A study of two variables gives the two regression lines 6x - 5y + 10 = 0 and 10x - 3y - 26 = 0…

Step 1. The means. Solving $6x - 5y = -10$ and $10x - 3y = 26$ gives $x = 5$ and $y = 8$. So $\bar{x} = 5$ and $\bar{y} = 8$.

Step 2. Try an assignment. Take the first line as $y$ on $x$. Then $y = 1.2x + 2$ and $b_{yx} = 1.2$. The second line is then $x$ on $y$: $x = 0.3y + 2.6$, so $b_{xy} = 0.3$. The product is $0.36 \le 1$, so this assignment is possible.

Step 3. The other assignment. It would give $b_{yx} = \dfrac{10}{3}$ and $b_{xy} = \dfrac{5}{6}$, with product $2.78 > 1$. This is impossible.

Step 4. The correlation. $r = +\sqrt{0.36} = 0.6$, positive because both coefficients are positive.

Step 5. The spread of $y$. $b_{yx} = r\,\dfrac{\sigma_y}{\sigma_x}$ gives $1.2 = 0.6 \times \dfrac{\sigma_y}{4}$, so $\sigma_y = 8$.