anonymous
  • anonymous
the length of a rectangle is 15 inches longer than twice its width. if the perimeter is 78 inches, find the length and width of the rectangle.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
P = 2(l + w) l = 2w + 15 P = 2(2w + 15 + w) P = 4w + 30 + w P = 5w + 30 w = (P - 30) / 5 w = 48 / 5 l = 2w + 15 l = 2(48 / 5) + 15 l = 171/5
anonymous
  • anonymous
eh? I got something different. Here's what I got: Let the length be x and the width be y x = 2y + 15 ...(1) 2x + 2y = 78 ...(2) From (2): x + y = 39 x = 39 - y --> (1): 39 - y = 2y + 15 24 = 3y y = 8 --> (1): x = 2(8) + 15 x = 16 + 15 x = 31 So the length is 31 inches and the width is 8 inches. Checking: the length is 15 inches longer than twice its width 31 = 15 + 2(8) 31 = 15 + 16 Correct. the perimeter is 78 inches 2(31) + 2(8) = 62 + 16 = 78 Also correct.
anonymous
  • anonymous
Oh I found where you went wrong, Fruitless. P = 2(2w + 15 + w) P = 4w + 30 + w You expanded the bracket wrong. Shoudl've been 4w + 30 + 2w, which is 6w + 30 And then 6w + 30 = 78 6w = 48 w = 8

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