If the two lines of regression are `2x-y-4=0` and `9x-2y-38=0`, then the means of x and y variates respectively are

Options

A. 8, 6
B. 6, 8
C. 7 , 10
D. 5, 6

Answer

To solve the problem, we need to find the means of the x and y variates from the given lines of regression. The two lines of regression are:

1. \( 2x - y - 4 = 0 \)
2. \( 9x - 2y - 38 = 0 \)

### Step 1: Rewrite the equations in slope-intercept form
We can rewrite both equations to express y in terms of x.

For the first equation:
\[ 2x - y - 4 = 0 \implies y = 2x - 4 \]

For the second equation:
\[ 9x - 2y - 38 = 0 \implies 2y = 9x - 38 \implies y = \frac{9}{2}x - 19 \]

### Step 2: Substitute \( x = \bar{x} \) and \( y = \bar{y} \)
The lines of regression pass through the point \( (\bar{x}, \bar{y}) \). We substitute \( \bar{x} \) and \( \bar{y} \) into both equations.

From the first equation:
\[ 2\bar{x} - \bar{y} - 4 = 0 \implies 2\bar{x} - \bar{y} = 4 \tag{1} \]

From the second equation:
\[ 9\bar{x} - 2\bar{y} - 38 = 0 \implies 9\bar{x} - 2\bar{y} = 38 \tag{2} \]

### Step 3: Solve the system of equations
Now we have a system of equations (1) and (2):
1. \( 2\bar{x} - \bar{y} = 4 \)
2. \( 9\bar{x} - 2\bar{y} = 38 \)

We can solve these equations using the elimination method.

From equation (1), we can express \( \bar{y} \):
\[ \bar{y} = 2\bar{x} - 4 \tag{3} \]

Substituting equation (3) into equation (2):
\[ 9\bar{x} - 2(2\bar{x} - 4) = 38 \]
\[ 9\bar{x} - 4\bar{x} + 8 = 38 \]
\[ 5\bar{x} + 8 = 38 \]
\[ 5\bar{x} = 30 \]
\[ \bar{x} = 6 \]

### Step 4: Find \( \bar{y} \)
Now substituting \( \bar{x} = 6 \) back into equation (3):
\[ \bar{y} = 2(6) - 4 \]
\[ \bar{y} = 12 - 4 \]
\[ \bar{y} = 8 \]

### Conclusion
The means of the x and y variates are:
\[ \bar{x} = 6, \quad \bar{y} = 8 \]

### Final Answer
Thus, the means of x and y variates respectively are \( (6, 8) \).

---