Solve algebraically: `4x + 3y = 14` and 3x - 4y = 23.

Answer

To solve the system of equations algebraically, we have the following two equations:

1. \( 4x + 3y = 14 \) (Equation 1)
2. \( 3x - 4y = 23 \) (Equation 2)

### Step 1: Equalize the coefficients
To eliminate one of the variables, we can equalize the coefficients of \( x \) or \( y \). Here, we will equalize the coefficients of \( x \). We can do this by multiplying Equation 1 by 3 and Equation 2 by 4.

- Multiply Equation 1 by 3:
\[ 3(4x + 3y) = 3(14) \implies 12x + 9y = 42 \quad \text{(Equation 3)} \]

- Multiply Equation 2 by 4:
\[ 4(3x - 4y) = 4(23) \implies 12x - 16y = 92 \quad \text{(Equation 4)} \]

### Step 2: Subtract the equations
Now we will subtract Equation 4 from Equation 3 to eliminate \( x \):

\[ (12x + 9y) - (12x - 16y) = 42 - 92 \]

This simplifies to:
\[ 9y + 16y = 42 - 92 \]
\[ 25y = -50 \]

### Step 3: Solve for \( y \)
Now, we can solve for \( y \):
\[ y = \frac{-50}{25} = -2 \]

### Step 4: Substitute \( y \) back into one of the original equations
Now that we have \( y = -2 \), we can substitute this value back into Equation 1 to find \( x \):

\[ 4x + 3(-2) = 14 \]
\[ 4x - 6 = 14 \]
\[ 4x = 14 + 6 \]
\[ 4x = 20 \]
\[ x = \frac{20}{4} = 5 \]

### Final Solution
The solution to the system of equations is:
\[ x = 5, \quad y = -2 \]