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Question Number 207205 by hardmath last updated on 09/May/24

$$\{\begin{array}{l}{\mathrm{x}}^2+\mathrm{xy}=4\mathrm{x}\\ {\mathrm{y}}^2+\mathrm{xy}=4\mathrm{y}\end{array}$$$$\mathrm{Find}:{\log }_{16}({\mathrm{x}}_1+{\mathrm{y}}_1+{\mathrm{x}}_2+{\mathrm{y}}_2)=?$$

Answered by A5T last updated on 09/May/24

$$x^2-y^2=4x-4y;x-y\ne 0\Rightarrow x+y=4$$$$x=y\Rightarrow 2x^2=4x\Rightarrow x^2-2x=0\Rightarrow x_1=0;x_2=2$$$$\Rightarrow {log}_{16}(x_1+x_2+y_1+y_2)={log}_{16}4=\frac{1}{2}$$$$Thereareinfinitelymanysolutionsotherwise.$$

Commented by hardmath last updated on 09/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

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