Question

If $4x - 5y + 33 = 0$ and $20x - 9y - 107 = 0$ are two lines of regression, then what are the values of $x$ and $y$ respectively.

Answer 1

First we have to know what are meant by the line of regression, A regression line is just a single line which best fits the data such that the overall distance from the line to the points (variable values) plotted on a graph is the smallest. The formula of the regression line or a best-fitting line is $y = mx + b$, here m is the slope of the line and b is the y-intercept of the line. In short a regression line is used to minimize the squared deviations of proportions called as the regression line.

Complete step-by-step solution:
Here given the equations of the lines:
$\Rightarrow 4x - 5y + 33 = 0$; and
$\Rightarrow 20x - 9y - 107 = 0$
Let us suppose $4x - 5y + 33 = 0$, as the equation 1.
And let $20x - 9y - 107 = 0$, be the equation 2.
Now to solve for x and y by the method of solving the two equations, subtraction of equation 1 and equation 2.
Now multiplying the equation 1 with 5, so as to match the coefficient of x in the equation 2, which is 20 :
$\Rightarrow 5(4x - 5y + 33 = 0)$
$\Rightarrow 20x - 25y + 165 = 0$
Now the coefficient of x in equation 1 is 20, which is the same as the coefficient of x in equation 2 which is also 20.
Now subtracting the equation 2 from equation 1, to get the value of y :
$20x - 25y + 165 = 0$
$20x - 9y - 107 = 0$
$0 - 25y + 9y + 165 + 107 = 0$
Now the obtained equation is :
$\Rightarrow - 16y + 272 = 0$
$\Rightarrow 16y = 272$

$\Rightarrow y = 17$
$\therefore y = 17$
To find the value of x, substitute the solved value of $y = 17$in the equation 1 :
$\Rightarrow 4x - 5y + 33 = 0$
$\Rightarrow 4x - 5(17) + 33 = 0$
$\Rightarrow 4x - 85 + 33 = 0$
$\Rightarrow 4x = 52$
$\Rightarrow x = 13$
$\therefore x = 13$
$\therefore$The values of x and y are 13 and 17 respectively.

The values of $x = 13$ and $y = 17$.

Note: Here while solving for x and y equation 2 is subtracted from equation 1 which is multiplied with 5, so as to match the coefficients of x, to find the value of y, it can be in another method also where instead solving for y first, can solve for x, by multiplying the equation 1 with 9/5 so as to match the coefficients of y, where now x can be extracted by subtracting the equations. Either of the methods finally give the same solution.