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2xy-x-y-i-6xz-6z-2x-ii-3yz-3y-4z-iii-Solve-for-x-y-and-z-




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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