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Gloria had a rectangle garden plot last year with an area of 60 square feet This year Glorias garden plot is one foot wider and three feet shorter than last years garden but it has the same area What?

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10/07/2007

Here's the 5th grade answer: Write down all factors of 60 - 1,2,3,4,5,6,10,12,15,20,30,60. Find a factor in the list where there is an available number that is 3 lower. That would be 4 (1) ,5 (2) ,6 (3),and 15 (12).

Which of these fit the puzzle? Only 15, which you can figure out by trying the 4 possible combinations. For example, if the original garden was 5X12, what happens if you reduce 5 by 3 (to 2)? Doesn't work, because the width can only be increased by 1, and in this attempt, you would have to increase the original side (12) by 30 (2X30=60)! You only have to try one of the smaller numbers to realize that you need as large a width as available. That leaves 15. If you assume a length of 15 and width of 4, you can see from the list of factors that 15 can be reduced by 3 to 12. 60/12 =5... 1 more than the original width of 4. For the 9th grade answer, see below. Last year, the garden was 4 ft by 15 ft: 4 x 15 = 60 This year, the garden is 1 ft wider (5 ft) and 3 ft shorter (12 ft): 5 x 12 = 60 In both cases, the total area of the garden is 60 square feet. To figure it out for yourself, you have to solve a simultaneous system of 2 equations: 1. a X b = 60 (the original area of the garden, with width expressed as a and length expressed as b) and 2. (a + 1) X (b - 3) = 60 (the new area of the garden, still 60, with the original width a increased by 1 and the original length b decreased by 3.) Solve equation 1 for b by dividing both sides by a: 3. b = 60/a Since the two sides of this equation 3 are equal, you can replace one for the other in equation 2. Replace b with 60/a so you only have one variable to deal with: 4. (a + 1) X (60/a - 3) = 60 or, dropping the X, which is not needed to signify multiplication in Algebra: 4'. (a + 1) (60/a - 3) = 60 Use the FOIL method to multiply the two terms in parantheses on the left side of the equation to each other: 5. 60 - 3a + 60/a - 3 = 60 Subtract 60 from both sides to simplify the equation, and to make the right side equal zero, for reasons that will become obvious shortly: 6. - 3a + 60/a - 3 = 0 Multiply both sides by a to get rid of the fraction (note that this does not affect the right side because zero times any other number, even an unknown number, is still zero): 7. - 3a2 + 60 - 3a = 0 Divide both sides by -3 to make the largest power of a have a coefficient of 1 (again, the right side is unaffected): 8. a2 - 20 + a = 0 (Note that the last two steps can be combined by just multiplying both sides by -a/3) Rearrange terms to form a "proper" quadratic equation (terms listed in order of decreasing powers of the variable): 9. a2 + a - 20 = 0 Then factor (using the quadratic formula if necessary) this quadratic equation into: 10. (a + 5) (a - 4) = 0 (After factoring, check the result by multiplying the two factors to each other, using the FOIL method, and make sure you get the same thing you had in equation 9.) Now, the only time a product of two numbers can be zero is when one of those two numbers is itself zero. Therefore, this equation 10 can only be true if either a - 4 = 0 or a + 5 = 0. Solving these two possibilities for a results in two possible solutions, a = 4, and a = -5. We can discard the latter solution because a negative distance is meaningless (at least in this context), so the solution is a = 4. Inserting this back into equation 3 results in b = 60/a = 60/4 = 15. So the original width and length of the garden were 4 and 15, respectively. The new width is a+1 = 4+1 = 5, and the new length is b-3 = 15-3 = 12.