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Question Number 95608 by bobhans last updated on 26/May/20

Commented by john santu last updated on 26/May/20

$$\mathrm{another}\mathrm{way}$$$$\mathrm{set}x+y=handxy=g$$$$\Rightarrow xy(x+y)=30\mathrm{\&}g+h=11$$$$\Rightarrow gh=30;g=11-h$$$$\Rightarrow 11h-h^2=30;h^2-11h+30=0$$$$(h-6)(h-5)=0$$$$$$$$$$

Commented by bobhans last updated on 26/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}$$

Answered by john santu last updated on 26/May/20

$$\Rightarrow y=\frac{11-x}{x+1}$$$$x^2\left(\frac{11-x}{x+1}\right)+x{\left(\frac{11-x}{x+1}\right)}^2=30$$$$11x^3-11x^2-x^4+121x=30x^2+60x+30$$$$x_1=1\mathrm{\&}{\mathrm{y}}_1=5$$$$x_2=2\mathrm{\&}{\mathrm{y}}_2=3$$$$x_3=3\mathrm{\&}{\mathrm{y}}_3=2$$$$x_4=5\mathrm{\&}{\mathrm{y}}_4=1$$

Answered by MJS last updated on 26/May/20

$$\mathrm{let}x=p-q\wedge y=p+q$$$$\{\begin{array}{l}{p}^2+2p-q^2-11=0\\ 2p^3-2{pq}^2-30=0\end{array}$$$$\{\begin{array}{l}{q}^2=p^2+2p-11\\ q^2=\frac{p^3-15}{p}\end{array}$$$$p^2+2p-11=\frac{p^3-15}{p}$$$$p^2-\frac{11}{2}p+\frac{15}{2}=0$$$$p_1=\frac{5}{2}\wedge p_2=3$$$$\Rightarrow$$$$q_1=\pm \frac{1}{2}\wedge q_2=\pm 2$$$$\Rightarrow$$$$(x=2\wedge y=3)\vee (x=3\wedge y=2)\vee$$$$(x=1\wedge y=5)\vee (x=5\wedge y=1)$$

Answered by behi83417@gmail.com last updated on 26/May/20

$$\mathrm{x}+\mathrm{y}=\mathrm{a},\mathrm{xy}=\mathrm{b}$$$$\Rightarrow \{\begin{array}{l}\mathrm{a}+\mathrm{b}=11\\ \mathrm{ab}=30\Rightarrow {\mathrm{z}}^2-11\mathrm{z}+30=0\Rightarrow \mathrm{z}=5,6\end{array}$$$$\Rightarrow \{\begin{array}{l}\mathrm{x}+\mathrm{y}=5\\ \mathrm{xy}=6\Rightarrow {\mathrm{x}}^2-5\mathrm{x}+6=0\Rightarrow \mathrm{x}\vee \mathrm{y}=2,3\end{array}$$$$\Rightarrow \{\begin{array}{l}\mathrm{x}+\mathrm{y}=6\\ \mathrm{xy}=5\Rightarrow {\mathrm{x}}^2-6\mathrm{x}+5=0\Rightarrow \mathrm{x}\vee \mathrm{y}=1,5\end{array}$$

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