is there an error in this problem 3(4+3a)=21a

$$\begin{array}{rll}
3(4+3a)&=&21a\\
3(4+3a)/3&=&21/3a\\
4+3a&=&7a\\
4+3a-3a&=&7a-3a\\
4&=&4a\\
a&=&1\\
\end{array}$$

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