Question Number 207611 by mr W last updated on 20/May/24
Answered by A5T last updated on 20/May/24
Commented by A5T last updated on 20/May/24
$$\mathrm{4}{x}+\mathrm{2}{y}={x}+\mathrm{5}{y}\Rightarrow{x}={y}\Rightarrow{s}=\mathrm{6}{x} \\ $$$$\sqrt{\mathrm{2}^{\mathrm{2}} −\left(\mathrm{2}{y}=\mathrm{2}{x}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}−\mathrm{4}{x}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{6}{x}}{\mathrm{6}{x}+\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\Rightarrow\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{4}{x}\Rightarrow\mathrm{4}{x}^{\mathrm{2}} =\mathrm{1}−{x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\Rightarrow{s}=\frac{\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{5}}\Rightarrow{s}^{\mathrm{2}} =\frac{\mathrm{36}}{\mathrm{5}}=\mathrm{7}.\mathrm{2} \\ $$
Commented by mr W last updated on 20/May/24
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