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If $\mathrm{f}(x)=\frac{3 x+4}{5 x-7}$ and $\mathrm{g}(x)=\frac{7 x+4}{5 x-3}$, then $\mathrm{f}(\mathrm{g}(x))=$
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$g(f(x))$
$\begin{aligned} f(g(x)) & =f\left(\frac{7 x+4}{5 x-3}\right) \\ & =\frac{3\left(\frac{7 x+4}{5 x-3}\right)+4}{5\left(\frac{7 x+4}{5 x-3}\right)-7} \\ & =\frac{21 x+12+20 x-12}{35 x+20-35 x+21} \\ & =\frac{41 x}{41} \\ & =x\end{aligned}$
$\begin{aligned} & \text { Now, } \mathrm{g}(\mathrm{f}(x))=\mathrm{g}\left(\frac{3 x+4}{5 x-7}\right) \\ &=\frac{7\left(\frac{3 x+4}{5 x-7}\right)+4}{5\left(\frac{3 x+4}{5 x-7}\right)-3} \\ &=\frac{21 x+28+20 x-28}{15 x+20-15 x+21} \\ &=\frac{41 x}{41} \\ &=x \\ & \mathrm{f}(\mathrm{g}(x))=\mathrm{g}(\mathrm{f}(x))\end{aligned}$
$\begin{aligned} & \text { Now, } \mathrm{g}(\mathrm{f}(x))=\mathrm{g}\left(\frac{3 x+4}{5 x-7}\right) \\ &=\frac{7\left(\frac{3 x+4}{5 x-7}\right)+4}{5\left(\frac{3 x+4}{5 x-7}\right)-3} \\ &=\frac{21 x+28+20 x-28}{15 x+20-15 x+21} \\ &=\frac{41 x}{41} \\ &=x \\ & \mathrm{f}(\mathrm{g}(x))=\mathrm{g}(\mathrm{f}(x))\end{aligned}$
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