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Question Number 10693 by ABD last updated on 23/Feb/17

f(x)=((5x−1)/4)  , f(a)+f(b)=(9/2)  ⇒a+b=?

$${f}\left({x}\right)=\frac{\mathrm{5}{x}−\mathrm{1}}{\mathrm{4}}\:\:,\:{f}\left({a}\right)+{f}\left({b}\right)=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\Rightarrow{a}+{b}=? \\ $$

Answered by nume1114 last updated on 23/Feb/17

f(x)=((5x−1)/4)  ⇒f(a)=((5a−1)/4),f(b)=((5b−1)/4)  ⇒f(a)+f(b)=((5a−1)/4)+((5b−1)/4)=(9/2)  ⇒((5(a+b))/4)−(1/2)=(9/2)  ⇒a+b=((20)/5)

$${f}\left({x}\right)=\frac{\mathrm{5}{x}−\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{f}\left({a}\right)=\frac{\mathrm{5}{a}−\mathrm{1}}{\mathrm{4}},{f}\left({b}\right)=\frac{\mathrm{5}{b}−\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{f}\left({a}\right)+{f}\left({b}\right)=\frac{\mathrm{5}{a}−\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{5}{b}−\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{5}\left({a}+{b}\right)}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\Rightarrow{a}+{b}=\frac{\mathrm{20}}{\mathrm{5}} \\ $$$$ \\ $$

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