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Question Number 170568 by mathlove last updated on 27/May/22

Answered by Rasheed.Sindhi last updated on 27/May/22

$$\overline{)\begin{array}{c}\underset{\mathrm{f}(\mathrm{x}-1)+\mathrm{g}(2\mathrm{x}+1)=2\mathrm{x}}{\mathrm{f}(2\mathrm{x}+2)+2\mathrm{g}(4\mathrm{x}+7)=\mathrm{x}-1}\end{array}}$$$$\left(1\right):2\mathrm{x}+2=\mathrm{y}\Rightarrow \mathrm{x}=\frac{\mathrm{y}-2}{2}:$$$$\mathrm{f}\left(\mathrm{y}\right)+2\mathrm{g}(4\cdot \frac{\mathrm{y}-2}{2}+7)=\frac{\mathrm{y}-2}{2}-1$$$$\Rightarrow \mathrm{f}\left(\mathrm{y}\right)+2\mathrm{g}(2\mathrm{y}+3)=\frac{\mathrm{y}-4}{2}...\left(3\right)$$$$\left(2\right):\mathrm{x}-1=\mathrm{y}\Rightarrow \mathrm{x}=\mathrm{y}+1$$$$\mathrm{f}\left(\mathrm{y}\right)+\mathrm{g}(2\mathrm{y}+3)=2(\mathrm{y}+1)...\left(4\right)$$$$\left(3\right)-\left(4\right):\mathrm{g}(2\mathrm{y}+3)=\frac{\mathrm{y}-4}{2}-2\mathrm{y}-2=\frac{\mathrm{y}-4-4\mathrm{y}-4}{2}$$$$\mathrm{g}(2\mathrm{y}+3)=\frac{-3\mathrm{y}-8}{2}$$$$2\mathrm{y}+3=\mathrm{x}\Rightarrow \mathrm{y}=\frac{\mathrm{x}-3}{2}$$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{-3\left(\frac{\mathrm{x}-3}{2}\right)-8}{2}=\frac{-3\mathrm{x}+9-16}{4}$$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{-3\mathrm{x}-7}{4}$$$$\left(3\right):\mathrm{f}\left(\mathrm{y}\right)+2\mathrm{g}(2\mathrm{y}+3)=\frac{\mathrm{y}-4}{2}$$$$\mathrm{f}\left(\mathrm{x}\right)+2\mathrm{g}(2\mathrm{x}+3)=\frac{\mathrm{x}-4}{2}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}-4}{2}-2\mathrm{g}(2\mathrm{x}+3)$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}-4}{2}-2\left(\frac{-3(2\mathrm{x}+3)-7}{4}\right)$$$$=\frac{\mathrm{x}-4}{2}-\frac{-6\mathrm{x}-9-7}{2}$$$$=\frac{\mathrm{x}-4}{2}+\frac{6\mathrm{x}+16}{2}$$$$=\frac{7\mathrm{x}+12}{2}$$$$$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{7\mathrm{x}+12}{2},\mathrm{g}\left(\mathrm{x}\right)=\frac{-3\mathrm{x}-7}{4}$$$$\mathrm{Pl}\mathrm{confirm}\mathrm{the}\mathrm{answers}$$

Commented by Rasheed.Sindhi last updated on 27/May/22

$$\mathcal{VERIFICATION}:$$$$\overline{)\begin{array}{c}\underset{\mathrm{f}(\mathrm{x}-1)+\mathrm{g}(2\mathrm{x}+1)=2\mathrm{x}}{\mathrm{f}(2\mathrm{x}+2)+2\mathrm{g}(4\mathrm{x}+7)=\mathrm{x}-1}\end{array}}$$$$\bullet \mathrm{f}(2\mathrm{x}+2)+2\mathrm{g}(4\mathrm{x}+7)=\mathrm{x}-1$$$$\frac{7(2\mathrm{x}+2)+12}{2}+2\left(\frac{-3(4\mathrm{x}+7)-7}{4}\right)=\mathrm{x}-1$$$$\frac{14\mathrm{x}+14+12}{2}+2\left(\frac{-12\mathrm{x}-21-7}{4}\right)=\mathrm{x}-1$$$$\frac{14\mathrm{x}+26}{2}+\frac{-12\mathrm{x}-28}{2}=\mathrm{x}-1$$$$7\mathrm{x}+13-6\mathrm{x}-14=\mathrm{x}-1$$$$\mathrm{x}-1=\mathrm{x}-1✓$$$$\bullet \mathrm{f}(\mathrm{x}-1)+\mathrm{g}(2\mathrm{x}+1)=2\mathrm{x}$$$$\frac{7(\mathrm{x}-1)+12}{2}+\frac{-3(2\mathrm{x}+1)-7}{4}=2\mathrm{x}$$$$\frac{7\mathrm{x}-7+12}{2}+\frac{-6\mathrm{x}-3-7}{4}=2\mathrm{x}$$$$\frac{7\mathrm{x}+5}{2}+\frac{-6\mathrm{x}-10}{4}=2\mathrm{x}$$$$\frac{7\mathrm{x}+5}{2}+\frac{-3\mathrm{x}-5}{2}=2\mathrm{x}$$$$\frac{4\mathrm{x}}{2}=2\mathrm{x}$$$$2\mathrm{x}=2\mathrm{x}✓$$

Commented by SLVR last updated on 31/May/22

$$dearsirhowyvariablehas$$$$2notationy=2x+2and$$$$y=x-1atonego...kindly$$$$explain$$

Commented by Rasheed.Sindhi last updated on 31/May/22

$$\mathit{s}\mathit{i}\mathit{r}\mathit{m}\mathit{r}\mathit{W}canbetterexplain.$$$$\bullet \mathrm{f}(2\mathrm{x}+2)+2\mathrm{g}(4\mathrm{x}+7)=\mathrm{x}-1$$$$\bullet \mathrm{f}(\mathrm{x}-1)+\mathrm{g}(2\mathrm{x}+1)=2\mathrm{x}$$$$\mathcal{T}hetwoequationscanbeconsiderd$$$$separatelywithrespectto\mathrm{x}.Actually$$$$\mathrm{x}inoneequationisnothingtodo$$$$with\mathrm{x}inotherequation.(\mathcal{T}{hey}^{\prime }re$$$$simultaneousonlywithrespect$$$$to\mathrm{f}and\mathrm{g}only).Albeitallxinoneequation$$$$arerelatedtoeachother$$$$andallxinotherequationarerelatedto$$$$toeachother.$$

Commented by Rasheed.Sindhi last updated on 31/May/22

$$\mathrm{These}\mathrm{equatios}\mathrm{are}\mathbf{separate}\mathrm{\&}$$$$\mathbf{simultaneous}\mathrm{simultaneously}!:)$$

Commented by SLVR last updated on 31/May/22

$$understoodwellsir..Sokind$$

Commented by Rasheed.Sindhi last updated on 31/May/22

$${You}^{\prime }rewelcomesir!$$

Commented by mathlove last updated on 02/Jun/22

$$thankssir$$

Commented by mathlove last updated on 02/Jun/22

$$thanksir$$

Commented by Tawa11 last updated on 08/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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