MATH SOLVE

4 months ago

Q:
# the length of a rectangle is 8 in more than its width the area of a rectangle is equal to 3 in less than three times the perimeter find the length and width of the rectangle

Accepted Solution

A:

This is the sort of question that is easily solved graphically.

The length is 17 inches; the width is 9 inches.

_____

Let x and y represent the width and length, respectively.

y = x +8 . . . . . the length is 8 more than the width

xy = 3(2(x +y)) -3 . . . . . the area is 3 less than 3 times the perimeter*

You can use the first expression for y to substitute into the second equation to get a quadratic in x. Only the positive solution is of interest.

x(x +8) = 6(2x +8) -3

x^2 -4x -45 = 0

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

___

* This part of the problem statement is nonsensical. Area is in square inches; perimeter is in inches. You cannot compare these quantities; you can only compare their numerical values. Subtracting 3 inches from some number of square inches cannot be done.

The length is 17 inches; the width is 9 inches.

_____

Let x and y represent the width and length, respectively.

y = x +8 . . . . . the length is 8 more than the width

xy = 3(2(x +y)) -3 . . . . . the area is 3 less than 3 times the perimeter*

You can use the first expression for y to substitute into the second equation to get a quadratic in x. Only the positive solution is of interest.

x(x +8) = 6(2x +8) -3

x^2 -4x -45 = 0

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

___

* This part of the problem statement is nonsensical. Area is in square inches; perimeter is in inches. You cannot compare these quantities; you can only compare their numerical values. Subtracting 3 inches from some number of square inches cannot be done.