1. Erika bought 25 kilograms of meat consisting of beef and pork. She bought P3,500 worth of beef and P3,900 worth of pork. If the cost per kilogram of beef is P90 more than that of pork, how many kilograms of each meat did she buy?

Solution and answer:

Let's define the following variables:
- Let \( x \) be the number of kilograms of beef Erika bought.
- Let \( y \) be the number of kilograms of pork Erika bought.

From the problem, we know the following:
1. Erika bought a total of 25 kg of meat:
   \[
   x + y = 25 \quad \text{(Equation 1)}
   \]

2. The total cost of the beef is P3,500, and the total cost of the pork is P3,900. Let the cost per kilogram of pork be \( p \). Since the cost per kilogram of beef is P90 more than the cost per kilogram of pork, the cost per kilogram of beef is \( p + 90 \). Therefore, the cost equations are:
   \[
   x(p + 90) = 3,500 \quad \text{(Equation 2)}
   \]
   \[
   y(p) = 3,900 \quad \text{(Equation 3)}
   \]

### Step 1: Solve for \( p \) (Cost per kilogram of pork)

From Equation 3:
\[
y \cdot p = 3,900
\]
Solving for \( p \):
\[
p = \frac{3,900}{y}
\]

### Step 2: Substitute into Equation 2

Substitute \( p = \frac{3,900}{y} \) into Equation 2:
\[
x \left( \frac{3,900}{y} + 90 \right) = 3,500
\]
Simplify:
\[
x \left( \frac{3,900 + 90y}{y} \right) = 3,500
\]
Multiply both sides by \( y \) to eliminate the denominator:
\[
x(3,900 + 90y) = 3,500y
\]
Expand:
\[
x \cdot 3,900 + 90xy = 3,500y
\]

### Step 3: Solve for \( x \) and \( y \)

Now we have the system of equations to solve for \( x \) and \( y \). I'll solve it and provide the answer.The solution to the system of equations gives the following feasible result:

- Erika bought **10 kilograms of beef** and **15 kilograms of pork**.

Thus, she purchased 10 kg of beef and 15 kg of pork.

Would you like more details on any steps? Here are five related questions you could explore:

1. What if the price per kilogram of beef was P100 more than pork instead?
2. How much would it cost if Erika bought only pork?
3. How would the solution change if the total weight was 30 kg instead of 25 kg?
4. How would you solve the problem if the cost of pork was not given?
5. How would the equations change if there was a third type of meat?

**Tip:** When solving systems of equations, try substitution if one equation can easily express a variable in terms of another.

Erika bought 25 kilograms of meat consisting of beef and pork. She bought P3,500 worth of beef and P3,900 worth of pork. If the cost per kilogram of beef is P90 more than that of pork, how many kilograms of each meat did she buy?

This problem explores solving a system of linear equations involving the purchase of 25 kilograms of meat consisting of beef and pork, with prices given and a cost difference per kilogram. Learn how to use algebraic methods to solve for the weight of each meat. Suitable for students in Grades 9-11.

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