Question-145801

Question Number 145801 by mathdanisur last updated on 08/Jul/21

Answered by ajfour last updated on 08/Jul/21

xy+tz=a solving for x,y in terms of t,z from 2nd and 3rd eqs. x=((bz−ct)/(z^2 −t^2 )) ; y=((bt−cz)/(t^2 −z^2 )) (((bz−ct)(cz−bt))/((z^2 −t^2 )^2 ))+tz=a ..(i) x+y+z+t=((b+c)/(z+t))+z+t=Q if at all z≠t ((bc(z+t)^2 −(b+c)^2 tz)/((z+t)^2 {(z+t)^2 −4tz}))+tz=a ......

$${xy}+{tz}={a} \\ $$$${solving}\:{for}\:{x},{y}\:{in}\:{terms}\:{of} \\ $$$${t},{z}\:{from}\:\mathrm{2}{nd}\:{and}\:\mathrm{3}{rd}\:{eqs}. \\ $$$${x}=\frac{{bz}−{ct}}{{z}^{\mathrm{2}} −{t}^{\mathrm{2}} }\:\:;\:\:{y}=\frac{{bt}−{cz}}{{t}^{\mathrm{2}} −{z}^{\mathrm{2}} } \\ $$$$\frac{\left({bz}−{ct}\right)\left({cz}−{bt}\right)}{\left({z}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)^{\mathrm{2}} }+{tz}={a}\:\:..\left({i}\right) \\ $$$${x}+{y}+{z}+{t}=\frac{{b}+{c}}{{z}+{t}}+{z}+{t}={Q} \\ $$$${if}\:{at}\:{all}\:{z}\neq{t} \\ $$$$\frac{{bc}\left({z}+{t}\right)^{\mathrm{2}} −\left({b}+{c}\right)^{\mathrm{2}} {tz}}{\left({z}+{t}\right)^{\mathrm{2}} \left\{\left({z}+{t}\right)^{\mathrm{2}} −\mathrm{4}{tz}\right\}}+{tz}={a} \\ $$$$…… \\ $$

Commented by mathdanisur last updated on 08/Jul/21

$${Thankyou}\:{Ser} \\ $$

Answered by Rasheed.Sindhi last updated on 08/Jul/21

{ ((xy+zt=38.............(i))),((xz+yt=34.............(ii))),((xt+yz=43.............(iii))) :} (ii)+(iii)⇒ (x+y)(z+t)=77 77=1×77=7×11 { ((x+y=1∧z+t=77⇒x+y+z+t=78)),((x+y=7∧z+t=11⇒x+y+z+t=18)) :} Possible values for x+y+z+t {18,78} (iii)+(i):(x+z)(y+t)=81 81=1×81=3×27=9×9 Possible values for x+y+z+t {82,30,18} (i)+(ii)⇒(x+t)(y+z)=72 72=1×72=2×36=3×24=4×18 =6×12=8×9 Possible values for x+y+z+t {73,38,27,22,18,17} In all the three cases 18 is common. • •^(•) x+y+z+t=18

$$\begin{cases}{{xy}+{zt}=\mathrm{38}………….\left({i}\right)}\\{{xz}+{yt}=\mathrm{34}………….\left({ii}\right)}\\{{xt}+{yz}=\mathrm{43}………….\left({iii}\right)}\end{cases} \\ $$$$\left({ii}\right)+\left({iii}\right)\Rightarrow\:\left({x}+{y}\right)\left({z}+{t}\right)=\mathrm{77} \\ $$$$\mathrm{77}=\mathrm{1}×\mathrm{77}=\mathrm{7}×\mathrm{11} \\ $$$$\begin{cases}{{x}+{y}=\mathrm{1}\wedge{z}+{t}=\mathrm{77}\Rightarrow{x}+{y}+{z}+{t}=\mathrm{78}}\\{{x}+{y}=\mathrm{7}\wedge{z}+{t}=\mathrm{11}\Rightarrow{x}+{y}+{z}+{t}=\mathrm{18}}\end{cases} \\ $$$${Possible}\:{values}\:{for}\:{x}+{y}+{z}+{t} \\ $$$$\left\{\mathrm{18},\mathrm{78}\right\} \\ $$$$\left({iii}\right)+\left({i}\right):\left({x}+{z}\right)\left({y}+{t}\right)=\mathrm{81} \\ $$$$\mathrm{81}=\mathrm{1}×\mathrm{81}=\mathrm{3}×\mathrm{27}=\mathrm{9}×\mathrm{9} \\ $$$${Possible}\:{values}\:{for}\:{x}+{y}+{z}+{t} \\ $$$$\left\{\mathrm{82},\mathrm{30},\mathrm{18}\right\} \\ $$$$\left({i}\right)+\left({ii}\right)\Rightarrow\left({x}+{t}\right)\left({y}+{z}\right)=\mathrm{72} \\ $$$$\mathrm{72}=\mathrm{1}×\mathrm{72}=\mathrm{2}×\mathrm{36}=\mathrm{3}×\mathrm{24}=\mathrm{4}×\mathrm{18} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{6}×\mathrm{12}=\mathrm{8}×\mathrm{9} \\ $$$${Possible}\:{values}\:{for}\:{x}+{y}+{z}+{t} \\ $$$$\left\{\mathrm{73},\mathrm{38},\mathrm{27},\mathrm{22},\mathrm{18},\mathrm{17}\right\} \\ $$$${In}\:{all}\:{the}\:{three}\:{cases}\:\mathrm{18}\:{is}\:{common}. \\ $$$$\overset{\bullet} {\bullet\:\:\:\bullet}\:\:{x}+{y}+{z}+{t}=\mathrm{18}\: \\ $$

Commented by Rasheed.Sindhi last updated on 08/Jul/21

$$\mathcal{T}{he}\:{answer}\:{has}\:{been}\:{revised}\:{now}. \\ $$

Commented by mathdanisur last updated on 08/Jul/21

$${Thankyou}\:{Ser} \\ $$

Commented by mathdanisur last updated on 08/Jul/21

$${answer}:\:\mathrm{18}\:{Ser} \\ $$

Commented by Rasheed.Sindhi last updated on 08/Jul/21

Sorry I didn′t notice: x,y,z,t∈Z^+ Two answers 18 & −18 are in case x,y,z,t∈Z

$${Sorry}\:{I}\:{didn}'{t}\:{notice}: \\ $$$${x},{y},{z},{t}\in\mathbb{Z}^{+} \\ $$$$\:{Two}\:{answers}\:\mathrm{18}\:\&\:−\mathrm{18}\:{are} \\ $$$${in}\:{case}\:{x},{y},{z},{t}\in\mathbb{Z} \\ $$

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