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Question Number 83373 by Power last updated on 01/Mar/20

Commented by jagoll last updated on 02/Mar/20

$$(\mathrm{ax}+\mathrm{by})(\mathrm{x}+\mathrm{y})=5(\mathrm{x}+\mathrm{y})$$$${\mathrm{ax}}^2+{\mathrm{by}}^2+\mathrm{axy}+\mathrm{bxy}=5(\mathrm{x}+\mathrm{y})$$$$10+\mathrm{xy}(\mathrm{a}+\mathrm{b})=5(\mathrm{x}+\mathrm{y})\left(\mathrm{i}\right)$$$$({\mathrm{ax}}^2+{\mathrm{by}}^2)(\mathrm{x}+\mathrm{y})=10(\mathrm{x}+\mathrm{y})$$$${\mathrm{ax}}^3+{\mathrm{by}}^3+{\mathrm{ax}}^2\mathrm{y}+{\mathrm{bxy}}^2=10(\mathrm{x}+\mathrm{y})$$$$50+\mathrm{xy}(\mathrm{ax}+\mathrm{by})=10(\mathrm{x}+\mathrm{y})$$$$50+5\mathrm{xy}=10(\mathrm{x}+\mathrm{y})\Rightarrow 10+\mathrm{xy}=\mathrm{x}+\mathrm{y}\left(\mathrm{ii}\right)$$$$({\mathrm{ax}}^3+{\mathrm{by}}^3)(\mathrm{x}+\mathrm{y})=50(\mathrm{x}+\mathrm{y})$$$${\mathrm{ax}}^4+{\mathrm{by}}^4+{\mathrm{ax}}^3\mathrm{y}+{\mathrm{bxy}}^3=50(\mathrm{x}+\mathrm{y})$$$$130+\mathrm{xy}({\mathrm{ax}}^2+{\mathrm{by}}^2)=50(\mathrm{x}+\mathrm{y})$$$$130+10\mathrm{xy}=50(\mathrm{x}+\mathrm{y})$$$$13+\mathrm{xy}=5(\mathrm{x}+\mathrm{y})\left(\mathrm{iii}\right)$$$$\mathrm{so}\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{result}$$

Commented by jagoll last updated on 02/Mar/20

$$\left(\mathrm{ii}\right)-\left(\mathrm{iii}\right)\Rightarrow 3=4(\mathrm{x}+\mathrm{y}),\mathrm{x}+\mathrm{y}=\frac{3}{4}$$$$\Rightarrow \mathrm{xy}=\frac{3}{4}-10=\frac{-37}{4}$$$$\Rightarrow 10+(-\frac{37}{4})(\mathrm{a}+\mathrm{b})=5\times \frac{3}{4}$$$$\Rightarrow 40-37(\mathrm{a}+\mathrm{b})=15$$$$\Rightarrow \mathrm{a}+\mathrm{b}=\frac{25}{37}$$

Commented by Power last updated on 02/Mar/20

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

Commented by mr W last updated on 02/Mar/20

$$nicesolution!$$

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