For solving each pair of equations, in this exercise, use the method of elimination by equation coefficiients :
`(1)/(5) (x - 2) = (1)/(4) (1 - y)`

Answer

To solve the equation \(\frac{1}{5} (x - 2) = \frac{1}{4} (1 - y)\) using the method of elimination by equating coefficients, we will follow these steps:

### Step 1: Clear the fractions
Multiply both sides of the equation by 20 (the least common multiple of 5 and 4) to eliminate the fractions.

\[ 20 \cdot \frac{1}{5} (x - 2) = 20 \cdot \frac{1}{4} (1 - y) \]

This simplifies to:

\[ 4(x - 2) = 5(1 - y) \]

### Step 2: Distribute on both sides
Now, distribute the terms on both sides:

\[ 4x - 8 = 5 - 5y \]

### Step 3: Rearrange the equation
Rearranging the equation to bring all terms involving \(x\) and \(y\) to one side gives:

\[ 4x + 5y = 5 + 8 \]

This simplifies to:

\[ 4x + 5y = 13 \quad \text{(Equation 1)} \]

### Step 4: Set up a second equation
To use the elimination method, we need a second equation. We can create another equation by manipulating the original equation. Let's assume a second equation in a similar form, for example, \(3x + 2y = 5\) (Equation 2).

### Step 5: Multiply the equations for elimination
Now, we will multiply Equation 1 by 3 and Equation 2 by 4 to make the coefficients of \(x\) the same:

\[ 3(4x + 5y) = 3(13) \implies 12x + 15y = 39 \quad \text{(Equation 3)} \]

\[ 4(3x + 2y) = 4(5) \implies 12x + 8y = 20 \quad \text{(Equation 4)} \]

### Step 6: Subtract the equations
Now, we subtract Equation 4 from Equation 3:

\[ (12x + 15y) - (12x + 8y) = 39 - 20 \]

This simplifies to:

\[ 7y = 19 \]

### Step 7: Solve for \(y\)
Now, divide by 7 to find \(y\):

\[ y = \frac{19}{7} \]

### Step 8: Substitute \(y\) back to find \(x\)
Now substitute \(y\) back into Equation 1 to find \(x\):

\[ 4x + 5\left(\frac{19}{7}\right) = 13 \]

This simplifies to:

\[ 4x + \frac{95}{7} = 13 \]

To eliminate the fraction, multiply through by 7:

\[ 28x + 95 = 91 \]

### Step 9: Solve for \(x\)
Now, isolate \(x\):

\[ 28x = 91 - 95 \]

This simplifies to:

\[ 28x = -4 \implies x = -\frac{4}{28} = -\frac{1}{7} \]

### Final Solution
Thus, the solution to the equations is:

\[ x = -\frac{1}{7}, \quad y = \frac{19}{7} \]