` 2x + 3y + 1 = 0 ` ,
` ( 7- 4x )/(3) = y`

Options

A. ` x = 4, y = - 3`
B. ` x = 2, y = - 3`
C. ` x = 4, y = - 4`
D. ` x = 5, y = - 3`

Explanation

To solve the given system of linear equations, we have:

1. \( 2x + 3y + 1 = 0 \) (Equation 1)
2. \( \frac{7 - 4x}{3} = y \) (Equation 2)

### Step 1: Rewrite Equation 2 in standard form
First, we will rewrite Equation 2 to make it easier to work with. We can multiply both sides by 3 to eliminate the fraction:

\[ 7 - 4x = 3y \]

Rearranging this gives us:

\[ 4x + 3y - 7 = 0 \quad \text{(Equation 3)} \]

### Step 2: Set up the equations for elimination
Now we have two equations:

1. \( 2x + 3y + 1 = 0 \) (Equation 1)
2. \( 4x + 3y - 7 = 0 \) (Equation 3)

Next, we can use the elimination method to eliminate \( y \). We will subtract Equation 1 from Equation 3.

### Step 3: Subtract Equation 1 from Equation 3
Subtracting Equation 1 from Equation 3 gives:

\[ (4x + 3y - 7) - (2x + 3y + 1) = 0 \]

This simplifies to:

\[ 4x + 3y - 7 - 2x - 3y - 1 = 0 \]

Combining like terms results in:

\[ 2x - 8 = 0 \]

### Step 4: Solve for \( x \)
Now, we can solve for \( x \):

\[ 2x = 8 \implies x = 4 \]

### Step 5: Substitute \( x \) back into one of the equations to find \( y \)
We can substitute \( x = 4 \) back into Equation 2 to find \( y \):

\[ y = \frac{7 - 4(4)}{3} \]

Calculating this gives:

\[ y = \frac{7 - 16}{3} = \frac{-9}{3} = -3 \]

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

\[ x = 4, \quad y = -3 \]