Solve-simultaneously-xy-x-y-23-i-xz-x-z-41-ii-yz-y-z-27-iii-

Question Number 9312 by tawakalitu last updated on 29/Nov/16

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{xy}\:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{23}\:\:\:\:…….\:\left(\mathrm{i}\right) \\ $$$$\mathrm{xz}\:+\:\mathrm{x}\:+\:\mathrm{z}\:=\:\mathrm{41}\:\:\:\:……..\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{yz}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{27}\:\:\:\:\:……..\:\left(\mathrm{iii}\right) \\ $$

Commented by RasheedSoomro last updated on 30/Nov/16

xy + x + y = 23 ....... (i) xz + x + z = 41 ........ (ii) yz + y + z = 27 ........ (iii) (i)⇒x(y+1)=23−y⇒x=((23−y)/(y+1)).......(iv) (ii)⇒x(z+1)=41−z⇒x=((41−z)/(z+1)).........(v) (iv),(v)⇒((23−y)/(y+1))=((41−z)/(z+1)) Let y+1=Y and z+1=Z ((23−(Y−1))/Y)=((41−(Z−1))/Z) ((24−Y)/Y)=((42−Z)/Z) 24Z−YZ=42Y−YZ 24Z=42Y Z=((7Y)/4) Continue

$$\mathrm{xy}\:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{23}\:\:\:\:…….\:\left(\mathrm{i}\right) \\ $$$$\mathrm{xz}\:+\:\mathrm{x}\:+\:\mathrm{z}\:=\:\mathrm{41}\:\:\:\:……..\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{yz}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{27}\:\:\:\:\:……..\:\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{i}\right)\Rightarrow\mathrm{x}\left(\mathrm{y}+\mathrm{1}\right)=\mathrm{23}−\mathrm{y}\Rightarrow\mathrm{x}=\frac{\mathrm{23}−\mathrm{y}}{\mathrm{y}+\mathrm{1}}…….\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{ii}\right)\Rightarrow\mathrm{x}\left(\mathrm{z}+\mathrm{1}\right)=\mathrm{41}−\mathrm{z}\Rightarrow\mathrm{x}=\frac{\mathrm{41}−\mathrm{z}}{\mathrm{z}+\mathrm{1}}………\left(\mathrm{v}\right) \\ $$$$\left(\mathrm{iv}\right),\left(\mathrm{v}\right)\Rightarrow\frac{\mathrm{23}−\mathrm{y}}{\mathrm{y}+\mathrm{1}}=\frac{\mathrm{41}−\mathrm{z}}{\mathrm{z}+\mathrm{1}} \\ $$$$\mathrm{Let}\:\mathrm{y}+\mathrm{1}=\mathrm{Y}\:\:\:\mathrm{and}\:\:\mathrm{z}+\mathrm{1}=\mathrm{Z} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{23}−\left(\mathrm{Y}−\mathrm{1}\right)}{\mathrm{Y}}=\frac{\mathrm{41}−\left(\mathrm{Z}−\mathrm{1}\right)}{\mathrm{Z}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{24}−\mathrm{Y}}{\mathrm{Y}}=\frac{\mathrm{42}−\mathrm{Z}}{\mathrm{Z}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{24Z}−\mathrm{YZ}=\mathrm{42Y}−\mathrm{YZ} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{24Z}=\mathrm{42Y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Z}=\frac{\mathrm{7Y}}{\mathrm{4}} \\ $$$$\mathrm{Continue} \\ $$

Commented by sou1618 last updated on 30/Nov/16

(i) ⇒(x+1)(y+1)=24 (ii) ⇒(x+1)(z+1)=42 (iii)⇒(y+1)(z+1)=28 (X,Y,Z)=(x+1,y+1,z+1) { ((XY=24)),((XZ=42)),((YZ=28)) :} X,Y,Z≠0 XY×XZ×YZ=24×42×28 (XYZ)^2 =(8×3)×(2×3×7)×(4×7)=2^6 ×3^2 ×7^2 XYZ=±2^3 ×3×7 ∗if XYZ>0 X=((XYZ)/(YZ))=((2^3 ×3×7)/(28))=6 Y=((XYZ)/(XZ))=4 Z=((XYZ)/(XY))=7 (X,Y,Z)=(6,4,7) ∗if XYZ<0 (X,Y,Z)=(−6,−4,−7) −−−−−−−−−−−−−−− (X,Y,Z)=(x+1,y+1,z+1)=(±6,±4,±7) ⇒ (x,y,z)=(5,3,6),(−7,−5,−8).

$$\left({i}\right)\:\:\:\:\Rightarrow\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)=\mathrm{24} \\ $$$$\left({ii}\right)\:\:\Rightarrow\left({x}+\mathrm{1}\right)\left({z}+\mathrm{1}\right)=\mathrm{42} \\ $$$$\left({iii}\right)\Rightarrow\left({y}+\mathrm{1}\right)\left({z}+\mathrm{1}\right)=\mathrm{28} \\ $$$$\left({X},{Y},{Z}\right)=\left({x}+\mathrm{1},{y}+\mathrm{1},{z}+\mathrm{1}\right) \\ $$$$\begin{cases}{{XY}=\mathrm{24}}\\{{XZ}=\mathrm{42}}\\{{YZ}=\mathrm{28}}\end{cases} \\ $$$${X},{Y},{Z}\neq\mathrm{0} \\ $$$${XY}×{XZ}×{YZ}=\mathrm{24}×\mathrm{42}×\mathrm{28} \\ $$$$\:\:\:\:\:\:\left({XYZ}\right)^{\mathrm{2}} =\left(\mathrm{8}×\mathrm{3}\right)×\left(\mathrm{2}×\mathrm{3}×\mathrm{7}\right)×\left(\mathrm{4}×\mathrm{7}\right)=\mathrm{2}^{\mathrm{6}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{7}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{XYZ}=\pm\mathrm{2}^{\mathrm{3}} ×\mathrm{3}×\mathrm{7} \\ $$$$\ast{if}\:{XYZ}>\mathrm{0} \\ $$$${X}=\frac{{XYZ}}{{YZ}}=\frac{\mathrm{2}^{\mathrm{3}} ×\mathrm{3}×\mathrm{7}}{\mathrm{28}}=\mathrm{6} \\ $$$${Y}=\frac{{XYZ}}{{XZ}}=\mathrm{4} \\ $$$${Z}=\frac{{XYZ}}{{XY}}=\mathrm{7} \\ $$$$\left({X},{Y},{Z}\right)=\left(\mathrm{6},\mathrm{4},\mathrm{7}\right) \\ $$$$ \\ $$$$\ast{if}\:{XYZ}<\mathrm{0} \\ $$$$\left({X},{Y},{Z}\right)=\left(−\mathrm{6},−\mathrm{4},−\mathrm{7}\right) \\ $$$$−−−−−−−−−−−−−−− \\ $$$$\left({X},{Y},{Z}\right)=\left({x}+\mathrm{1},{y}+\mathrm{1},{z}+\mathrm{1}\right)=\left(\pm\mathrm{6},\pm\mathrm{4},\pm\mathrm{7}\right) \\ $$$$\Rightarrow \\ $$$$\left({x},{y},{z}\right)=\left(\mathrm{5},\mathrm{3},\mathrm{6}\right),\left(−\mathrm{7},−\mathrm{5},−\mathrm{8}\right). \\ $$$$ \\ $$

Commented by RasheedSoomro last updated on 30/Nov/16