Answer :
To solve the problem about the rectangle, we need to find the values of [tex]x[/tex] and [tex]y[/tex] given the conditions of perimeter and area.
Step 1: Calculate using the perimeter.
The formula for the perimeter [tex]P[/tex] of a rectangle is given by:
[tex]P = 2 \times L + 2 \times W[/tex]
Here, the length [tex]L = y - 3[/tex] and the width [tex]W = x + 2[/tex], and the perimeter [tex]P = 24 \, \text{metres}[/tex].
So, the equation becomes:
[tex]2(y - 3) + 2(x + 2) = 24[/tex]
Simplifying, this gives:
[tex]2y - 6 + 2x + 4 = 24[/tex]
[tex]2x + 2y - 2 = 24[/tex]
[tex]2x + 2y = 26[/tex]
[tex]x + y = 13 \quad \text{(Equation 1)}[/tex]
Step 2: Calculate using the area.
The formula for the area [tex]A[/tex] of a rectangle is:
[tex]A = L \times W[/tex]
Given [tex]A = 32 \, \text{square metres}[/tex], the equation becomes:
[tex](y - 3)(x + 2) = 32[/tex]
Expanding this, we get:
[tex]yx + 2y - 3x - 6 = 32[/tex]
[tex]yx + 2y - 3x = 38 \quad \text{(Equation 2)}[/tex]
Step 3: Solve the system of equations.
From Equation 1, we have:
[tex]y = 13 - x[/tex]
Substitute [tex]y = 13 - x[/tex] into Equation 2:
[tex](13 - x)x + 2(13 - x) - 3x = 38[/tex]
Expanding this gives:
[tex]13x - x^2 + 26 - 2x - 3x = 38[/tex]
[tex]-x^2 + 8x + 26 = 38[/tex]
[tex]-x^2 + 8x = 12[/tex]
[tex]x^2 - 8x + 12 = 0[/tex]
Solve this quadratic equation using the quadratic formula:
[tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
Where [tex]a = 1, b = -8, c = 12[/tex].
[tex]x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 1 \times 12}}{2 \times 1}[/tex]
[tex]x = \frac{8 \pm \sqrt{64 - 48}}{2}[/tex]
[tex]x = \frac{8 \pm \sqrt{16}}{2}[/tex]
[tex]x = \frac{8 \pm 4}{2}[/tex]
This gives two possible solutions:
[tex]x = 6 \quad \text{or} \quad x = 2[/tex]
Substituting back into [tex]y = 13 - x[/tex]:
- If [tex]x = 6,[/tex] then [tex]y = 13 - 6 = 7[/tex]
- If [tex]x = 2,[/tex] then [tex]y = 13 - 2 = 11[/tex]
Hence, the possible solutions for [tex](x, y)[/tex] are:
[tex](6, 7) \quad \text{or} \quad (2, 11)[/tex]
Now let's prove that [tex](1 + x^m + x^{-n})^2 - (1 - x^m - x^{-n})^2[/tex] is divisible by 2 for all real values of [tex]m[/tex] and [tex]n[/tex].
Step 4: Simplify and Factoring.
The expression is:
[tex](1 + x^m + x^{-n})^2 - (1 - x^m - x^{-n})^2[/tex]
This can be rewritten using the identity [tex](a + b)^2 = a^2 + 2ab + b^2[/tex]:
[tex]=[(1 + x^m + x^{-n}) + (1 - x^m - x^{-n})][(1 + x^m + x^{-n}) - (1 - x^m - x^{-n})][/tex]
[tex]= 2(1)(2x^m + 2x^{-n})[/tex]
[tex]= 4(x^m + x^{-n})[/tex]
Since 4, an even number, multiplies [tex](x^m + x^{-n})[/tex], the whole expression is divisible by 2.
Thus, we have shown that it is always divisible by 2 for all real [tex]m[/tex] and [tex]n[/tex].
