The regression equation of x on y is 40x-18y = 214… (i)
The regression equation of y on x is 8x -10y +66 =0….. (ii)
Solving equations i and ii,
`barx=square`
`bary=square`
`:. b_(yx) = (square)/(square)`
`:. b_(xy) = (square)/(square)`
`:. r = square`

Answer

To solve the problem, we need to find the values of \( \bar{x} \), \( \bar{y} \), \( b_{yx} \), \( b_{xy} \), and \( r \) using the given regression equations.

### Step 1: Solve the regression equations for \( \bar{x} \) and \( \bar{y} \)

We have the following regression equations:
1. \( 40x - 18y = 214 \) (Equation i)
2. \( 8x - 10y + 66 = 0 \) (Equation ii)

First, we can rearrange Equation ii:
\[ 8x - 10y = -66 \quad \text{(Rearranging)} \]

Now we have:
1. \( 40x - 18y = 214 \)
2. \( 8x - 10y = -66 \)

### Step 2: Simplify the equations

Let's simplify both equations. We can divide the first equation by 2:
\[ 20x - 9y = 107 \quad \text{(Equation i simplified)} \]

And divide the second equation by 2:
\[ 4x - 5y = -33 \quad \text{(Equation ii simplified)} \]

### Step 3: Solve the system of equations

Now we can solve these two equations simultaneously. We can express \( x \) in terms of \( y \) from the second equation:
\[ 4x = 5y - 33 \implies x = \frac{5y - 33}{4} \]

Now substitute \( x \) in the first equation:
\[ 20\left(\frac{5y - 33}{4}\right) - 9y = 107 \]

Multiply through by 4 to eliminate the fraction:
\[ 20(5y - 33) - 36y = 428 \]
\[ 100y - 660 - 36y = 428 \]
\[ 64y - 660 = 428 \]
\[ 64y = 428 + 660 \]
\[ 64y = 1088 \]
\[ y = \frac{1088}{64} = 17 \]

### Step 4: Find \( x \)

Now substitute \( y = 17 \) back into the equation for \( x \):
\[ 4x = 5(17) - 33 \]
\[ 4x = 85 - 33 \]
\[ 4x = 52 \implies x = \frac{52}{4} = 13 \]

Thus, we have:
\[ \bar{x} = 13, \quad \bar{y} = 17 \]

### Step 5: Find the regression coefficients

Next, we find the regression coefficients \( b_{xy} \) and \( b_{yx} \).

**For \( b_{xy} \)** (from equation i):
The regression equation can be written as:
\[ x - \bar{x} = b_{xy}(y - \bar{y}) \]
Rearranging gives:
\[ 40x - 18y = 214 \implies x - 13 = \frac{18}{40}(y - 17) \]
Thus,
\[ b_{xy} = \frac{18}{40} = \frac{9}{20} \]

**For \( b_{yx} \)** (from equation ii):
The regression equation can be written as:
\[ y - \bar{y} = b_{yx}(x - \bar{x}) \]
Rearranging gives:
\[ 8x - 10y + 66 = 0 \implies y - 17 = \frac{8}{10}(x - 13) \]
Thus,
\[ b_{yx} = \frac{8}{10} = \frac{4}{5} \]

### Step 6: Find the correlation coefficient \( r \)

The correlation coefficient \( r \) can be calculated using:
\[ r = \sqrt{b_{yx} \cdot b_{xy}} = \sqrt{\left(\frac{4}{5}\right) \cdot \left(\frac{9}{20}\right)} \]
Calculating this gives:
\[ r = \sqrt{\frac{36}{100}} = \frac{6}{10} = 0.6 \]

### Summary of Results
- \( \bar{x} = 13 \)
- \( \bar{y} = 17 \)
- \( b_{yx} = \frac{4}{5} \)
- \( b_{xy} = \frac{9}{20} \)
- \( r = 0.6 \)