Question Number 90842 by jagoll last updated on 26/Apr/20
$${x}^{\mathrm{2}} −\left({y}−{z}\right)^{\mathrm{2}} \:=\:\mathrm{3} \\ $$$${y}^{\mathrm{2}} \:−\:\left({z}−{x}\right)^{\mathrm{2}} \:=\:\mathrm{5} \\ $$$${z}^{\mathrm{2}} \:−\:\left({x}−{y}\right)^{\mathrm{2}} \:=\:\mathrm{12} \\ $$
Commented by john santu last updated on 26/Apr/20
$$\left(\mathrm{1}\right)\:\left({x}+{y}−{z}\right)\left({x}−{y}+{z}\right)=\mathrm{3} \\ $$$$\left(\mathrm{2}\right)\:\left({y}+{z}−{x}\right)\left({y}−{z}+{x}\right)\:=\:\mathrm{5} \\ $$$$\left(\mathrm{3}\right)\:\left({z}+{x}−{y}\right)\left({z}−{x}+{y}\right)\:=\:\mathrm{12} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:−\left\{{y}^{\mathrm{2}} −\mathrm{2}{yz}+{z}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{xz}+{x}^{\mathrm{2}} \right. \\ $$$$\left.+{x}^{\mathrm{2}} −\mathrm{2}{xy}+{y}^{\mathrm{2}} \right\}\:=\:\mathrm{20} \\ $$$$−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+\mathrm{2}{xy}+\mathrm{2}{yz}+\mathrm{2}{xz}\:=\:\mathrm{20} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\mathrm{2}\left({xy}+{xz}+{yz}\right)\:=\:−\mathrm{20} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} −\mathrm{4}\left({xy}+{xz}+{yz}\right)\:=\:−\mathrm{20} \\ $$$$ \\ $$
Commented by jagoll last updated on 26/Apr/20
$${by}\:{observe}\:{i}\:{got}\: \\ $$$${x}\:=\:\mathrm{2}\:,\:{y}\:=\:\mathrm{3}\:,\:{z}\:=\:\mathrm{4}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{4}−\left(\mathrm{3}−\mathrm{4}\right)^{\mathrm{2}} \:=\:\mathrm{3}\: \\ $$$$\left(\mathrm{2}\right)\:\mathrm{9}−\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{5}\: \\ $$$$\left(\mathrm{3}\right)\:\mathrm{16}−\left(\mathrm{2}−\mathrm{3}\right)^{\mathrm{2}} \:\neq\:\mathrm{12} \\ $$$${i}\:{think}\:{it}\:{wrong}\: \\ $$
Commented by Prithwish Sen 1 last updated on 26/Apr/20
$$\mathrm{after}\:\mathrm{mr}.\:\mathrm{john}\:\mathrm{santu}\:\mathrm{i}\:\mathrm{think} \\ $$$$\frac{\left(\mathrm{1}\right)}{\left(\mathrm{2}\right)}\:\:\mathrm{we}\:\mathrm{get}\:\Rightarrow\:\frac{\mathrm{x}−\mathrm{y}+\mathrm{z}}{\mathrm{y}+\mathrm{z}−\mathrm{x}}\:\:=\:\frac{\mathrm{3}}{\mathrm{5}}\:….\left(\mathrm{4}\right) \\ $$$$\frac{\left(\mathrm{2}\right)}{\left(\mathrm{3}\right)}\:\Rightarrow\:\frac{\mathrm{x}+\mathrm{y}−\mathrm{z}}{\mathrm{x}−\mathrm{y}+\mathrm{z}}\:=\:\frac{\mathrm{5}}{\mathrm{12}}\:……\left(\mathrm{5}\right) \\ $$$$\frac{\left(\mathrm{1}\right)}{\left(\mathrm{3}\right)}\:\Rightarrow\:\frac{\mathrm{x}+\mathrm{y}−\mathrm{z}}{\mathrm{y}−\mathrm{x}+\mathrm{z}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}….\left(\mathrm{6}\right) \\ $$$$\mathrm{now}\:\mathrm{by}\:\left(\mathrm{4}\right)×\left(\mathrm{3}\right)\Rightarrow\left(\mathrm{x}−\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{36}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{x}−\mathrm{y}+\mathrm{z}\:=\:\pm\frac{\mathrm{6}}{\:\sqrt{\mathrm{5}}}\:….\left(\mathrm{7}\right) \\ $$$$\mathrm{and}\:\mathrm{by}\:\left(\mathrm{5}\right)×\left(\mathrm{1}\right)\:\Rightarrow\:\left(\mathrm{x}+\mathrm{y}−\mathrm{z}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{x}+\mathrm{y}−\mathrm{z}\:=\:\pm\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}…….\left(\mathrm{8}\right) \\ $$$$\boldsymbol{\mathrm{considering}}\:\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{value}} \\ $$$$\boldsymbol{\mathrm{by}}\:\left(\mathrm{7}\right)\:+\:\left(\mathrm{8}\right)\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{\mathrm{x}}\:=\:\frac{\mathrm{6}}{\:\sqrt{\mathrm{5}}}\:+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\Rightarrow\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{17}}{\mathrm{2}\sqrt{\mathrm{5}}}\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{on}}. \\ $$
Commented by jagoll last updated on 28/Apr/20
$${waw}..{great}\:{sir}.\:{thank}\:{you} \\ $$
Answered by ajfour last updated on 26/Apr/20
$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} −{z}^{\mathrm{2}} =\mathrm{3}−\mathrm{2}{yz}\:\:\:..\left({i}\right) \\ $$$${y}^{\mathrm{2}} −{z}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{5}−\mathrm{2}{xz}\:\:\:\:\:…\left({ii}\right) \\ $$$${z}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{12}−\mathrm{2}{xy}\:\:\:\:\:….\left({iii}\right) \\ $$$${Adding}\:\:{all} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\mathrm{2}\left({xy}+{yz}+{zx}\right)=−\mathrm{20} \\ $$$$\Rightarrow\:\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{z}\left({x}+{y}\right)−\mathrm{4}{xy}+{z}^{\mathrm{2}} =−\mathrm{20} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\left({I}\right) \\ $$$$\left({i}\right)+\left({ii}\right) \\ $$$$\mathrm{2}{z}\left({x}+{y}\right)−\mathrm{2}{z}^{\mathrm{2}} =\mathrm{8}\:\:\: \\ $$$$\Rightarrow\:\:{x}+{y}={z}+\frac{\mathrm{4}}{{z}} \\ $$$${from}\:\left({III}\right)\: \\ $$$$\Rightarrow\:\:\left({x}−{y}\right)^{\mathrm{2}} ={z}^{\mathrm{2}} −\mathrm{12} \\ $$$$\left({i}\right)−\left({ii}\right) \\ $$$$\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)−{z}\left({x}−{y}\right)=−\mathrm{1} \\ $$$$\Rightarrow\:\left({x}−{y}\right)\left({x}+{y}−{z}\right)=−\mathrm{1} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} \left\{\left({x}+{y}\right)^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{z}\left({x}+{y}\right)\right\}=\mathrm{1} \\ $$$$\left({z}^{\mathrm{2}} −\mathrm{12}\right)\left\{\left({z}+\frac{\mathrm{4}}{{z}}\right)^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{z}\left({z}+\frac{\mathrm{4}}{{z}}\right)\right\}=\mathrm{1} \\ $$$$\left({z}^{\mathrm{2}} −\mathrm{12}\right)\left(\frac{\mathrm{16}}{{z}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$$\mathrm{16}{z}^{\mathrm{2}} −\mathrm{192}={z}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{z}^{\mathrm{2}} =\frac{\mathrm{192}}{\mathrm{15}}=\frac{\mathrm{64}}{\mathrm{5}} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\frac{\mathrm{64}}{\mathrm{5}}+\frac{\mathrm{16}×\mathrm{5}}{\mathrm{64}}+\mathrm{8}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{64}}{\mathrm{5}}+\frac{\mathrm{5}}{\mathrm{4}}+\mathrm{8}\:=\:\frac{\mathrm{441}}{\mathrm{20}} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} ={z}^{\mathrm{2}} −\mathrm{12}\:=\frac{\mathrm{64}}{\mathrm{5}}−\mathrm{12}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${let}\:{x}>{y} \\ $$$$\Rightarrow\:\:\:{x}−{y}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{20}}}\:\:,\:\:{x}+{y}=\frac{\mathrm{21}}{\:\sqrt{\mathrm{20}}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{25}}{\mathrm{4}\sqrt{\mathrm{5}}}\:,\:\:{y}=\frac{\mathrm{17}}{\mathrm{4}\sqrt{\mathrm{5}}}\:,\:\:{z}=\frac{\mathrm{32}}{\mathrm{4}\sqrt{\mathrm{5}}}\:. \\ $$$${these}\:{dont}\:{fit}\:{very}\:{well},\:{so}\:{let} \\ $$$$\:\:\:\:\:\:{x}<{y}\:\:\Rightarrow\:\:{y}−{x}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{20}}} \\ $$$$\Rightarrow\:\:{x}=\frac{\mathrm{17}}{\mathrm{4}\sqrt{\mathrm{5}}}\:,\:\:{y}=\frac{\mathrm{25}}{\mathrm{4}\sqrt{\mathrm{5}}}\:,\:\:{z}=\frac{\mathrm{32}}{\mathrm{4}\sqrt{\mathrm{5}}} \\ $$
Commented by ajfour last updated on 26/Apr/20
$${i}\:{checked}\:{them}\:{even},\:{hurray}! \\ $$
Commented by jagoll last updated on 26/Apr/20
$${hHaha}\:{hurray}\:{sir}\: \\ $$
Answered by mr W last updated on 26/Apr/20
$$\left({x}+{y}−{z}\right)\left({x}−{y}+{z}\right)=\mathrm{3} \\ $$$$\left({x}+{y}+{z}−\mathrm{2}{z}\right)\left({x}+{y}+{z}−\mathrm{2}{y}\right)=\mathrm{3}\:\:\:…\left({i}\right) \\ $$$$\left({y}+{z}−{x}\right)\left({y}−{z}+{x}\right)=\mathrm{5} \\ $$$$\left({x}+{y}+{z}−\mathrm{2}{x}\right)\left({x}+{y}+{z}−\mathrm{2}{z}\right)=\mathrm{5}\:\:\:…\left({ii}\right) \\ $$$$\left({z}+{x}−{y}\right)\left({z}−{x}+{y}\right)=\mathrm{12} \\ $$$$\left({x}+{y}+{z}−\mathrm{2}{y}\right)\left({x}+{y}+{z}−\mathrm{2}{x}\right)=\mathrm{12}\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$$\left({i}\right)×\left({ii}\right)×\left({iii}\right): \\ $$$$\left({x}+{y}+{z}−\mathrm{2}{x}\right)^{\mathrm{2}} \left({x}+{y}+{z}−\mathrm{2}{y}\right)^{\mathrm{2}} \left({x}+{y}+{z}−\mathrm{2}{z}\right)^{\mathrm{2}} =\mathrm{3}×\mathrm{5}×\mathrm{12}=\mathrm{180} \\ $$$$\left({x}+{y}+{z}−\mathrm{2}{x}\right)\left({x}+{y}+{z}−\mathrm{2}{y}\right)\left({x}+{y}+{z}−\mathrm{2}{z}\right)=\pm\mathrm{6}\sqrt{\mathrm{5}}\:\:\:…\left({iv}\right) \\ $$$$ \\ $$$$\left({iv}\right)/\left({i}\right)\:{and}\:{similarly}: \\ $$$${x}+{y}+{z}−\mathrm{2}{x}=\pm\frac{\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{3}}\:\:\:…\:{I} \\ $$$${x}+{y}+{z}−\mathrm{2}{y}=\pm\frac{\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{5}}\:\:\:…{II} \\ $$$${x}+{y}+{z}−\mathrm{2}{z}=\pm\frac{\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{12}}\:\:\:…{III} \\ $$$${I}+{II}+{III}: \\ $$$${x}+{y}+{z}=\pm\mathrm{6}\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{12}}\right)\:\:\:…{IV} \\ $$$$ \\ $$$${IV}−{I}\:{and}\:{similarly}: \\ $$$$\mathrm{2}{x}=\pm\mathrm{6}\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{12}}\right)\:\Rightarrow{x}=\pm\frac{\mathrm{17}\sqrt{\mathrm{5}}}{\mathrm{20}} \\ $$$$\mathrm{2}{y}=\pm\mathrm{6}\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{12}}\right)\:\Rightarrow{y}=\pm\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{2}{z}=\pm\mathrm{6}\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}\right)\:\Rightarrow{z}=\pm\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$
Commented by mr W last updated on 26/Apr/20
$${yes}!\:{it}\:{means}\:{all}\:+\:{or}\:{all}\:−. \\ $$$${we}\:{see}\:{if}\:{x}={a},{y}={b},{z}={c}\:{is}\:{a}\:{solution},\:{then} \\ $$$${x}=−{a},{y}=−{b},{z}=−{c}\:{is}\:{also}\:{a}\:{solution}, \\ $$
Commented by ajfour last updated on 26/Apr/20
$${all}\:\pm\:{may}\:{not}\:{be}\:{valid},\:{Sir},\:{i}\:{think}.. \\ $$
Commented by mr W last updated on 26/Apr/20
$${to}\:{be}\:{exact},\:{the}\:{solutions}\:{are}: \\ $$$${x}=\frac{\mathrm{17}\sqrt{\mathrm{5}}}{\mathrm{20}} \\ $$$${y}=\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${z}=\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$${or} \\ $$$${x}=−\frac{\mathrm{17}\sqrt{\mathrm{5}}}{\mathrm{20}} \\ $$$${y}=−\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${z}=−\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$
Commented by ajfour last updated on 26/Apr/20
$${Thanks}\:{for}\:{clarifying}\:{Sir}! \\ $$
Commented by jagoll last updated on 26/Apr/20
$${great}…. \\ $$
Commented by I want to learn more last updated on 27/Apr/20
$$\mathrm{Nice}. \\ $$
Answered by MWSuSon last updated on 26/Apr/20
$$\left({x}+{y}−{z}\right)\left({x}−{y}+{z}\right)=\mathrm{3}\ldots\left(\mathrm{1}\right) \\ $$$$\left({y}−{z}+{x}\right)\left({y}+{z}−{x}\right)=\mathrm{5}\ldots\left(\mathrm{2}\right) \\ $$$$\left({z}−{x}+{y}\right)\left({z}+{x}−{y}\right)=\mathrm{12}\ldots\left(\mathrm{3}\right) \\ $$$$\frac{{eqn}\left(\mathrm{1}\right)}{{eqn}\left(\mathrm{2}\right)}\:\:\:\frac{\left({x}−{y}+{z}\right)}{\left({y}+{z}−{x}\right)}=\frac{\mathrm{3}}{\mathrm{5}}\ldots\left(\mathrm{4}\right) \\ $$$$\frac{{eqn}\left(\mathrm{3}\right)}{{eqn}\left(\mathrm{2}\right)}\:\:\:\frac{\left({z}+{x}−{y}\right)}{\left({y}−{z}+{x}\right)}=\frac{\mathrm{12}}{\mathrm{5}}\ldots\left(\mathrm{5}\right) \\ $$$$\frac{{eqn}\left(\mathrm{1}\right)}{{eqn}\left(\mathrm{3}\right)}\:\:\:\frac{\left({x}+{y}−{z}\right)}{\left({z}−{x}+{y}\right)}=\frac{\mathrm{3}}{\mathrm{12}}\ldots\left(\mathrm{6}\right) \\ $$$${from}\:{eqn}\left(\mathrm{4}\right)\:\mathrm{3}{y}+\mathrm{3}{z}−\mathrm{3}{x}=\mathrm{5}{x}−\mathrm{5}{y}+\mathrm{5}{z} \\ $$$$\mathrm{8}{y}−\mathrm{2}{z}−\mathrm{8}{x}=\mathrm{0} \\ $$$$\left[{x}−{y}=\frac{−{z}}{\mathrm{4}}\right] \\ $$$${from}\:{eqn}\left(\mathrm{5}\right)\:\mathrm{5}{z}+\mathrm{5}{x}−\mathrm{5}{y}=\mathrm{12}{y}−\mathrm{12}{z}+\mathrm{12}{x} \\ $$$$\mathrm{17}{z}−\mathrm{7}{x}−\mathrm{17}{y}=\mathrm{0} \\ $$$$\left[{y}−{z}=\frac{−\mathrm{7}{x}}{\mathrm{17}}\right] \\ $$$${from}\:{eqn}\left(\mathrm{6}\right)\:\mathrm{3}{z}−\mathrm{3}{x}+\mathrm{3}{y}=\mathrm{12}{x}+\mathrm{12}{y}−\mathrm{12}{z} \\ $$$$\left[{z}−{x}=\frac{\mathrm{9}{y}}{\mathrm{15}}\right] \\ $$$${input}\:{the}\:{value}\:{of}\:\left({x}−{y}\right)\:{into}\:{eqn}\left(\mathrm{3}\right) \\ $$$${input}\:{the}\:{value}\:{of}\:\left({y}−{z}\right)\:{into}\:{eqn}\left(\mathrm{1}\right) \\ $$$${input}\:{the}\:{value}\:{of}\:\left({z}−{x}\right)\:{into}\:{eqn}\left(\mathrm{2}\right) \\ $$$${and}\:{you}'{ll}\:{have}\:{the}\:{answer}\:{Mr}\:{W}\:{had}. \\ $$