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Question Number 161564 by HongKing last updated on 19/Dec/21

Find all values x;y;z>0 such that: { ((x + y + 2z = 6)),(((3/y)∙((2/x) + (1/y)) = 4∙((2/(x + y)) + (1/(2y)))^2 )),((x + 2^y + log_2 z = 4)) :}

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{values}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{such}\:\mathrm{that}: \\ $$ $$\begin{cases}{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{2z}\:=\:\mathrm{6}}\\{\frac{\mathrm{3}}{\mathrm{y}}\centerdot\left(\frac{\mathrm{2}}{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{y}}\right)\:=\:\mathrm{4}\centerdot\left(\frac{\mathrm{2}}{\mathrm{x}\:+\:\mathrm{y}}\:+\:\frac{\mathrm{1}}{\mathrm{2y}}\right)^{\mathrm{2}} }\\{\mathrm{x}\:+\:\mathrm{2}^{\boldsymbol{\mathrm{y}}} \:+\:\mathrm{log}_{\mathrm{2}} \boldsymbol{\mathrm{z}}\:=\:\mathrm{4}}\end{cases} \\ $$

Commented byRasheed.Sindhi last updated on 20/Dec/21

(x,y,z)=(1,1,2)

$$\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{1},\mathrm{1},\mathrm{2}\right) \\ $$

Answered by Rasheed.Sindhi last updated on 20/Dec/21

Assuming x,y,z∈Z^+ x+y+2z=6 x+y=6−2z⇒z=1,2 x+y=2(3−z)⇒x+y is even. x+y=2=1+1 ∣ x+y=4=2+2=1+3=3+1 (x,y,z)=(1,1,2) , (2,2,1),(1,3,1),(3,1,1) (1,1,2) satisfying all the three equations. ∴ (x,y,z)=(1,1,2)

$$\underline{\boldsymbol{{Assuming}}\:\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\in\mathbb{Z}^{+} } \\ $$ $$\mathrm{x}+\mathrm{y}+\mathrm{2z}=\mathrm{6} \\ $$ $$\mathrm{x}+\mathrm{y}=\mathrm{6}−\mathrm{2z}\Rightarrow\mathrm{z}=\mathrm{1},\mathrm{2} \\ $$ $$\mathrm{x}+\mathrm{y}=\mathrm{2}\left(\mathrm{3}−\mathrm{z}\right)\Rightarrow\mathrm{x}+\mathrm{y}\:{is}\:{even}. \\ $$ $$\mathrm{x}+\mathrm{y}=\mathrm{2}=\mathrm{1}+\mathrm{1}\:\mid\:\mathrm{x}+\mathrm{y}=\mathrm{4}=\mathrm{2}+\mathrm{2}=\mathrm{1}+\mathrm{3}=\mathrm{3}+\mathrm{1} \\ $$ $$\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{1},\mathrm{1},\mathrm{2}\right)\:,\:\left(\mathrm{2},\mathrm{2},\mathrm{1}\right),\left(\mathrm{1},\mathrm{3},\mathrm{1}\right),\left(\mathrm{3},\mathrm{1},\mathrm{1}\right) \\ $$ $$\left(\mathrm{1},\mathrm{1},\mathrm{2}\right)\:{satisfying}\:{all}\:{the}\:{three} \\ $$ $${equations}. \\ $$ $$\therefore\:\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{1},\mathrm{1},\mathrm{2}\right) \\ $$

Commented byHongKing last updated on 20/Dec/21

very nice dear Sir thank you

$$\mathrm{very}\:\mathrm{nice}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

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