Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

Solve simultaneously xy + x + y = 23 ....... (i) xz + x + z = 41 ........ (ii) yz + y + z = 27 ........ (iii)

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

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

G^( OO) D Approach!

$$\mathcal{G}^{\:\mathcal{OO}} \mathcal{D}\:\:\:\mathcal{A}{pproach}! \\ $$

Commented by tawakalitu last updated on 30/Nov/16

Thank you sir. God bless you.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Commented by tawakalitu last updated on 30/Nov/16

i really appreciate your effort.

$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}. \\ $$

Commented by tawakalitu last updated on 30/Nov/16

God bless you

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com