The general term in the expansion of \((2 + ax)^7\) is given by:

\(T_k = \binom{7}{k} (2)^{7-k} (ax)^k\).

For the coefficient of \(x\), set \(k = 1\):

\(T_1 = \binom{7}{1} (2)^6 (ax)^1 = 7 \times 2^6 \times a \times x\).

The coefficient of \(x\) is \(7 \times 2^6 \times a\).

For the coefficient of \(x^2\), set \(k = 2\):

\(T_2 = \binom{7}{2} (2)^5 (ax)^2 = 21 \times 2^5 \times a^2\).

The coefficient of \(x^2\) is \(21 \times 2^5 \times a^2\).

Equating the coefficients:

\(7 \times 2^6 \times a = 21 \times 2^5 \times a^2\).

Divide both sides by \(2^5\):

\(7 \times 2 \times a = 21 \times a^2\).

Divide both sides by \(a\) (since \(a \neq 0\)):

\(14 = 21a\).

Solve for \(a\):

\(a = \frac{14}{21} = \frac{2}{3}\).