Q.Using $\bar{x}=3$, $\bar{y}=4$, $b_{yx}=0.6$ and $b_{xy}=1.0$ from Worked Examples 1, 2 and 5, write down the two regression line equations and verify that they intersect at $(\bar{x},\bar{y})$.

Substitute the known means and coefficients directly into the two regression-line formulas, then solve them together to check where they cross.

Step 1 — Write the regression line of $y$ on $x$. $y - \bar{y} = b_{yx}(x-\bar{x})$ gives $y - 4 = 0.6(x-3)$, i.e. $y = 0.6x - 1.8 + 4 = 0.6x + 2.2$.

Step 2 — Write the regression line of $x$ on $y$. $x - \bar{x} = b_{xy}(y-\bar{y})$ gives $x - 3 = 1.0(y-4)$, i.e. $x = y - 1$, which rearranges to $y = x + 1$.

Step 3 — Solve the two equations simultaneously to find their intersection. Setting the two expressions for $y$ equal: $0.6x + 2.2 = x + 1$. Rearranging, $2.2 - 1 = x - 0.6x$, so $1.2 = 0.4x$, giving $x = 3$. Substituting back, $y = x+1 = 3+1 = 4$.

Step 4 — Confirm this matches the known means. The intersection point found, $(3,4)$, is exactly $(\bar{x},\bar{y}) = (3,4)$ — confirming, as the theory predicts, that both regression lines pass through the point of means. …

The zero-substitution check in the Independent check above (plugging $x=\bar{x}, y=\bar{y}$ directly into the un-rearranged deviation-form equations) is a faster way to verify the intersection than solving the system from scratch, and …

Forgetting to rearrange $x-\bar{x}=b_{xy}(y-\bar{y})$ into a $y=\ldots$ form before equating it with the other line (or equating the two equations in inconsistent forms) is the usual algebra slip. It is also easy to mix up which coefficient, $b_{yx}$ or $b_{xy}$, belongs in which equatio …