To get mean values we must solve the given lines.
4X – 5Y = -33 … (1)
20X – 9Y = 107 … (2)
(1) × 5 ⇒ 20X – 25Y = -165
20X – 9Y = 107
Subtracting (1) and (2), -16Y = -272
Y = \(\frac{272}{16}\) = 17
i.e., \(\bar{Y}\) = 17
Using Y = 17 in (1) we get, 4X – 85 = -33
4X = 85 – 33
4X = 52
X = 13
i.e., \(\bar{X}\) = 13
Mean values are \(\bar{X}\) = 13, \(\bar{Y}\) = 17,
Let regression line of Y on X be
4X – 5Y + 33 = 0
5Y = 4X + 33
Y = (4X + 33)
Y = \(\frac{1}{5}\)(4x + 33)
Y = \(\frac{4}{5}X+\frac{33}{5}\)
Y = 0.8X + 6.6
∴ byx = 0.8
Let regression line of X on Y be
20X – 9Y – 107 = 0
20X = 9Y + 107
X = \(\frac{1}{20}\)(9Y + 107)
X = \(\frac{9}{20}Y+\frac{107}{20}\)
X = 0.45Y + 5.35
∴ bxy = 0.45
Coefficient of correlation between X and Y is

= ±0.6
= 0.6
Both byx and bxy is positive take positive sign.