Question Number 187754 by 073 last updated on 21/Feb/23
Answered by horsebrand11 last updated on 21/Feb/23
$$\:{let}\:{f}\left({x}\right)={ax}+{b} \\ $$$$\:{f}\left(\mathrm{3}{x}+\mathrm{1}\right)=\mathrm{3}{ax}+{a}+{b} \\ $$$$\:{f}\left({x}−\mathrm{2}\right)={ax}−\mathrm{2}{a}+{b} \\ $$$$\:{f}\left(\mathrm{3}{x}+\mathrm{1}\right)+{f}\left({x}−\mathrm{2}\right)=\mathrm{4}{ax}−{a}+\mathrm{2}{b}=\mathrm{10}{x} \\ $$$$\:\begin{cases}{\mathrm{4}{a}=\mathrm{10}\Rightarrow{a}=\frac{\mathrm{5}}{\mathrm{2}}}\\{−{a}+\mathrm{2}{b}=\mathrm{0}\Rightarrow{b}=\frac{\mathrm{5}}{\mathrm{4}}}\end{cases} \\ $$$$\:\therefore\:{f}\left({x}\right)=\frac{\mathrm{5}}{\mathrm{2}}{x}+\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{3}\right)=\frac{\mathrm{15}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{4}}=\frac{\mathrm{35}}{\mathrm{4}} \\ $$