Question Number 220391 by Lekhraj last updated on 12/May/25
Answered by MathematicalUser2357 last updated on 13/May/25
$${yes} \\ $$
Answered by Rasheed.Sindhi last updated on 12/May/25
$$\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{1}}={y}\Rightarrow\mathrm{2}{x}−\mathrm{1}={xy}+{y} \\ $$$$\mathrm{2}{x}−{xy}={y}+\mathrm{1} \\ $$$${x}\left(\mathrm{2}−{y}\right)={y}+\mathrm{1} \\ $$$${x}=\frac{{y}+\mathrm{1}}{\mathrm{2}−{y}} \\ $$$${f}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)=\frac{\mathrm{3}{x}}{\mathrm{2}{x}+\mathrm{2}} \\ $$$$\Rightarrow{f}\left({y}\right)=\frac{\mathrm{3}\left(\frac{{y}+\mathrm{1}}{\mathrm{2}−{y}}\right)}{\mathrm{2}\left(\frac{{y}+\mathrm{1}}{\mathrm{2}−{y}}\right)+\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}{y}+\mathrm{3}}{\mathrm{2}{y}+\mathrm{2}+\mathrm{4}−\mathrm{2}{y}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}\left({y}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{y}+\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left({x}\right)=\frac{{x}+\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Lekhraj last updated on 12/May/25
$$\mathrm{Thank}\:\mathrm{you} \\ $$
Answered by Ghisom last updated on 12/May/25
$$\mathrm{the}\:\mathrm{question}\:\mathrm{is} \\ $$$$“\mathrm{can}\:\mathrm{you}\:\mathrm{solve}\:{f}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)=\frac{\mathrm{3}{x}}{\mathrm{2}{x}+\mathrm{2}}?'' \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{correct}\:\mathrm{answer}\:\left(\mathrm{at}\:\mathrm{least}\:\mathrm{for}\:\mathrm{me}\right)\:\mathrm{is} \\ $$$$\mathrm{yes} \\ $$