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Let the width of the rectangle = x

Therefore length of the rectangle = x + 4

(length exceed its width by 4 feet)

Area of the rectangle = 45

Step 2:

Area of the rectangle = Length * Width

=> 45 = x(x + 4)

=> 45 = $x^2$ + 4x

=> $x^2$ + 4x - 45 = 0

Step 3:

Solve for x,

=> $x^2$ + 4x - 45 = 0

=> $x^2$ + 9x - 5x - 45 = 0

=> x(x + 9) - 5(x + 9) = 0

=> (x - 5)(x + 9) = 0

Step 4:

either x - 5 = 0 or x + 9 = 0

=> x = 5 or x = - 9

The solution x = - 9 must be discarded because a rectangle can't have a negative width.

=> x = 5, is the width of the rectangle.

Step 5:

Length of the rectangle = x + 4 = 5 + 4 = 9

Hence the dimensions of the rectangle are: Length = 9 feet and Width = 5 feet. answer

Let the three consecutive odd integer are, x, x + 2 and x + 4.

That is,

First integer = x

Second integer = x + 2

Third integer = x + 4

Step 1:

The problem states that:

Sum of the second and third integers = 5(first integer) - 9.

=> (x + 2) + (x + 4) = 5x - 9

=> x + 2 + x + 4 = 5x - 9

Step 2:

Combine the like terms:

=> 2x + 6 = 5x - 9

Subtract 2x from both side

=> 2x - 2x + 6 = 5x - 9 - 2x

=> 6 = 3x - 9

Subtract 6 from both sides

=> 6 - 6 = 3x - 9 - 6

That is,

First integer = x

Second integer = x + 2

Third integer = x + 4

Step 1:

The problem states that:

Sum of the second and third integers = 5(first integer) - 9.

=> (x + 2) + (x + 4) = 5x - 9

=> x + 2 + x + 4 = 5x - 9

Step 2:

Combine the like terms:

=> 2x + 6 = 5x - 9

Subtract 2x from both side

=> 2x - 2x + 6 = 5x - 9 - 2x

=> 6 = 3x - 9

Subtract 6 from both sides

=> 6 - 6 = 3x - 9 - 6

=> 0 = 3x - 15

=> 3x = 15

Divide each side by 3

=> $\frac{3x}{3} = \frac{15}{3}$

=> x = 5, first integer.

Step 3:

Largest or third integer = x + 4 = 5 + 4 = 9

Hence the largest integer is 9. answer