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Question Number 91378 by jagoll last updated on 30/Apr/20

Commented by jagoll last updated on 30/Apr/20

$$ihavenoideatosolvethis$$$$question$$

Commented by MJS last updated on 30/Apr/20

$$\mathrm{solve}\mathrm{the}\mathrm{first}\mathrm{one}\mathrm{for}x$$$$\mathrm{then}\mathrm{insert}\mathrm{in}{2}^{\mathrm{nd}}\mathrm{one}$$$$\mathrm{or}$$$$\mathrm{let}\sqrt{x-2y}=t\iff x=t^2+2y\wedge t⩾0$$$$$$$$\mathrm{then}\mathrm{it}\mathrm{should}\mathrm{not}\mathrm{be}\mathrm{too}\mathrm{hard}$$

Commented by MJS last updated on 01/May/20

$$\mathrm{I}\mathrm{get}4\mathrm{solutions}$$$$x_1\approx 2.20819\wedge y_1\approx .533931$$$$x_2=\frac{8}{3}\wedge y_2=\frac{4}{9}$$$$x_3\approx 8.79946\wedge y_3\approx -1.10613$$$$x_4=12\wedge y_4=-2$$

Commented by jagoll last updated on 01/May/20

$$mywaysame,likethisbut$$$$i^{\prime }mstuck$$

Commented by jagoll last updated on 01/May/20

$$sirmjs.canyoupostthere$$$$yourworkingsir.please$$

Commented by Prithwish Sen 1 last updated on 01/May/20

$$2+6\mathrm{y}=\frac{\mathrm{x}}{\mathrm{y}}-\sqrt{\mathrm{x}-2\mathrm{y}}$$$$2\mathrm{y}+{6\mathrm{y}}^2=\mathrm{x}-\mathrm{y}\sqrt{\mathrm{x}-2\mathrm{y}}$$$${6\mathrm{y}}^2+\mathrm{y}\sqrt{\mathrm{x}-2\mathrm{y}}-(\mathrm{x}-2\mathrm{y})=0$$$$\therefore \mathrm{y}=\frac{-\sqrt{\mathrm{x}-2\mathrm{y}}\pm 5\sqrt{\mathrm{x}-2\mathrm{y}}}{12}$$$$\Rightarrow 3\mathbf{y}=\sqrt{\mathbf{x}-2\mathbf{y}}\mathbf{or}2\mathbf{y}=-\sqrt{\mathbf{x}-2\mathbf{y}}$$$$\mathbf{now}\mathbf{considering}\mathbf{the}{1}^{\mathbf{st}}\mathbf{relation},\mathbf{we}\mathbf{get}\mathbf{from}\mathbf{eqn}.2$$$$\sqrt{\mathrm{x}+\sqrt{\mathrm{x}-2\mathrm{y}}}=\mathrm{x}+3\mathrm{y}-2$$$$\sqrt{\mathrm{x}+3\mathrm{y}}=\mathrm{x}+3\mathrm{y}-2$$$$\mathbf{now}\mathbf{you}\mathbf{can}\mathbf{get}\mathbf{it}.$$

Answered by john santu last updated on 01/May/20

Commented by Prithwish Sen 1 last updated on 01/May/20

$$\mathrm{Nice}.$$

Commented by MJS last updated on 01/May/20

$$\mathrm{but}\mathrm{you}\mathrm{have}\mathrm{to}\mathrm{check}\mathrm{each}\mathrm{solution},\mathrm{not}\mathrm{all}$$$$\mathrm{of}\mathrm{them}\mathrm{solve}\mathrm{the}\mathrm{given}\mathrm{equations}$$

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