Question

The width and the length of a rectangle are consecutive even integers. If the width is decreased by 3 inches, then the area of the resulting rectangle is 24 square inches. What is the area of the original rectangle

Answer 1

Original Area : 48 square inches

Explanation:

Let's say that the width of this rectangle is x, and the length of this rectangular is y. Remember that they are even consecutive integers as well, so let's make the length 2 greater than the width.

x = width,

x + 2 = length

And remember that the width decreases by 3 inches, to result in a rectangle of area 24. So the width will be x - 3, and the area will be (x - 3)(x + 2). Therefore we have the quadratic equation (x - 3)(x + 2) = 24, in which we can solve for x and y to determine the area of the original rectangle.

Step # 1 : Expand the expression (x - 3)(x + 2),

x² - x - 6 = 24

Step # 2 : Simplify and factor the expression x² - x - 6

x² - x - 30 = 0 ⇒ (x + 5)(x - 6) = 0

Step # 3 : Apply the Zero Factor Principle

x + 5 = 0, and x - 6 = 0

x = - 5, x = 6

Now remember that the width can't be negative, so it will be 6 inches long. That would make the length 6 + 2 = 8 inches. The area of the original rectangular will be 6
`*` 8 = 48 square inches.

Answer 2

48 square units

Explanation:

Let L and W represent the length and width of the rectangle.

L = W +2 . . . . . . assuming the length is more than the width

L(W -3) = 24 . . . . the area after the reduction in width

__

(W +2)(W -3) = 24

So, we're looking for factors of 24 that differ by 5. We know that 24=8·3, so we have ...

W+2 = 8

W = 6 . . . . . . . matches W-3=3, also

Then the length is ...

L = W+2 = 8

and the area of the original rectangle is ...

A = LW = (8 inches)(6 inches) = 48 square inches