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Question Number 164770 by cortano1 last updated on 21/Jan/22

$$\{\begin{array}{l}f(3x-1)+g(6x-1)=3x\\ f(x+1)+xg(2x+3)=2x^2+x\end{array}$$$$f\left(x\right)=?$$

Answered by blackmamba last updated on 21/Jan/22

$$\mathrm{let}\{\begin{array}{l}f\left(x\right)=px+q\\ g\left(x\right)=ax+b\end{array}$$$$\{\begin{array}{l}f(3x-1)=3px+q-p\\ g(6x-1)=6ax+b-a\end{array}\Rightarrow (3p+6a)x+b+q-(a+p)=3x$$$$\{\begin{array}{l}3p+6a=3\\ b+q=a+p\end{array}$$$$\{\begin{array}{l}f(x+1)=px+q+p\\ g(2x+3)=2ax+b+3a\end{array}$$$$\Rightarrow px+q+p+2{ax}^2+(b+3a)x=2x^2+x$$$$\Rightarrow 2{ax}^2+(b+3a+p)x+p+q=2x^2+x$$$$\{\begin{array}{l}p+q=0\Rightarrow p=-q\\ 2a=2\Rightarrow a=1;b+3a+p=1;b+p=-2\end{array}$$$$\{\begin{array}{l}3p+6=3\Rightarrow p=-1\wedge q=1\\ b+1=1+(-1)\Rightarrow b=-1\end{array}$$$$\therefore \{\begin{array}{l}f\left(x\right)=-x+1\\ g\left(x\right)=x-1\end{array}$$

Answered by TheSupreme last updated on 21/Jan/22

$$6x-1=t$$$$f\left(\frac{t-1}{2}\right)+g\left(t\right)=\frac{t+1}{2}$$$$2x+3=t$$$$f\left(\frac{t-1}{2}\right)+\frac{t-3}{2}g\left(t\right)=\left(\frac{t-3}{2}\right)(t-2)$$$$g\left(t\right)(1-\frac{t-3}{2})=\frac{t-3}{2}(t-2)-\frac{t+1}{2}$$$$g\left(t\right)=\frac{\frac{(t-3)(t-2)}{2}-\frac{t+1}{2}}{1-\frac{t-3}{2}}=\frac{t^2-6t+5}{5-t}=\frac{(t-5)(t-1)}{-(t-5)}=1-t$$$$f(3x+1)+2-6x=3x$$$$f(3x+1)=9x-2$$$$f\left(u\right)=3u+1$$

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