Solving Simultaneous Equations and Factoring
Date: 3/18/96 at 21:52:36 From: Mr. Michael J. Fitzpatrick Subject: Math Problem How would you use factoring to solve these problems? It's 9th grade level, using one variable. Vanessa built a rectangular pen for her dogs. She used an outside wall of the garage for one of the sides of the pen. She had to buy 20 m of fencing in order to build the other sides of the pen. Find the dimensions of the pen if its area is 48 m^2. The Parkhursts used 160 yd of fencing to enclose a rectangular corral and to divide it into two parts by a fence parallel to one of the shorter sides. Find the dimensions of the corral if its area is 1000 yd^2. Each edge of one cube is 2 cm longer than each edge of another cube. The volumes of the cubes differ by 98cm^3. Find the lengths of the edges of each cube. Thank you.
Date: 3/19/96 at 11:41:52 From: Doctor Elise Subject: Re: Math Problem Hi! Let's start with the dog pen. >Vanessa built a rectangular pen for her dogs... To start out with, we're going to need 2 variables, one for the length of the pen, and one for the width. Let's call the length L, and say that it's the length of the garage wall and the side opposite it. Then let's call the width W (real original, huh?) for the other two walls. Okay, so now we have a rectangle, with sides L and W. We know that the total perimeter is the length of fence bought, 20m, plus the length of the garage wall, L. And we know how to calculate the perimeter of a rectangle, so our first equation is: L + W + L + W = 20 + L 2L + 2W = 20 + L L + 2W = 20 L = 20 - 2W Okay? Now we use the other bit of information, that the area is 48m^2. Well, the area of a rectangle is length times width, right? So now we have L*W = 48 And we can substitute for L from the first equation... (20 - 2W)* W = 48 20W - 2W^2 = 48 I hate having a negative squared term, (it's harder to factor) so I'm going to multiply everything by -1: -20W + 2W^2 = -48 Now let's get everything together on one side of the = sign so that we can start factoring: 2W^2 - 20W + 48 = 0 Now let's divide by 2 W^2 - 10W + 24 = 0 Can you factor that? Let's see. If I factor 24, I get either 3 * 8 or 4 * 6, and I have to be able to add or subtract factors to get 10. So it has to be 6 and 4. And they both have to be the same sign to get positive 24, but the 10 is negative so they have to be negative. (W - 6)(W - 4) = 0 Looks like this is our answer. The width is either 6 or 4 meters. Now let's solve for the length and make sure both of these numbers work in the real world: L = 20 - 2W L = 20 - 12 = 8 so if the width is 6, the length is 8 or L = 20 - 8 = 12 or if the width is 4 then the length is 12. Both of these combinations give us 20 meters of fencing and an area of 48 square meters. Let's look at the Parkhursts. >The Parkhursts used 160 yd of fencing to enclose a rectangular >corral and to divide it into two parts by a fence parallel to one of >the shorter sides. Find the dimensions of the corral if its area is >1000 yd^2. Let's use our old friends L and W again, and let's have W be the short side. If you draw the corral, you'll see that instead of the perimeter, we have the perimeter plus one extra short side's worth of fencing adding up to 160. L + W + L + W + W = 160 2L + 3W = 260 L = 1/2(160 - 3W) And we know the total area, our old friend L * W, is 1000. L * W = 1000 And we substitute for L again, 1/2(160 - 320 W) * W = 1000 Let's get rid of the 1/2, it won't help us factor! (160 - 3W) * W = 2000 3W^2 - 160W + 2000 = 0 Okay, we need to factor this mess. Both factors have to be subtracted, because the 2000 is positive, (so they're the same - plus times plus = plus, or neg times neg = plus) and the 260 is negative, so they're both negative. And we have a 3 thrown in: (3W - ?)(W - ??) Okay, 160. 3 times something, plus something else, equals 160. Let's start with 120 + 40. 120/3 = 40. Does 40 * 40 = 2000? Nope, let's try again. How about 60 and 100? Yes! 60/3 = 20, 20 * 100 is 2000! (3W - 100)(W - 20) = 0 So our fence has a width of either 20 or 100/3. If we plug W back into our first equation, we get L = 50 if W = 20, or L = 30 if W = 100/3. Phew! I had to do that one twice because I wrote the numbers wrong! Finally, Cubes. >Each edge of one cube is 2 cm longer than each edge of another >cube. The volumes of the cubes differ by 98cm^3. Find the lengths >of the edges of each cube. Okay, we have one cube that is L centimeters long on a side, and one that is L + 2 cm on a side. The volume of the first one is L^3, and the volume of the second one is 98 + L^3, right? so (L + 2)^3 = 98 + L^3 (L^2 + 4L + 4)(L + 2) = 98 + L^3 L^3 + 4L^2 + 4L + 2L^2 + 8L + 8 = 98 + L^3 And the L^3 terms cancel out! 6L^2 + 12L + 8 = 98 6L^2 + 12L - 90 = 0 L^2 + 2L - 15 = 0 (L + ?)(L - ??) = 0 15 = 3 * 5, 5 > 3, (L + 5)(L - 3) = 0 The possible values for L are L = -5 or L = 3. In the real world, we aren't going to have a cube with a side of length -5, so we throw that solution out as mathematically possible, but not physically possible. So L = 3. One cube is 3 cm on a side, and one cube is L + 2 or 5 cm on a side. I hope this helps! Word problems can be tricky to set up mathematical equations for, but I think they're the real life problems math was invented to solve in the first place! -Doctor Elise, The Math Forum
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