pictured below Simplify your answer as much as possible W L W L s s s s

Pictured below simplify your answer as much as

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pictured below. Simplify your answer as much as possible. W L W L s s s s Solution. The perimeter of any polygonal figure is the sum of the lengths of its sides. a) To find the perimeter P of the rectangle, sum its four sides. P = L + W + L + W. Combine like terms. P = 2 L + 2 W. b) To find the perimeter P of the square, sum its four sides. P = s + s + s + s. Combine like terms. P = 4 s. Answer: P = 6 x Sometimes it is useful to replace a variable with an expression containing another variable. You Try It! EXAMPLE 11. The length of a rectangle is three feet longer than twice its The length L of a rectangle is 5 meters longer than twice its width W . Find the perimeter P of the rectangle in terms of its width W . width. Find the perimeter P of the rectangle in terms of its width alone. Solution. From the previous problem, the perimeter of the rectangle is given by P = 2 L + 2 W, (3.1) where L and W are the length and width of the rectangle, respectively. This equation gives the perimeter in terms of its length and width, but we’re asked to get the perimeter in terms of the width alone. However, we’re also given the fact that the length is three feet longer than twice the width.
204 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA Length is Three Feet longer than Twice the Width L = 3 + 2 W Because L = 3+2 W , we can replace L with 3+2 W in the perimeter equation 3.1 . P = 2 L + 2 W P = 2(3 + 2 W ) + 2 W Use the distributive property, then combine like terms. P = 6 + 4 W + 2 W P = 6 + 6 W. This last equation gives the perimeter P in terms of the width W alone. Answer: P = 6 W + 10 You Try It! EXAMPLE 12. The width of a rectangle is two feet less than its length. The width W of a rectangle is 5 feet less than twice its width L . Find the perimeter P of the rectangle in terms of its length L . Find the perimeter P of the rectangle in terms of its length alone. Solution. Again, the perimeter of a rectangle is given by the equation P = 2 L + 2 W, (3.2) where L and W are the length and width of the rectangle, respectively. This equation gives the perimeter in terms of its length and width, but we’re asked to get the perimeter in terms of the length alone. However, we’re also given the fact that the width is two feet less than the length. Width is Length minus Two feet W = L - 2 Because W = L - 2, we can replace W with L - 2 in the perimeter equation 3.2 . P = 2 L + 2 W P = 2 L + 2( L - 2) Use the distributive property, then combine like terms. P = 2 L + 2 L - 4 P = 4 L - 4 . This last equation gives the perimeter P in terms of the length L alone. Answer: P = 6 L - 10
3.4. COMBINING LIKE TERMS 205 Exercises In Exercises 1 - 16 , combine like terms by first using the distributive property to factor out the common variable part, and then simplifying. 1 . 17 xy 2 + 18 xy 2 + 20 xy 2 2 . 13 xy - 3 xy + xy 3 . - 8 xy 2 - 3 xy 2 - 10 xy 2 4 . - 12 xy - 2 xy + 10 xy 5 . 4 xy - 20 xy 6 . - 7 y 3 + 15 y 3 7 . 12 r - 12 r 8 . 16 s - 5 s 9 . - 11 x - 13 x + 8 x 10 . - 9 r - 10 r + 3 r 11 . - 5 q + 7 q 12 . 17 n + 15 n 13 . r - 13 r - 7 r 14 . 19 m + m + 15 m 15 . 3 x 3 - 18 x 3 16 . 13 x 2 y + 2 x 2 y In Exercises 17 - 32 , combine like terms by first rearranging the terms, then using the distributive property to factor out the common variable part, and then simplifying.

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