Strategy: Finding a Formula
Date: 11/14/97 at 09:43:43 From: Betsy Subject: Strategy: using a formula Jill needs three pieces of lumber a total of 27 ft. long for a dog pen. The second piece has to be 4 ft. longer than the first, and the third piece has to be 4ft. longer than the second. How long is each piece of lumber?
Date: 11/25/97 at 11:46:38 From: Doctor Allan Subject: Re: Strategy: using a formula Hi Betsy, You need to translate your word problem into math. I will give you a suggestion for how to proceed, and then you can see if this helps you understand. You have three pieces of lumber, whose total length is 27 ft. Let's call the length of the first piece of lumber x, the length of the second piece y, and the length of the third piece z. Since they add up to a total of 27 ft. we have: (1) x + y + z = 27 Do you understand this? Then we are told that the second piece of lumber is 4 ft. longer than the first piece. This means that if we take the length of the first piece and add 4 ft. we will get the length of the second piece: (2) x + 4 = y Finally we are told that the length of the third piece of lumber is 4 ft. longer than the second piece. As before, this yields: (3) y + 4 = z Now we have to do the math part. We need to find values for x, y and z such that the above equations become true statements. From equation (3) we get (4) y = z - 4 and since there is a y in equation 2 we can write z - 4 there instead. Equation (2) then becomes x + 4 = z - 4 But if we isolate x in this last equation we have (5) x = z - 8 Do you understand what I have done so far? In (4) and (5) you have new expressions for x and y, and you should substitute these into equation (1). Then you will find a value for z, and once you have this you can immediately find the values for y and z by putting your value for z into equations (4) and (5). Try to do this part yourself and let me know if you have problems. Since x and y and z were the lengths of pieces 1, 2, and 3 you will have found the lengths once you have found the values for x, y and z. I hope this helps. -Doctor Allan, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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