If-xy-1-3-xz-3-8-and-yz-1-2-For-x-y-and-z-compare-

Question Number 207197 by hardmath last updated on 09/May/24

$$\mathrm{If}\:\:\:\mathrm{xy}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:,\:\:\:\mathrm{xz}\:=\:\frac{\mathrm{3}}{\mathrm{8}}\:\:\:\mathrm{and}\:\:\:\mathrm{yz}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{For}\:\:\:\mathrm{x}\:,\:\mathrm{y}\:\mathrm{and}\:\mathrm{z}\:\:\:\mathrm{compare}. \\ $$

Answered by A5T last updated on 09/May/24

x=(1/(3y));z=(1/(2y)) xz=(3/8)⇒(1/(6y^2 ))=(3/8)⇒y^2 =(4/9)⇒y=+_− (2/3) ⇒x=+_− (1/2);z=+_− (3/4)

$${x}=\frac{\mathrm{1}}{\mathrm{3}{y}};{z}=\frac{\mathrm{1}}{\mathrm{2}{y}} \\ $$$${xz}=\frac{\mathrm{3}}{\mathrm{8}}\Rightarrow\frac{\mathrm{1}}{\mathrm{6}{y}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{8}}\Rightarrow{y}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{9}}\Rightarrow{y}=\underset{−} {+}\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{x}=\underset{−} {+}\frac{\mathrm{1}}{\mathrm{2}};{z}=\underset{−} {+}\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Answered by mr W last updated on 09/May/24

(i)×(ii)×(iii): (xyz)^2 =(1/3)×(3/8)×(1/2)=(1/(16)) ⇒xyz=±(1/4) ...(iv) (iv)/(i): z=±(3/4) (iv)/(ii): y=±(2/3) (iv)/(iii): x=±(1/2)

$$\left({i}\right)×\left({ii}\right)×\left({iii}\right): \\ $$$$\left({xyz}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{8}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\Rightarrow{xyz}=\pm\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:…\left({iv}\right) \\ $$$$\left({iv}\right)/\left({i}\right): \\ $$$${z}=\pm\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left({iv}\right)/\left({ii}\right): \\ $$$${y}=\pm\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left({iv}\right)/\left({iii}\right): \\ $$$${x}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by hardmath last updated on 09/May/24