If the price of 1 kg of potatoes is 10 rupees, 1 kg of onions is 0.50 rupees, and 1 kg of garlic is 0.25 rupees, and I purchase a total of 10 kg of these three items (potatoes, onions, and garlic) for a total of 10 rupees, how many kilograms of each item did I buy?

Let the quantities of potatoes, onions, and garlic that you bought be denoted as:

- \( x \) kg of potatoes
- \( y \) kg of onions
- \( z \) kg of garlic

The problem gives us the following conditions:

1. The total weight is 10 kg:
   \[
   x + y + z = 10
   \]
   
2. The total cost is 10 rupees, with the cost of 1 kg of potatoes being 10 rupees, 1 kg of onions being 0.50 rupees, and 1 kg of garlic being 0.25 rupees:
   \[
   10x + 0.50y + 0.25z = 10
   \]

We now have a system of two equations:
1. \( x + y + z = 10 \)
2. \( 10x + 0.50y + 0.25z = 10 \)

We can solve this system of equations for \( x \), \( y \), and \( z \).

Let's proceed to solve these equations.

First, multiply the second equation by 4 to eliminate the decimals:
\[
40x + 2y + z = 40
\]

Now we have the system:
1. \( x + y + z = 10 \)
2. \( 40x + 2y + z = 40 \)

We will solve these equations by substitution or elimination method.The solution obtained is expressed in terms of \( z \) (the amount of garlic). To understand it better:

- \( x = \frac{z}{38} + \frac{10}{19} \) (amount of potatoes)
- \( y = \frac{180}{19} - \frac{39z}{38} \) (amount of onions)

This solution indicates that the values of \( x \), \( y \), and \( z \) depend on each other. To find realistic values for the weights (since they must be non-negative and should fit the total of 10 kg), we will now compute a particular set of values for \( z \) and check whether they satisfy the system.

Let me compute specific possible values for \( z \).Here are the possible solutions, showing the quantities of potatoes, onions, and garlic (in kg) that satisfy the conditions:

1. \( x = 0.53 \) kg of potatoes, \( y = 9.47 \) kg of onions, \( z = 0 \) kg of garlic
2. \( x = 0.58 \) kg of potatoes, \( y = 7.42 \) kg of onions, \( z = 2 \) kg of garlic
3. \( x = 0.61 \) kg of potatoes, \( y = 6.39 \) kg of onions, \( z = 3 \) kg of garlic
4. \( x = 0.63 \) kg of potatoes, \( y = 5.37 \) kg of onions, \( z = 4 \) kg of garlic
5. \( x = 0.66 \) kg of potatoes, \( y = 4.34 \) kg of onions, \( z = 5 \) kg of garlic
6. \( x = 0.68 \) kg of potatoes, \( y = 3.32 \) kg of onions, \( z = 6 \) kg of garlic
7. \( x = 0.71 \) kg of potatoes, \( y = 2.29 \) kg of onions, \( z = 7 \) kg of garlic
8. \( x = 0.76 \) kg of potatoes, \( y = 0.24 \) kg of onions, \( z = 9 \) kg of garlic

Do you have any preferences, or would you like to explore one of these solutions in more detail?

Here are 5 related questions to expand on this topic:

1. How would the solution change if the price of onions increased to 1 rupee per kg?
2. What would the total cost be if you bought 15 kg of these items with the same price ratios?
3. Can you determine how much you would spend if you bought 5 kg of garlic only?
4. How can you generalize this problem for any number of different goods with given prices?
5. How do systems of linear equations help in solving real-world cost and weight problems?

**Tip:** When solving systems of equations, substitution or elimination methods work best when you have two variables. Keep the system balanced by ensuring the units of measurement and coefficients align.

Discover how to solve a system of linear equations to determine how many kilograms of potatoes, onions, and garlic you can buy with 10 rupees and a total weight of 10 kg. The solution involves key algebraic concepts such as substitution and elimination methods. Suitable for students in Grades 9-11.

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