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Question Number 13986 by RasheedSindhi last updated on 26/May/17

Solve the following system of equations. (x^2 /(√x))+((√y)/y^2 )=((1729)/(64)) (y^2 /(√x))−((√y)/x^2 )=((6908)/(81))

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{system} \\ $$$$\mathrm{of}\:\mathrm{equations}. \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\sqrt{\mathrm{x}}}+\frac{\sqrt{\mathrm{y}}}{\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{1729}}{\mathrm{64}} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{y}^{\mathrm{2}} }{\sqrt{\mathrm{x}}}−\frac{\sqrt{\mathrm{y}}}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{6908}}{\mathrm{81}} \\ $$$$ \\ $$

Commented by ajfour last updated on 27/May/17

((1729)/(64))=((1728)/(64))+(1/(64)) =(((12)^3 )/((4)^3 ))+(((1)^3 )/((4)^3 )) = (3)^3 +(1/((4)^3 )) =(9)^(3/2) +(1/((16)^(3/2) )) =(((9)^2 )/(√9))+((√(16))/((16)^2 )) =(x^2 /(√x))+((√y)/y^2 ) by comparing, x=9 and y=16 ((6908)/(81))=((6912)/(81))−(4/(81)) =((256)/3)−(4/(81)) =(((16)^2 )/(√9))−((√(16))/((9)^2 ))=(y^2 /(√x))−((√y)/x^2 ) again on comparison, x=9 and y=16 .

$$\frac{\mathrm{1729}}{\mathrm{64}}=\frac{\mathrm{1728}}{\mathrm{64}}+\frac{\mathrm{1}}{\mathrm{64}} \\ $$$$\:\:=\frac{\left(\mathrm{12}\right)^{\mathrm{3}} }{\left(\mathrm{4}\right)^{\mathrm{3}} }+\frac{\left(\mathrm{1}\right)^{\mathrm{3}} }{\left(\mathrm{4}\right)^{\mathrm{3}} }\:=\:\:\:\left(\mathrm{3}\right)^{\mathrm{3}} +\frac{\mathrm{1}}{\left(\mathrm{4}\right)^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{9}\right)^{\mathrm{3}/\mathrm{2}} +\frac{\mathrm{1}}{\left(\mathrm{16}\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{9}\right)^{\mathrm{2}} }{\sqrt{\mathrm{9}}}+\frac{\sqrt{\mathrm{16}}}{\left(\mathrm{16}\right)^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} }{\sqrt{{x}}}+\frac{\sqrt{{y}}}{{y}^{\mathrm{2}} } \\ $$$${by}\:{comparing},\:\:\boldsymbol{{x}}=\mathrm{9}\:\:\:{and}\:\:\boldsymbol{{y}}=\mathrm{16}\: \\ $$$$ \\ $$$$\frac{\mathrm{6908}}{\mathrm{81}}=\frac{\mathrm{6912}}{\mathrm{81}}−\frac{\mathrm{4}}{\mathrm{81}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{256}}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{81}}\: \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{16}\right)^{\mathrm{2}} }{\sqrt{\mathrm{9}}}−\frac{\sqrt{\mathrm{16}}}{\left(\mathrm{9}\right)^{\mathrm{2}} }=\frac{{y}^{\mathrm{2}} }{\sqrt{{x}}}−\frac{\sqrt{{y}}}{{x}^{\mathrm{2}} }\: \\ $$$${again}\:{on}\:{comparison}, \\ $$$$\:\:\:\:\boldsymbol{{x}}=\mathrm{9}\:\:{and}\:\:\boldsymbol{{y}}=\mathrm{16}\:\:. \\ $$$$ \\ $$

Commented by RasheedSindhi last updated on 27/May/17

=(((12)^3 )/((4)^3 ))+(((1)^3 )/((4)^3 )) =(((9)^2 )/(√9))+((√(16))/((16)^2 )) How? pl insert some steps

$$\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{12}\right)^{\mathrm{3}} }{\left(\mathrm{4}\right)^{\mathrm{3}} }+\frac{\left(\mathrm{1}\right)^{\mathrm{3}} }{\left(\mathrm{4}\right)^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{9}\right)^{\mathrm{2}} }{\sqrt{\mathrm{9}}}+\frac{\sqrt{\mathrm{16}}}{\left(\mathrm{16}\right)^{\mathrm{2}} }\:\:\mathrm{How}? \\ $$$$\mathrm{pl}\:\mathrm{insert}\:\mathrm{some}\:\mathrm{steps} \\ $$$$ \\ $$

Commented by ajfour last updated on 27/May/17

did . Sir !

$${did}\:.\:{Sir}\:! \\ $$

Commented by RasheedSindhi last updated on 27/May/17

θαnk you!

$$\theta\alpha{nk}\:{you}! \\ $$

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