Solve the following pair of linear (simultaneous) equations using method of substitution :
`6x = 7y + y`
` 7y-x = 8`

Answer

To solve the given pair of linear equations using the method of substitution, we will follow these steps:

### Given Equations:
1. \( 6x = 7y + y \)
2. \( 7y - x = 8 \)

### Step 1: Simplify the First Equation
First, we simplify the first equation:
\[ 6x = 7y + y \]
Combine like terms:
\[ 6x = 8y \]
Now, we can express \( x \) in terms of \( y \):
\[ x = \frac{8y}{6} = \frac{4y}{3} \]
Let's label this as Equation (1).

### Step 2: Substitute \( x \) in the Second Equation
Now, we will substitute the value of \( x \) from Equation (1) into the second equation:
\[ 7y - x = 8 \]
Substituting \( x \):
\[ 7y - \frac{4y}{3} = 8 \]

### Step 3: Solve for \( y \)
To eliminate the fraction, we can multiply the entire equation by 3 (the denominator):
\[ 3(7y) - 3\left(\frac{4y}{3}\right) = 3(8) \]
This simplifies to:
\[ 21y - 4y = 24 \]
Combine like terms:
\[ 17y = 24 \]
Now, solve for \( y \):
\[ y = \frac{24}{17} \]

### Step 4: Substitute \( y \) Back to Find \( x \)
Now that we have \( y \), we will substitute it back into Equation (1) to find \( x \):
\[ x = \frac{4y}{3} \]
Substituting \( y \):
\[ x = \frac{4 \cdot \frac{24}{17}}{3} \]
This simplifies to:
\[ x = \frac{96}{51} = \frac{32}{17} \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = \frac{32}{17}, \quad y = \frac{24}{17} \]