For solving each pair of equations, in this exercise, use the method of elimination by equation coefficiients :
`(5y)/(2) - (x)/(3) =8`
`(y)/(2) + (5x)/(3) = 12`

Answer

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

### Step 1: Write the equations in standard form
The given equations are:
1. \(\frac{5y}{2} - \frac{x}{3} = 8\)
2. \(\frac{y}{2} + \frac{5x}{3} = 12\)

To eliminate the fractions, we can multiply each equation by the least common multiple (LCM) of the denominators. The LCM of 2 and 3 is 6.

**Equation 1:**
Multiply by 6:
\[ 6 \left(\frac{5y}{2}\right) - 6 \left(\frac{x}{3}\right) = 6 \cdot 8 \]
This simplifies to:
\[ 15y - 2x = 48 \quad \text{(Equation 1')} \]

**Equation 2:**
Multiply by 6:
\[ 6 \left(\frac{y}{2}\right) + 6 \left(\frac{5x}{3}\right) = 6 \cdot 12 \]
This simplifies to:
\[ 3y + 10x = 72 \quad \text{(Equation 2')} \]

### Step 2: Rearrange the equations
Now we have:
1. \(15y - 2x = 48\) (Equation 1')
2. \(3y + 10x = 72\) (Equation 2')

### Step 3: Eliminate one variable
To eliminate \(x\), we can manipulate the equations. We will multiply Equation 1' by 5 to match the coefficient of \(x\) in Equation 2'.

Multiply Equation 1' by 5:
\[ 5(15y - 2x) = 5(48) \]
This gives:
\[ 75y - 10x = 240 \quad \text{(Equation 3)} \]

Now we have:
1. \(75y - 10x = 240\) (Equation 3)
2. \(3y + 10x = 72\) (Equation 2')

### Step 4: Add the equations
Now, add Equation 3 and Equation 2':
\[ (75y - 10x) + (3y + 10x) = 240 + 72 \]
This simplifies to:
\[ 78y = 312 \]

### Step 5: Solve for \(y\)
Now, divide both sides by 78:
\[ y = \frac{312}{78} = 4 \]

### Step 6: Substitute \(y\) back to find \(x\)
Now substitute \(y = 4\) back into one of the original equations. We can use Equation 1':
\[ 15(4) - 2x = 48 \]
This simplifies to:
\[ 60 - 2x = 48 \]
Now, solve for \(x\):
\[ -2x = 48 - 60 \]
\[ -2x = -12 \]
\[ x = 6 \]

### Final Solution
Thus, the solution to the equations is:
\[ x = 6, \quad y = 4 \]