Solve the following pair of linear equations by the substitution method:
`(3x)/(2)- (5y)/(3)= -2`
`(x)/(3) + (y)/(2)= (13)/(6)`

Answer

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

### Given Equations:
1. \(\frac{3x}{2} - \frac{5y}{3} = -2\)
2. \(\frac{x}{3} + \frac{y}{2} = \frac{13}{6}\)

### Step 1: Convert the equations to standard form
First, we will eliminate the fractions in both equations.

**For the first equation:**
Multiply through by 6 (the least common multiple of 2 and 3):
\[ 6 \left(\frac{3x}{2}\right) - 6 \left(\frac{5y}{3}\right) = 6(-2) \]
This simplifies to:
\[ 9x - 10y = -12 \quad \text{(Equation 1)} \]

**For the second equation:**
Multiply through by 6:
\[ 6 \left(\frac{x}{3}\right) + 6 \left(\frac{y}{2}\right) = 6 \left(\frac{13}{6}\right) \]
This simplifies to:
\[ 2x + 3y = 13 \quad \text{(Equation 2)} \]

### Step 2: Solve one equation for one variable
Let's solve Equation 2 for \(x\):
\[ 2x + 3y = 13 \]
Rearranging gives:
\[ 2x = 13 - 3y \]
Dividing by 2:
\[ x = \frac{13 - 3y}{2} \quad \text{(Equation 3)} \]

### Step 3: Substitute into the other equation
Now, substitute Equation 3 into Equation 1:
\[ 9\left(\frac{13 - 3y}{2}\right) - 10y = -12 \]
Multiply through by 2 to eliminate the fraction:
\[ 9(13 - 3y) - 20y = -24 \]
Distributing gives:
\[ 117 - 27y - 20y = -24 \]
Combine like terms:
\[ 117 - 47y = -24 \]
Rearranging gives:
\[ -47y = -24 - 117 \]
\[ -47y = -141 \]
Dividing by -47:
\[ y = \frac{141}{47} = 3 \]

### Step 4: Substitute back to find \(x\)
Now substitute \(y = 3\) back into Equation 3:
\[ x = \frac{13 - 3(3)}{2} \]
This simplifies to:
\[ x = \frac{13 - 9}{2} = \frac{4}{2} = 2 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = 2, \quad y = 3 \]