Solve the following pair of linear equations using method of elimination by equating coefficients .
`{:((x-y)/(6)=2(4-x)),(2x+4=3(x-4)):}`

Answer

To solve the given pair of linear equations using the method of elimination by equating coefficients, we will follow these steps:

### Step 1: Write the equations in standard form

The given equations are:
1. \(\frac{x - y}{6} = 2(4 - x)\)
2. \(2x + 4 = 3(x - 4)\)

Let's start with the first equation:

\[ \frac{x - y}{6} = 2(4 - x) \]

Multiply both sides by 6 to eliminate the fraction:

\[ x - y = 12(4 - x) \]

Now, distribute on the right side:

\[ x - y = 48 - 12x \]

Rearranging gives:

\[ x + 12x - y = 48 \]

This simplifies to:

\[ 13x - y = 48 \quad \text{(Equation 1)} \]

Now, for the second equation:

\[ 2x + 4 = 3(x - 4) \]

Distributing on the right side gives:

\[ 2x + 4 = 3x - 12 \]

Rearranging gives:

\[ 2x - 3x + 4 + 12 = 0 \]

This simplifies to:

\[ -x + 16 = 0 \]

Or:

\[ x = 16 \quad \text{(Equation 2)} \]

### Step 2: Substitute the value of \(x\) into Equation 1

Now we have two equations:
1. \(13x - y = 48\)
2. \(x = 16\)

Substituting \(x = 16\) into Equation 1:

\[ 13(16) - y = 48 \]

Calculating \(13 \times 16\):

\[ 208 - y = 48 \]

### Step 3: Solve for \(y\)

Rearranging gives:

\[ y = 208 - 48 \]

Calculating the right side:

\[ y = 160 \]

### Final Solution

Thus, the solution to the system of equations is:

\[ x = 16, \quad y = 160 \]