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The problem is about a closed cuboid box with the following dimensions and information:

- Width \( x \) cm
- Length \( y \) cm
- Height \( 5 \) cm (given)
- Volume of the box is \( 40 \, \text{cm}^3 \)
- External surface area of the box is \( 76 \, \text{cm}^2 \)

We are asked to form two equations using this information and solve for \( x \) and \( y \).

### Step 1: Volume Equation
The volume \( V \) of a cuboid is given by:
\[
V = x \times y \times 5
\]
Since the volume is \( 40 \, \text{cm}^3 \), we have:
\[
x \times y \times 5 = 40
\]
\[
x \times y = \frac{40}{5} = 8 \quad \text{(Equation 1)}
\]

### Step 2: Surface Area Equation
The surface area \( S \) of a cuboid is given by:
\[
S = 2(xy + xh + yh)
\]
Substituting \( h = 5 \) and the given surface area \( S = 76 \, \text{cm}^2 \):
\[
76 = 2(xy + 5x + 5y)
\]
\[
38 = xy + 5x + 5y
\]
Using \( xy = 8 \) from Equation 1, we can substitute \( xy = 8 \) into this:
\[
38 = 8 + 5x + 5y
\]
\[
30 = 5x + 5y
\]
\[
6 = x + y \quad \text{(Equation 2)}
\]

### Step 3: Solving the System of Equations
Now, we have two equations:
1. \( xy = 8 \)
2. \( x + y = 6 \)

To solve this system:
From Equation 2, solve for \( y \) in terms of \( x \):
\[
y = 6 - x
\]
Substitute this into Equation 1:
\[
x(6 - x) = 8
\]
\[
6x - x^2 = 8
\]
\[
x^2 - 6x + 8 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)}
\]
\[
x = \frac{6 \pm \sqrt{36 - 32}}{2}
\]
\[
x = \frac{6 \pm \sqrt{4}}{2}
\]
\[
x = \frac{6 \pm 2}{2}
\]
So, the two possible values for \( x \) are:
\[
x = \frac{6 + 2}{2} = 4 \quad \text{or} \quad x = \frac{6 - 2}{2} = 2
\]

### Step 4: Find Corresponding Values of \( y \)
For \( x = 4 \):
\[
y = 6 - 4 = 2
\]

For \( x = 2 \):
\[
y = 6 - 2 = 4
\]

### Final Solution:
The two possible sets of dimensions are:
1. \( x = 4 \, \text{cm}, y = 2 \, \text{cm}, h = 5 \, \text{cm} \)
2. \( x = 2 \, \text{cm}, y = 4 \, \text{cm}, h = 5 \, \text{cm} \)

Would you like further details or explanations on any part of this process?

### Related Questions:
1. How is the surface area of a cuboid calculated for different dimensions?
2. What is the general formula for the volume of a cuboid?
3. How can the quadratic formula be derived and applied to different types of equations?
4. What are the practical applications of cuboids in real-world packaging or construction?
5. How would the process change if the height of the box was unknown?

### Tip:
When solving systems of equations, always check for possible simplifications early, like isolating one variable, which can make the overall process easier.

A closed box is in the shape of a cuboid with width x cm, length y cm, and height 5 cm. It has a volume of 40 cm³ and an external surface area of 76 cm². Form two equations and solve them to find two possible sets of dimensions for the box.

Learn how to solve for the dimensions of a cuboid box given its volume and surface area. This step-by-step solution involves solving a system of equations using algebraic methods and quadratic equations. Suitable for students in Grades 9-10.

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