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Question Number 31286 by 6123 last updated on 05/Mar/18
Find all set of ordered triple/s (x,y,z), x,y,z∈ℜ, such that x−y=1−z 3(x^2 −y^2 )=5(1−z^2 ) 7(x^3 −y^3 )=19(1−z^3 ). Please show your solution.
$${Find}\:{all}\:{set}\:{of}\:{ordered}\:{triple}/{s}\:\left({x},{y},{z}\right),\:\:{x},{y},{z}\in\Re,\:{such}\:{that} \\ $$$${x}−{y}=\mathrm{1}−{z} \\ $$$$\mathrm{3}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{5}\left(\mathrm{1}−{z}^{\mathrm{2}} \right) \\ $$$$\mathrm{7}\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right)=\mathrm{19}\left(\mathrm{1}−{z}^{\mathrm{3}} \right). \\ $$$${Please}\:{show}\:{your}\:{solution}. \\ $$
Commented by 6123 last updated on 06/Mar/18
Thank you!!
$${Thank}\:{you}!! \\ $$
Answered by MJS last updated on 05/Mar/18
obviously all triples (x,y,1) with x=y∈R solve the given equations 3(x^2 −y^2 )=5(1−z)(1+z) from (i): (1−z)=(x−y) 3(x−y)(x+y)=5(x−y)(1+z) 3(x+y)=5(1+z) x=(5/3)(1+z)−y into (i): (5/3)(1+z)−2y=1−z y=(1/3)+(4/3)z x=(4/3)+(1/3)z (from above) into (iii): 7(((4/3)+(1/3)z)^3 −((1/3)+(4/3)z)^3 )=19−19z^3 z^3 −(7/2)z^2 +(7/2)z−1=0 (z−1)(z^2 −(5/2)z+1)=0 z_1 =1; z_2 =(1/2); z_3 =2 x_1 =(5/3); x_2 =(3/2); x_3 =2 y_1 =(5/3); y_2 =1; y_3 =3 so we have (r;r;1) with r∈R ((3/2);1;(1/2)) (2;3;2)
$$\mathrm{obviously}\:\mathrm{all}\:\mathrm{triples} \\ $$$$\left({x},{y},\mathrm{1}\right)\:\mathrm{with}\:{x}={y}\in\mathbb{R}\:\mathrm{solve}\:\mathrm{the} \\ $$$$\mathrm{given}\:\mathrm{equations} \\ $$$$ \\ $$$$\mathrm{3}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{5}\left(\mathrm{1}−{z}\right)\left(\mathrm{1}+{z}\right) \\ $$$$\mathrm{from}\:\left({i}\right):\:\left(\mathrm{1}−{z}\right)=\left({x}−{y}\right) \\ $$$$\mathrm{3}\left({x}−{y}\right)\left({x}+{y}\right)=\mathrm{5}\left({x}−{y}\right)\left(\mathrm{1}+{z}\right) \\ $$$$\mathrm{3}\left({x}+{y}\right)=\mathrm{5}\left(\mathrm{1}+{z}\right) \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{3}}\left(\mathrm{1}+{z}\right)−{y} \\ $$$$\mathrm{into}\:\left({i}\right):\:\frac{\mathrm{5}}{\mathrm{3}}\left(\mathrm{1}+{z}\right)−\mathrm{2}{y}=\mathrm{1}−{z} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{3}}{z} \\ $$$${x}=\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}{z}\:\left(\mathrm{from}\:\mathrm{above}\right) \\ $$$$\mathrm{into}\:\left({iii}\right): \\ $$$$\mathrm{7}\left(\left(\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}{z}\right)^{\mathrm{3}} −\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{3}}{z}\right)^{\mathrm{3}} \right)=\mathrm{19}−\mathrm{19}{z}^{\mathrm{3}} \\ $$$${z}^{\mathrm{3}} −\frac{\mathrm{7}}{\mathrm{2}}{z}^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{2}}{z}−\mathrm{1}=\mathrm{0} \\ $$$$\left({z}−\mathrm{1}\right)\left({z}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{2}}{z}+\mathrm{1}\right)=\mathrm{0} \\ $$$${z}_{\mathrm{1}} =\mathrm{1};\:{z}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}};\:{z}_{\mathrm{3}} =\mathrm{2} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{5}}{\mathrm{3}};\:{x}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}};\:{x}_{\mathrm{3}} =\mathrm{2} \\ $$$${y}_{\mathrm{1}} =\frac{\mathrm{5}}{\mathrm{3}};\:{y}_{\mathrm{2}} =\mathrm{1};\:{y}_{\mathrm{3}} =\mathrm{3} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\left({r};{r};\mathrm{1}\right)\:\mathrm{with}\:{r}\in\mathbb{R} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}};\mathrm{1};\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2};\mathrm{3};\mathrm{2}\right) \\ $$
Commented by 6123 last updated on 06/Mar/18
Thank you!!!
$${Thank}\:{you}!!! \\ $$

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