In the book of sakurai Modern Quantum Mechanics they have this $$ \begin{aligned} & \sqrt{(j \mp m)(j \pm m+1)}\left\langle j_1 j_2 ; m_1 m_2 \mid j_1 j_2 ; j, m \pm 1\right\rangle \\ & =\sqrt{\left(j_1 \mp m_1+1\right)\left(j_1 \pm m_1\right)}\left\langle j_1 j_2 ; m_1 \mp 1, m_2 \mid j_1 j_2 ; j m\right\rangle \\ & \quad+\sqrt{\left(j_2 \mp m_2+1\right)\left(j_2 \pm m_2\right)}\left\langle j_1 j_2 ; m_1, m_2 \mp 1 \mid j_1 j_2 ; j m\right\rangle . \end{aligned} \tag 1$$ with $m_1+m_2=m \pm 1$. Then he derive $$ \begin{aligned} & \sqrt{\left(j^{\prime} \pm m^{\prime}\right)\left(j^{\prime} \mp m^{\prime}+1\right)}\left\langle\alpha^{\prime}, j^{\prime}, m^{\prime} \mp 1\right| T_q^{(k)}|\alpha, j m\rangle \\ &=\sqrt{(j \mp m)(j \pm m+1)}\left\langle\alpha^{\prime}, j^{\prime} m^{\prime}\right| T_q^{(k)}|\alpha, j, m \pm 1\rangle \\ &+\sqrt{(k \mp q)(k \pm q+1)}\left\langle\alpha^{\prime}, j^{\prime} m^{\prime}\right| T_{q \pm 1}^{(k)}|\alpha, j m\rangle . \end{aligned} $$
And continue Compare this with the recursion relation for the Clebsch-Gordan coefficient (1). Note the striking similarity if we substitute $j^{\prime} \rightarrow j$, $m^{\prime} \rightarrow m$, $j \rightarrow j_1$, $m \rightarrow m_1$, $k \rightarrow j_2$, and $q \rightarrow m_2$. Both recursion relations are of the form $\sum_j a_{i j} x_j=0$, that is, first-order linear homogeneous equations with the same coefficients $a_{i j}$. Whenever we have
$$ \sum_j a_{i j} x_j=0, \quad \sum_j a_{i j} y_j=0, $$
we cannot solve for the $x_j$ (or $y_j$ ) individually but we can solve for the ratios; so
$$ \frac{x_j}{x_k}=\frac{y_j}{y_k} \quad \text { or } \quad x_j=c y_j. $$
How did he obtain $$ \frac{x_j}{x_k}=\frac{y_j}{y_k} \ ? $$
