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Answer :
- Set up equations for the perimeter and area of the rectangle.
- Simplify the perimeter equation and solve for $y$ in terms of $x$: $y = 13 - x$.
- Substitute this expression for $y$ into the area equation, resulting in a quadratic equation in $x$.
- Solve the quadratic equation to find the possible values of $x$, then find the corresponding values of $y$. The solutions are $x=2, y=11$ and $x=6, y=7$. The final answer is $\boxed{x=2, y=11 \text{ or } x=6, y=7}$
### Explanation
1. Problem Setup
We are given a rectangle with sides $(y-3)$ and $(x+2)$, a perimeter of 24 metres, and an area of 32 square metres. Our goal is to find the values of $x$ and $y$.
2. Perimeter Equation
The perimeter of a rectangle is given by $2(\text{length} + \text{width})$. Therefore, we have the equation:
$$2((y-3) + (x+2)) = 24$$
3. Area Equation
The area of a rectangle is given by $\text{length} \times \text{width}$. Therefore, we have the equation:
$$(y-3)(x+2) = 32$$
4. Simplifying Perimeter Equation
Let's simplify the perimeter equation:
$$2(y - 3 + x + 2) = 24$$
$$2(x + y - 1) = 24$$
$$x + y - 1 = 12$$
$$x + y = 13$$
5. Solving for y
Now, solve for $y$ in terms of $x$:
$$y = 13 - x$$
6. Substituting into Area Equation
Substitute $y = 13 - x$ into the area equation:
$$(13 - x - 3)(x+2) = 32$$
$$(10 - x)(x+2) = 32$$
7. Expanding Area Equation
Expand the area equation:
$$10x + 20 - x^2 - 2x = 32$$
$$-x^2 + 8x + 20 = 32$$
8. Quadratic Equation
Rearrange the equation to form a quadratic equation:
$$x^2 - 8x + 12 = 0$$
9. Solving for x
Solve the quadratic equation for $x$. We can factor the quadratic equation as follows:
$$(x - 2)(x - 6) = 0$$
Thus, $x = 2$ or $x = 6$.
10. Solving for y
Now, substitute the values of $x$ back into $y = 13 - x$ to find the corresponding values of $y$:
If $x = 2$, then $y = 13 - 2 = 11$.
If $x = 6$, then $y = 13 - 6 = 7$.
11. Final Answer
Therefore, the values of $x$ and $y$ are either $x=2$ and $y=11$, or $x=6$ and $y=7$.
### Examples
Understanding how to solve problems involving the perimeter and area of rectangles is useful in many real-world scenarios. For example, if you're designing a rectangular garden with a specific area and want to minimize the amount of fencing needed (perimeter), you would use these concepts. Similarly, architects and engineers use these calculations to optimize space and material usage in building designs. This type of problem also applies to optimizing the layout of rooms in a house or planning the arrangement of furniture.
- Simplify the perimeter equation and solve for $y$ in terms of $x$: $y = 13 - x$.
- Substitute this expression for $y$ into the area equation, resulting in a quadratic equation in $x$.
- Solve the quadratic equation to find the possible values of $x$, then find the corresponding values of $y$. The solutions are $x=2, y=11$ and $x=6, y=7$. The final answer is $\boxed{x=2, y=11 \text{ or } x=6, y=7}$
### Explanation
1. Problem Setup
We are given a rectangle with sides $(y-3)$ and $(x+2)$, a perimeter of 24 metres, and an area of 32 square metres. Our goal is to find the values of $x$ and $y$.
2. Perimeter Equation
The perimeter of a rectangle is given by $2(\text{length} + \text{width})$. Therefore, we have the equation:
$$2((y-3) + (x+2)) = 24$$
3. Area Equation
The area of a rectangle is given by $\text{length} \times \text{width}$. Therefore, we have the equation:
$$(y-3)(x+2) = 32$$
4. Simplifying Perimeter Equation
Let's simplify the perimeter equation:
$$2(y - 3 + x + 2) = 24$$
$$2(x + y - 1) = 24$$
$$x + y - 1 = 12$$
$$x + y = 13$$
5. Solving for y
Now, solve for $y$ in terms of $x$:
$$y = 13 - x$$
6. Substituting into Area Equation
Substitute $y = 13 - x$ into the area equation:
$$(13 - x - 3)(x+2) = 32$$
$$(10 - x)(x+2) = 32$$
7. Expanding Area Equation
Expand the area equation:
$$10x + 20 - x^2 - 2x = 32$$
$$-x^2 + 8x + 20 = 32$$
8. Quadratic Equation
Rearrange the equation to form a quadratic equation:
$$x^2 - 8x + 12 = 0$$
9. Solving for x
Solve the quadratic equation for $x$. We can factor the quadratic equation as follows:
$$(x - 2)(x - 6) = 0$$
Thus, $x = 2$ or $x = 6$.
10. Solving for y
Now, substitute the values of $x$ back into $y = 13 - x$ to find the corresponding values of $y$:
If $x = 2$, then $y = 13 - 2 = 11$.
If $x = 6$, then $y = 13 - 6 = 7$.
11. Final Answer
Therefore, the values of $x$ and $y$ are either $x=2$ and $y=11$, or $x=6$ and $y=7$.
### Examples
Understanding how to solve problems involving the perimeter and area of rectangles is useful in many real-world scenarios. For example, if you're designing a rectangular garden with a specific area and want to minimize the amount of fencing needed (perimeter), you would use these concepts. Similarly, architects and engineers use these calculations to optimize space and material usage in building designs. This type of problem also applies to optimizing the layout of rooms in a house or planning the arrangement of furniture.
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