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Question Number 9449 by tawakalitu last updated on 08/Dec/16

2xy = x + y ...... (i) 6xz = 6z − 2x ..... (ii) 3yz = 3y + 4z ....... (iii) Solve for x, y and z

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

Answered by mrW last updated on 09/Dec/16

(i)⇒(2x−1)y=x y=(x/(2x−1)) ....(iv) (ii)⇒3(x−1)z=−x z=((−x)/(3(x−1))) ....(v) (iii)⇒−((3x^2 )/(3(x−1)(2x−1)))=((3x)/(2x−1))−((4x)/(3(x−1))) −3x^2 =9x(x−1)−4x(2x−1) −3x^2 =9x^2 −9x−8x^2 +4x 4x^2 −5x=0 x(x−(5/4))=0 x_1 =0 x_2 =(5/4) y_1 =(0/(2×0−1))=0 y_2 =(5/(4×(2×(5/4)−1)))=(5/6) z_1 =−(0/(3×(0−1)))=0 z_2 =−(5/(4×3×((5/4)−1)))=−(5/3) (x,y,z)=(0,0,0) or ((5/4),(5/6),−(5/3))

$$\left(\mathrm{i}\right)\Rightarrow\left(\mathrm{2x}−\mathrm{1}\right)\mathrm{y}=\mathrm{x} \\ $$$$\mathrm{y}=\frac{\mathrm{x}}{\mathrm{2x}−\mathrm{1}}\:\:\:....\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{ii}\right)\Rightarrow\mathrm{3}\left(\mathrm{x}−\mathrm{1}\right)\mathrm{z}=−\mathrm{x} \\ $$$$\mathrm{z}=\frac{−\mathrm{x}}{\mathrm{3}\left(\mathrm{x}−\mathrm{1}\right)}\:\:\:\:....\left(\mathrm{v}\right) \\ $$$$\left(\mathrm{iii}\right)\Rightarrow−\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{3}\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{2x}−\mathrm{1}\right)}=\frac{\mathrm{3x}}{\mathrm{2x}−\mathrm{1}}−\frac{\mathrm{4x}}{\mathrm{3}\left(\mathrm{x}−\mathrm{1}\right)} \\ $$$$−\mathrm{3x}^{\mathrm{2}} =\mathrm{9x}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{4x}\left(\mathrm{2x}−\mathrm{1}\right) \\ $$$$−\mathrm{3x}^{\mathrm{2}} =\mathrm{9x}^{\mathrm{2}} −\mathrm{9x}−\mathrm{8x}^{\mathrm{2}} +\mathrm{4x} \\ $$$$\mathrm{4x}^{\mathrm{2}} −\mathrm{5x}=\mathrm{0} \\ $$$$\mathrm{x}\left(\mathrm{x}−\frac{\mathrm{5}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\mathrm{x}_{\mathrm{1}} =\mathrm{0} \\ $$$$\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{y}_{\mathrm{1}} =\frac{\mathrm{0}}{\mathrm{2}×\mathrm{0}−\mathrm{1}}=\mathrm{0} \\ $$$$\mathrm{y}_{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}×\left(\mathrm{2}×\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{1}\right)}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\mathrm{z}_{\mathrm{1}} =−\frac{\mathrm{0}}{\mathrm{3}×\left(\mathrm{0}−\mathrm{1}\right)}=\mathrm{0} \\ $$$$\mathrm{z}_{\mathrm{2}} =−\frac{\mathrm{5}}{\mathrm{4}×\mathrm{3}×\left(\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{1}\right)}=−\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:\mathrm{or}\:\left(\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{5}}{\mathrm{6}},−\frac{\mathrm{5}}{\mathrm{3}}\right) \\ $$

Commented by tawakalitu last updated on 08/Dec/16

i really appreciate sir. God bless you.

$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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