Problem:
Divide 56 into 2 parts such that 3 times the first part exceeds one third of the second by 48.

Solution:
Let's assume the two parts of 56 as x and y.

Step 1: Set up the equations:
According to the problem statement, we can form the following equation:
3x = (1/3)y + 48

Step 2: Simplify the equation:
To simplify the equation, we can multiply the entire equation by 3 to get rid of the fraction:
9x = y + 144

Step 3: Substitute the value of y:
Since we are given that the sum of the two parts is 56, we can substitute the value of y in terms of x in the equation:
9x = 56 - x

Step 4: Solve for x:
Simplifying the equation further, we get:
10x = 56

Dividing both sides by 10, we find:
x = 5.6

Step 5: Solve for y:
Substituting the value of x in the equation y = 56 - x, we find:
y = 56 - 5.6
y = 50.4

Step 6: Final answer:
Therefore, dividing 56 into two parts such that 3 times the first part exceeds one third of the second by 48, the two parts are 5.6 and 50.4.

Key Points:
- The problem requires dividing 56 into two parts.
- Let the two parts be x and y.
- According to the problem statement, 3 times the first part exceeds one third of the second by 48.
- Form the equation: 3x = (1/3)y + 48.
- Simplify the equation: 9x = y + 144.
- Substitute the value of y in terms of x: 9x = 56 - x.
- Solve for x: x = 5.6.
- Substitute the value of x to find y: y = 50.4.
- The two parts are 5.6 and 50.4.