Justin sells mango and papaya in his fruit stand. Each mango costs one amount and each papaya costs another amount. 6 mango + 1 papaya cost 80 peso. 3 mango and 4 papaya cost 89 peso. At these prices, how many pesos do 12 mango and 5 papaya?In this problem can you make 4 solution on how to solve this and can you solve this using table ( refer to the method in solving routine word problem

Let's analyze the problem and use different solution strategies to find the cost of \(12\) mangoes and \(5\) papayas.

### Given Information

1. The cost of one mango = \( x \) pesos.
2. The cost of one papaya = \( y \) pesos.
3. **Equations based on the problem:**
   - \(6x + y = 80\)  (Equation 1)
   - \(3x + 4y = 89\) (Equation 2)
4. **Goal:** Find the cost of \(12\) mangoes and \(5\) papayas, or \(12x + 5y\).

### Solution Approaches

#### Solution 1: Elimination Method

1. Start with the equations:
   \[
   6x + y = 80
   \]
   \[
   3x + 4y = 89
   \]
2. Multiply Equation 1 by \(4\) to align the \(y\)-terms:
   \[
   4(6x + y) = 4(80)
   \]
   \[
   24x + 4y = 320
   \]
3. Now subtract Equation 2 from the result:
   \[
   (24x + 4y) - (3x + 4y) = 320 - 89
   \]
   \[
   21x = 231
   \]
   \[
   x = 11
   \]
4. Substitute \(x = 11\) into Equation 1:
   \[
   6(11) + y = 80
   \]
   \[
   66 + y = 80
   \]
   \[
   y = 14
   \]
5. Calculate \(12x + 5y\):
   \[
   12(11) + 5(14) = 132 + 70 = 202
   \]
   **Answer:** \(12\) mangoes and \(5\) papayas cost \(202\) pesos.

---

#### Solution 2: Substitution Method

1. From Equation 1, solve for \(y\):
   \[
   y = 80 - 6x
   \]
2. Substitute \(y = 80 - 6x\) into Equation 2:
   \[
   3x + 4(80 - 6x) = 89
   \]
   \[
   3x + 320 - 24x = 89
   \]
   \[
   -21x = -231
   \]
   \[
   x = 11
   \]
3. Substitute \(x = 11\) back into \(y = 80 - 6x\):
   \[
   y = 80 - 6(11) = 80 - 66 = 14
   \]
4. Calculate \(12x + 5y\):
   \[
   12(11) + 5(14) = 132 + 70 = 202
   \]
   **Answer:** \(202\) pesos.

---

#### Solution 3: Matrix Method

1. Write the system in matrix form:
   \[
   \begin{pmatrix} 6 & 1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 80 \\ 89 \end{pmatrix}
   \]
2. Find the inverse of the coefficient matrix:
   \[
   \text{Determinant} = (6)(4) - (1)(3) = 24 - 3 = 21
   \]
   Inverse:
   \[
   \frac{1}{21} \begin{pmatrix} 4 & -1 \\ -3 & 6 \end{pmatrix}
   \]
3. Multiply the inverse by the constants to find \(x\) and \(y\):
   \[
   \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{21} \begin{pmatrix} 4 & -1 \\ -3 & 6 \end{pmatrix} \begin{pmatrix} 80 \\ 89 \end{pmatrix}
   \]
4. After computation, \(x = 11\) and \(y = 14\).
5. Calculate \(12x + 5y = 202\).

**Answer:** \(202\) pesos.

---

#### Solution 4: Tabular Method

We organize information in a table to make calculations easier.

| Item        | Mango Cost (Peso) | Papaya Cost (Peso) | Total Cost (Peso) |
|-------------|-------------------|--------------------|--------------------|
| \(6\) mango + \(1\) papaya | \(6x\)              | \(y\)             | \(80\)            |
| \(3\) mango + \(4\) papaya | \(3x\)              | \(4y\)            | \(89\)            |

Using elimination and substitution (as in Solution 1 or 2) we find:
- \(x = 11\), \(y = 14\)

Then, we calculate the total cost for \(12\) mangoes and \(5\) papayas:
- \(12x + 5y = 202\)

**Answer:** \(202\) pesos.

---

Do you want further clarification on any of these methods? Here are some related questions for additional practice:

1. What if the cost of each mango was 13 pesos? How would that affect the total?
2. How would we solve this if the number of mangoes and papayas were doubled?
3. How would a change in price for papayas alone affect the total?
4. How can we check the solution if we only have one equation?
5. How can we use graphing to solve this system?

**Tip:** When tackling word problems, translating the information into equations or tables can clarify the relationships and guide your solution path efficiently.

Justin sells mango and papaya in his fruit stand. Each mango costs one amount and each papaya costs another amount. 6 mango + 1 papaya cost 80 peso. 3 mango and 4 papaya cost 89 peso. At these prices, how many pesos do 12 mango and 5 papaya? In this problem can you make 4 solution on how to solve this and can you solve this using table?

Learn to calculate the total cost for mangoes and papayas at a fruit stand using algebraic methods. This problem explores the elimination, substitution, and matrix methods to solve a system of linear equations. Suitable for Grades 8-10 students.

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