Problem:
A shopkeeper buys a number of books for Rs. 80. If he had bought 4 more books for the same amount, each book would have cost Rs. 1 less. How many books did he buy?

Solution:

Given:
- Cost of books bought initially = Rs. 80
- Number of books bought initially = x

Step 1: Set up the equation for the given information
Let's assume the cost of each book initially is y.

- Cost of x books = Rs. 80 (Equation 1)
- Cost of (x + 4) books = Rs. 80 (Equation 2)
- Cost of each book if (x + 4) books were bought initially = y - 1 (Equation 3)

Step 2: Solve the equation
From Equation 1, we can write:
y * x = 80

From Equation 2, we can write:
(y - 1) * (x + 4) = 80

Step 3: Simplify the equation
Expanding Equation 2, we get:
yx + 4y - x - 4 = 80

Rearranging the terms, we get:
yx - x + 4y - 4 = 80

Simplifying further, we get:
yx - x + 4y = 84 (Equation 4)

Step 4: Substitute the value of yx from Equation 1 into Equation 4
Substituting the value of yx from Equation 1, we get:
80 - x + 4y = 84

Rearranging the terms, we get:
- x + 4y = 4 (Equation 5)

Step 5: Solve the system of equations
We now have two equations:
yx = 80 (Equation 1)
- x + 4y = 4 (Equation 5)

We can solve this system of equations using substitution or elimination method.

Step 6: Solve the system of equations using elimination method
Multiply Equation 5 by y:
- yx + 4y^2 = 4y

Substitute the value of yx from Equation 1 into the above equation:
- 80 + 4y^2 = 4y

Rearranging the terms, we get a quadratic equation:
4y^2 - 4y - 80 = 0

Simplifying the equation, we get:
y^2 - y - 20 = 0

Factoring the quadratic equation, we get:
(y - 5)(y + 4) = 0

So, y - 5 = 0 or y + 4 = 0
y = 5 or y = -4

Since the cost of a book cannot be negative, we consider y = 5.

Step 7: Find the number of books
Substituting the value of y = 5 into Equation 1, we get:
5x = 80

Solving for x, we get:
x = 80/5
x = 16

Therefore, the