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

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

Commented by ajfour last updated on 27/May/17

$$\frac{1729}{64}=\frac{1728}{64}+\frac{1}{64}$$$$=\frac{{\left(12\right)}^3}{{\left(4\right)}^3}+\frac{{\left(1\right)}^3}{{\left(4\right)}^3}={\left(3\right)}^3+\frac{1}{{\left(4\right)}^3}$$$$={\left(9\right)}^{3/2}+\frac{1}{{\left(16\right)}^{3/2}}$$$$=\frac{{\left(9\right)}^2}{\sqrt{9}}+\frac{\sqrt{16}}{{\left(16\right)}^2}=\frac{x^2}{\sqrt{x}}+\frac{\sqrt{y}}{y^2}$$$$bycomparing,\mathit{x}=9and\mathit{y}=16$$$$$$$$\frac{6908}{81}=\frac{6912}{81}-\frac{4}{81}$$$$=\frac{256}{3}-\frac{4}{81}$$$$=\frac{{\left(16\right)}^2}{\sqrt{9}}-\frac{\sqrt{16}}{{\left(9\right)}^2}=\frac{y^2}{\sqrt{x}}-\frac{\sqrt{y}}{x^2}$$$$againoncomparison,$$$$\mathit{x}=9and\mathit{y}=16.$$$$$$

Commented by RasheedSindhi last updated on 27/May/17

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

Commented by ajfour last updated on 27/May/17

$$did.Sir!$$

Commented by RasheedSindhi last updated on 27/May/17

$$\theta \alpha nkyou!$$

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