A bucket when filled `(5)/(7)` with water weights 12 kg and filled `(3)/(4)` of it with water weight 12.5 kg. What is the weight of the empty bucket?

Answer

To find the weight of the empty bucket, we will set up a system of equations based on the information given in the problem.

Let:
- \( x \) = weight of the empty bucket (in kg)
- \( y \) = weight of the water when the bucket is full (in kg)

### Step 1: Set up the equations

From the problem, we have two scenarios:

1. When the bucket is filled \( \frac{5}{7} \) with water, the total weight is 12 kg:
\[ x + \frac{5}{7}y = 12 \quad \text{(Equation 1)} \]

2. When the bucket is filled \( \frac{3}{4} \) with water, the total weight is 12.5 kg:
\[ x + \frac{3}{4}y = 12.5 \quad \text{(Equation 2)} \]

### Step 2: Clear the fractions

To eliminate the fractions, we can multiply each equation by a common denominator.

For Equation 1, multiply by 7:
\[ 7x + 5y = 84 \quad \text{(Equation 3)} \]

For Equation 2, multiply by 4:
\[ 4x + 3y = 50 \quad \text{(Equation 4)} \]

### Step 3: Solve the system of equations

Now we have the following two equations:
1. \( 7x + 5y = 84 \) (Equation 3)
2. \( 4x + 3y = 50 \) (Equation 4)

Next, we can solve these equations using the elimination method.

### Step 4: Eliminate \( y \)

To eliminate \( y \), we can multiply Equation 3 by 3 and Equation 4 by 5:
\[ 21x + 15y = 252 \quad \text{(Equation 5)} \]
\[ 20x + 15y = 250 \quad \text{(Equation 6)} \]

Now, subtract Equation 6 from Equation 5:
\[ (21x + 15y) - (20x + 15y) = 252 - 250 \]
This simplifies to:
\[ x = 2 \]

### Step 5: Find the weight of the empty bucket

Now that we have \( x = 2 \), we can conclude that the weight of the empty bucket is:
\[ \text{Weight of the empty bucket} = 2 \text{ kg} \]

### Summary

The weight of the empty bucket is \( 2 \) kg.

---