$$\mathrm{6xz}\:=\:\mathrm{6z}\:−\:\mathrm{2x}\:\:\:\:.......\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{3yz}\:=\:\mathrm{3y}\:+\:\mathrm{4z}\:\:\:\:.........\:\left(\mathrm{iii}\right) \\ $$$$ \\ $$$$\mathrm{That}\:\mathrm{is}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{question}\:\mathrm{sir}. \\ $$

Answered by mrW last updated on 08/Dec/16

(ii)⇒6xz=4x if x≠0 ⇒z=(2/3) (iii)⇒3y(z−1)=4z y=((4z)/(3(z−1)))=((4×2)/(3×((2/3)−1)×3))=(8/(−3))=−(8/3) (i)⇒(2y−1)x=y ⇒x=(y/(2y−1))=−(8/(3×(−2×(8/3)−1)))=(8/(19)) if x=0 (i)⇒y=0 (iii)⇒z=0 (x,y,z)=(0,0,0) or ((8/(19)),−(8/3),(3/2))

$$\left(\mathrm{ii}\right)\Rightarrow\mathrm{6xz}=\mathrm{4x} \\ $$$$\mathrm{if}\:\mathrm{x}\neq\mathrm{0} \\ $$$$\Rightarrow\mathrm{z}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\mathrm{iii}\right)\Rightarrow\mathrm{3y}\left(\mathrm{z}−\mathrm{1}\right)=\mathrm{4z} \\ $$$$\mathrm{y}=\frac{\mathrm{4z}}{\mathrm{3}\left(\mathrm{z}−\mathrm{1}\right)}=\frac{\mathrm{4}×\mathrm{2}}{\mathrm{3}×\left(\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}\right)×\mathrm{3}}=\frac{\mathrm{8}}{−\mathrm{3}}=−\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\left(\mathrm{i}\right)\Rightarrow\left(\mathrm{2y}−\mathrm{1}\right)\mathrm{x}=\mathrm{y} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{y}}{\mathrm{2y}−\mathrm{1}}=−\frac{\mathrm{8}}{\mathrm{3}×\left(−\mathrm{2}×\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{1}\right)}=\frac{\mathrm{8}}{\mathrm{19}} \\ $$$$\mathrm{if}\:\mathrm{x}=\mathrm{0} \\ $$$$\left(\mathrm{i}\right)\Rightarrow\mathrm{y}=\mathrm{0} \\ $$$$\left(\mathrm{iii}\right)\Rightarrow\mathrm{z}=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:\mathrm{or}\:\left(\frac{\mathrm{8}}{\mathrm{19}},−\frac{\mathrm{8}}{\mathrm{3}},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$

Commented by tawakalitu last updated on 08/Dec/16

Thank you sir. God bless you. i have edit the equation (ii) sir.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}.\:\mathrm{i}\:\mathrm{have}\:\mathrm{edit}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\left(\mathrm{ii}\right)\:\mathrm{sir}. \\ $$

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