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Question Number 58195 by salaw2000 last updated on 19/Apr/19

$$\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=\frac{3}{4}$$$$\frac{{\mathrm{x}}^2}{\mathrm{y}}+\frac{{\mathrm{y}}^2}{\mathrm{x}}=9$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y}$$

Answered by Kunal12588 last updated on 19/Apr/19

$$\frac{1}{x}+\frac{1}{y}=\frac{3}{4}$$$$\Rightarrow \frac{x+y}{xy}=\frac{3}{4}\left(1\right)$$$$\Rightarrow x+y=\frac{3xy}{4}\left(2\right)$$$$\frac{x^2}{y}+\frac{y^2}{x}=9$$$$\Rightarrow \frac{x+y}{xy}(x^2-xy+y^2)=9$$$$from\left(1\right)$$$$\Rightarrow x^2-xy+y^2=12$$$$\Rightarrow {(x+y)}^2-3xy=12\left(3\right)$$$$\Rightarrow {(x-y)}^2+xy=12\left(4\right)$$$$from\left(3\right)$$$$x+y=\pm \sqrt{3xy+12}$$$$butfrom\left(2\right)$$$$\frac{3xy}{4}=\sqrt{3xy+12}$$$$\Rightarrow 3\cdot 3x^2{y}^2=16\left[3(xy+4)\right]$$$$\Rightarrow 3x^2{y}^2-16xy-64=0$$$$\Rightarrow xy=\frac{16\pm \sqrt{{16}^2-4\cdot 3\cdot (-64)}}{2\cdot 3}$$$$\Rightarrow xy=\frac{48}{6},\frac{-16}{6}$$$$\Rightarrow xy=8,\frac{-8}{3}$$$$\left(i\right)xy=8$$$$puttingxyin\left(2\right)$$$$x+y=6\left(5\right)$$$$puttingxyin\left(4\right)$$$${(x-y)}^2=12-8$$$$\Rightarrow x-y=\pm \sqrt{4}$$$$\Rightarrow x-y=2orx-y=-2$$$$\left(a\right)takingx-y=2$$$$andcomparingwith\left(5\right)$$$$1^{st}answerx=4,y=2$$$$\left(b\right)takingx-y=-2$$$$andcomparingwith\left(5\right)$$$$2^{nd}answerx=2,y=4$$$$\left(ii\right)xy=\frac{-8}{3}$$$$puttingxyin\left(2\right)$$$$x+y=-2\left(6\right)$$$$puttingxyin\left(4\right)$$$${(x-y)}^2+xy=12$$$$3{(x-y)}^2-8=36$$$$x-y=\pm \sqrt{\frac{44}{3}}$$$$\left(a\right)takingx-y=\sqrt{\frac{44}{3}}$$$$andcomparingwith\left(6\right)$$$$3^{rd}answerx=-1+\sqrt{\frac{11}{3}},y=-1-\sqrt{\frac{11}{3}}$$$$\left(b\right)takingx-y=-\sqrt{\frac{44}{3}}$$$$andcomparingwith\left(6\right)$$$$4^{th}answerx=-1-\sqrt{\frac{11}{3}},y=-1+\sqrt{\frac{11}{3}}$$

Commented by salaw2000 last updated on 19/Apr/19

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}$$

Answered by MJS last updated on 19/Apr/19

$$\left(1\right)y=\frac{4x}{3x-4}$$$$\left(2\right)x^3+y^3-9xy=0$$$$\left(1\right)\mathrm{in}\left(2\right)\Rightarrow$$$$x^4-4x^3-\frac{20}{3}{x}^2+32x-\frac{64}{3}=0$$$$\mathrm{trying}\mathrm{factors}\mathrm{of}\pm \frac{64}{3}\mathrm{we}\mathrm{find}$$$$x_1=2;x_2=4$$$$\Rightarrow$$$$(x-2)(x-4)(x^2+2x-\frac{8}{3})=0$$$$\Rightarrow x_3=-1-\frac{\sqrt{33}}{3};x_4=-1+\frac{\sqrt{33}}{3}$$$$\Rightarrow \left(\begin{array}{c}x\\ y\end{array}\right)\in \{\left(\begin{array}{c}2\\ 4\end{array}\right),\left(\begin{array}{c}4\\ 2\end{array}\right),\left(\begin{array}{c}-1-\frac{\sqrt{33}}{3}\\ -1+\frac{\sqrt{33}}{3}\end{array}\right),\left(\begin{array}{c}-1-\frac{\sqrt{33}}{3}\\ -1+\frac{\sqrt{33}}{3}\end{array}\right)\}$$

Commented by Kunal12588 last updated on 19/Apr/19

Sir you are making me think why I did that ����

Commented by MJS last updated on 19/Apr/19

$$\mathrm{different}\mathrm{paths}\mathrm{lead}\mathrm{to}\mathrm{the}\mathrm{same}\mathrm{solutions}-$$$${\mathrm{that}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{good}\mathrm{thing}!$$

Commented by salaw2000 last updated on 19/Apr/19

$$\mathrm{a}\mathrm{very}\mathrm{big}\mathrm{thanks}\mathrm{to}\mathrm{you}$$$$$$$$$$$$$$

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