Express y in terms of x from the following equation.
3x - 2y + 6 = 0
and check whether `(-(5)/(2), - (3)/(4))` is a point on the given line.

Answer

To solve the problem, we will follow these steps:

### Step 1: Rearranging the equation to express y in terms of x
We start with the equation:
\[ 3x - 2y + 6 = 0 \]

To express \( y \) in terms of \( x \), we need to isolate \( y \). First, we can move the terms involving \( x \) and the constant to the right-hand side (RHS) of the equation:
\[ -2y = -3x - 6 \]

### Step 2: Dividing by -2
Next, we divide the entire equation by -2 to solve for \( y \):
\[ y = \frac{3x + 6}{2} \]

This can also be simplified:
\[ y = \frac{3}{2}x + 3 \]

### Step 3: Checking if the point \((-5/2, -3/4)\) lies on the line
Now we need to check if the point \((-5/2, -3/4)\) satisfies the original equation. We will substitute \( x = -\frac{5}{2} \) and \( y = -\frac{3}{4} \) into the left-hand side (LHS) of the equation:
\[ LHS = 3\left(-\frac{5}{2}\right) - 2\left(-\frac{3}{4}\right) + 6 \]

### Step 4: Calculating the LHS
Calculating each term:
1. \( 3 \times -\frac{5}{2} = -\frac{15}{2} \)
2. \( -2 \times -\frac{3}{4} = \frac{3}{2} \)
3. Adding these with 6:
\[ LHS = -\frac{15}{2} + \frac{3}{2} + 6 \]

Now, convert 6 to have a common denominator of 2:
\[ 6 = \frac{12}{2} \]
So,
\[ LHS = -\frac{15}{2} + \frac{3}{2} + \frac{12}{2} \]

Combining these:
\[ LHS = \frac{-15 + 3 + 12}{2} = \frac{0}{2} = 0 \]

### Step 5: Conclusion
Since \( LHS = 0 \), which is equal to the right-hand side (RHS) of the original equation, the point \((-5/2, -3/4)\) lies on the line.

### Final Answer
Thus, we have expressed \( y \) in terms of \( x \) as:
\[ y = \frac{3}{2}x + 3 \]
And verified that the point \((-5/2, -3/4)\) lies on the line.

---