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5/27/2022 5:41:04 PM [ARCHIVED THREAD] - Easy math question. That has me baffled.
 Posted: 10/30/2010 1:07:42 PM EDT The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle? I guess I've been in the sun too long. What would be the formula to solve this? Posted: 10/30/2010 1:10:41 PM EDT [#1] Quoted: The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle? I guess I've been in the sun too long. What would be the formula to solve this? perimeter = 2L + 2W . Write everything else out in algebraic form. Solve for L and W Posted: 10/30/2010 1:12:42 PM EDT [#2] 87 (sorry, couldn't help myself) Posted: 10/30/2010 1:15:38 PM EDT [#3] Quoted:The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?I guess I've been in the sun too long. What would be the formula to solve this?Units in centimeters.'P'erimeter = 25 + 'L'ength + 'W'idthFirst: P = 25 + L + WSecond: L = 5 + WKnowing: P = L + W + L + WThen: L + W = 25Or just looked at the question and said: "Oh, length and width are 25? That means it must be 10 by 15".  Posted: 10/30/2010 1:17:58 PM EDT [#4] 2L + 2W = 25+ (L+W)L=W+52(W+5)+2W=25+(W+5) +W2W+10+2W=25+2W+54W+10=30+2W2W=20W=102L+20=25+L+10L=15W is widthL is length Posted: 10/30/2010 1:18:28 PM EDT [#5] Quoted: The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle? I guess I've been in the sun too long. What would be the formula to solve this? P=2l+2w+25 l=5+w W=W Do the math from there. Posted: 10/30/2010 1:21:10 PM EDT [#6] Quoted:Quoted:The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?I guess I've been in the sun too long. What would be the formula to solve this?P=2l+2w+25l=5+wW=W Do the math from there.Really? Come on now...  Posted: 10/30/2010 1:25:43 PM EDT [#7] 2x + 2y = 25 + x + y Therefore x + y = 25 ... y - 5 = x .. Therefore y -5 + y = 25 Therefore 2y = 30 Therefore y = 15 Therefore x = 10  Posted: 10/30/2010 1:27:19 PM EDT [#8] Quoted:2x + 2y = 25 + x + yTherefore x + y = 25...y - 5 = x..Thereforey -5 + y = 25Therefore2y = 30Thereforey = 15Thereforex = 10 Yours is more elegant than mine. Posted: 10/30/2010 1:28:27 PM EDT [#9] Thanks guys! I was trying to help my ninth grader and I had figured out 15cm and 10cm but couldn't figure out how to write the steps to determine it. Posted: 10/30/2010 1:28:32 PM EDT [#10] Quoted:Quoted:Quoted:The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?I guess I've been in the sun too long. What would be the formula to solve this?P=2l+2w+25l=5+wW=W Do the math from there.Really? Come on now... I believe he was saying to use substitution. Posted: 10/30/2010 1:29:26 PM EDT [#11] P=25+2W+2(W+5) Posted: 10/30/2010 1:31:27 PM EDT [#12] Quoted:Quoted:Quoted:Quoted:The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?I guess I've been in the sun too long. What would be the formula to solve this?P=2l+2w+25l=5+wW=W Do the math from there.Really? Come on now... I believe he was saying to use substitution. You mean like substituting W for W? God, that's just brilliant. Or did you mean his first erroneous equation? Posted: 10/30/2010 1:31:33 PM EDT [#13] Quoted: 2L + 2W = 25+ (L+W) L=W+5 2(W+5)+2W=25+(W+5) +W 2W+10+2W=25+2W+5 4W+10=30+2W 2W=20 W=10 2L+20=25+L+10 L=15 W is widthL is length Correct. Posted: 10/30/2010 1:34:35 PM EDT [#14] Quoted: P=25+2W+2(W+5) That's was the line I was working. P=25+2W+2(W+5) P=25+2W+2W+10 P=35+4w P-35=35+4W-35 P-35=4W (which really doesn't work) Posted: 10/30/2010 1:36:23 PM EDT [#15] Quoted:P=25+2W+2(W+5)You too??O.M.G. The first sentence of the problem:The perimeter of a rectangle is 25 cm greater than the sum of the length and the width=Rectangle Perimeter is 25 greater than sum of Length and Width=Perimeter is 25 added to sum of Length and Width=Perimeter is 25 added to Length added to Width=Perimeter = 25 + Length + Width=P = 25 + L + W  Posted: 10/30/2010 1:36:33 PM EDT [#16] Quoted: 2L + 2W = 25+ (L+W) L=W+5 2(W+5)+2W=25+(W+5) +W 2W+10+2W=25+2W+5 4W+10=30+2W 2W=20 W=10 2L+20=25+L+10 L=15 W is widthL is length This is how I backed into the answer after my first attempt but I didn't keep the equations in order enough to duplicate it and be able to show the work. Posted: 10/30/2010 1:55:28 PM EDT [#17] Quoted: Quoted: Quoted: Quoted: The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle? I guess I've been in the sun too long. What would be the formula to solve this? P=2l+2w+25 l=5+w W=W Do the math from there.Really? Come on now... I believe he was saying to use substitution.   It was how I was taught to solve this problems. The book suggested deliniate x for w, but I use w to keep it simple. Posted: 10/30/2010 1:56:04 PM EDT [#18] Quoted: Quoted: Quoted: Quoted: Quoted: The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle? I guess I've been in the sun too long. What would be the formula to solve this?P=2l+2w+25 l=5+w W=W Do the math from there.Really? Come on now... I believe he was saying to use substitution. You mean like substituting W for W? God, that's just brilliant. Or did you mean his first erroneous equation?   I was establishing the numbers he needed then he does the distributive property and gets his answer. Posted: 10/30/2010 2:08:27 PM EDT [#19] Quoted:Quoted:Quoted:Quoted:Quoted:Quoted:The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?I guess I've been in the sun too long. What would be the formula to solve this?P=2l+2w+25l=5+wW=W Do the math from there.Really? Come on now... I believe he was saying to use substitution. You mean like substituting W for W? God, that's just brilliant. Or did you mean his first erroneous equation?  I was establishing the numbers he needed then he does the distributive property and gets his answer. The first part in red is wrong. Posted: 10/30/2010 2:12:34 PM EDT [#20] Quoted:Quoted:Quoted:Quoted:Quoted:The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?I guess I've been in the sun too long. What would be the formula to solve this?P=2l+2w+25l=5+wW=W Do the math from there.Really? Come on now... I believe he was saying to use substitution.  It was how I was taught to solve this problems. The book suggested deliniate x for w, but I use w to keep it simple.So... W = Width? W=W means nothing. Blue = Blue. F = F. Schizophrenic = Schizophrenic. x = x."Terms: P = perimeter; W = width; L = length"followed by the equations, then the written out solution.If that is what you meant, I can understand that. Maybe you just have a funny way of saying that.  Posted: 10/30/2010 2:12:36 PM EDT [#21] You just gave me a headache.  Posted: 10/30/2010 2:17:20 PM EDT [#22] Quoted:The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?I guess I've been in the sun too long. What would be the formula to solve this?P = PerimeterL = LengthW = WidthA rectangle has 4 sides: 2 length sides and 2 width sides. The perimeter of a rectangle is the sum of all sides, or (abbreviated) 2L + 2W.So far, we've arrived at:1.) P = 2L + 2WThe text says that the perimeter of this rectangle is 25cm greater than the sum of it's length and width.  Another way of phrasing this is "The perimeter is the length, plus the width, plus another 25". Or:2.) P = L + W + 25Now that we have a decent definition of what P is supposed to be, let's substitute it in formula #1 for calculating the perimeter of our hypothetical rectangle:3.) L + W + 25 = 2L + 2WHey look, now we're down to two variables: L and W. Sure would be nice if we could just get this down to one variable. Let's keep reading. "...the length is 5cm greater than the width"Sweet, I like more info. So, what they're saying is:4.) L = W + 5Awesome, now l know what L is supposed to be. Let's substitute L in formula #3, shall we?W + 5 + W + 25 = 2(W + 5) + 2WAnd let's simplify that expression:2W + 30 = 2(W + 5) + 2WSweet, now all we have to do is solve for W. W + 15 = W + 5 + W5 + W = 15W = 10Awesome. Now that we know the width is 10, we just need to find the length. You might recall in formula #4, that L = W + 5. Well...L = 10 + 5L = 15Therefore, the rectangle dimensions are 15cm long, and 10cm wide. How do I know this is right? Well, let's check our work. "The perimeter of a rectangle is 25 cm greater than the sum of the length and the width"What's the sum of the length and width? Well, it's 25 (L=15 and W=10, remember?).Is this 25cm greater than the perimeter?  Well....P = 2W + 2LP = 2(10) + 2(15)P = 50Is 50 25cm greater than the sum of the length and width, 25? I do believe it is.  Posted: 10/30/2010 2:27:08 PM EDT [#23] Quoted: Quoted: The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle? I guess I've been in the sun too long. What would be the formula to solve this? P = Perimeter L = Length W = Width A rectangle has 4 sides: 2 length sides and 2 width sides. The perimeter of a rectangle is the sum of all sides, or (abbreviated) 2L + 2W. So far, we've arrived at: 1.) P = 2L + 2W The text says that the perimeter of this rectangle is 25cm greater than the sum of it's length and width.  Another way of phrasing this is "The perimeter is the length, plus the width, plus another 25". Or: 2.) P = L + W + 25 Now that we have a decent definition of what P is supposed to be, let's substitute it in formula #1 for calculating the perimeter of our hypothetical rectangle: 3.) L + W + 25 = 2L + 2W Hey look, now we're down to two variables: L and W. Sure would be nice if we could just get this down to one variable. Let's keep reading. "...the length is 5cm greater than the width" Sweet, I like more info. So, what they're saying is: 4.) L = W + 5 Awesome, now l know what L is supposed to be. Let's substitute L in formula #3, shall we? (W + 5) + W + 25 = 2(W + 5) + 2W And let's simplify that expression: 2W + 30 = 2(W + 5) + 2W Sweet, now all we have to do is solve for W. W + 15 = W + 5 + W 5 + W = 15 W = 10 Awesome. Now that we know the width is 10, we just need to find the length. You might recall in formula #4, that L = W + 5. Well... L = 10 + 5 L = 15 Therefore, the rectangle dimensions are 15cm long, and 10cm wide. How do I know this is right? Well, let's check our work. "The perimeter of a rectangle is 25 cm greater than the sum of the length and the width" What's the sum of the length and width? Well, it's 25 (L=15 and W=10, remember?). Is this 25cm greater than the perimeter?  Well.... P = 2W + 2L P = 2(10) + 2(15) P = 50 Is 50 25cm greater than the sum of the length and width, 25? I do believe it is. Q. E. D.  Motherfuckers! Posted: 10/30/2010 2:28:42 PM EDT [#24] in my head in about 3 seconds, sounds like 10 & 15 Posted: 10/30/2010 2:32:01 PM EDT [#25] Quoted:Q. E. D.  Motherfuckers!Damn right. I'm full of shit until it's demonstrated.  Posted: 10/30/2010 2:34:15 PM EDT [#26] Just a running tally Quoted: P=2l+2w+25Quoted: P=25+2W+2(W+5)Quoted: That's was the line I was working. Posted: 10/30/2010 2:37:07 PM EDT [#27] As an engineer I am fapping hard to this thread!  Posted: 10/30/2010 2:59:34 PM EDT [#28] Quoted:You just gave me a headache.  How do I reech deese keeeeeds?!!! Posted: 10/30/2010 3:03:04 PM EDT [#29] Quoted: The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle? I guess I've been in the sun too long. What would be the formula to solve this? 2x + 2y - 25 = x + y x - 5 = y ETA: x = Length y = Width Posted: 10/30/2010 3:08:47 PM EDT [#30] Quoted:As an engineer I am fapping hard to this thread! I took the AP Calc exam my Senior year in HS (scored a 4), and stopped there. I never did go to college (well, I started but quit shortly thereafter), and thus, all personal math development ended right then and there. My brain is certainly wired for it, but...use it or lose it, as they say. The sad thing (it's been 14 years since I've sat in a math class), I doubt I could pass a Math 90 final if you handed one to me right now. I've forgotten a LOT of details. [ARCHIVED THREAD] - Easy math question. That has me baffled.   Win a FREE Membership!

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