#$&* course MTH 158 007. `* 7
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Given Solution: * * ** We factor the denominator to get first (x^2-4)(x^2+4), then (x-2)(x+2)(x^2+4). The numerator factors as (x+2)^2. So the fraction is • (x+2)(x+2)/[(x-2)(x+2)(x^2+4)], which reduces to • (x+2)/[(x-2)(x^2+4)]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.7.28 (was R.7.24). Show how you simplified[ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ]. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x-2 (12x) / 4x (x-2)(x-2) 12x (x-2) / 4x (x-2)(x-2) 3/ x-2 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: [ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ] = (x-2) * / 4x * 12 x / (x^2 - 4x + 4) = (x-2) * 12 x / [ 4x ( x^2 - 4x + 4) ] = 12 x (x-2) / [4x ( x-2) ( x-2) ] = 3/(x - 2) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.7.40 (was R.7.36). Show how you found and simplified the sum (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3x-5+x+4 / 3x+2 3x-1 / 3x+2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We have two like terms so we write • (2x-5)/(3x+2) + (x+4)/(3x+2) = [(2x-5)+(x+4)]/(3x+2). Simplifying the numerator we have • (3x-1)/(3x+2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.7.52 (was R.7.48). Show how you found and simplified the expression (x - 1) / x^3 + x / (x^2 + 1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x-1)(x2+1) / x3(x2+1) + x(x3)/ x3(x2+1) x3+x-x2-1+x4 / x3(x2+1) x4+x3-x2+x-1 / x3(x2+1) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Starting with (x-1)/x^3 + x/(x^2+1) we multiply the first term by (x^2 + 1) / (x^2 + 1) and the second by x^3 / x^3 to get a common denominator: [(x-1)/(x^3) * (x^2+1)/(x^2+1)]+[(x)/(x^2+1) * (x^3)/(x^3)], which simplifies to (x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)]. Since the denominator is common to both we combine numerators: (x^3+x-x^2-1+x^4) / [ (x^3)(x^2+1)] . We finally simplify to get (x^4 +x^3 - x^2+x-1) / [ (x^3)(x^2+1)] STUDENT QUESTION I got x^3 + x - x^2 - 1 + x^4 / (x^3)(x^2 + 1). Where did I go wrong? INSTRUCTOR RESPONSE As written your solution x^3 + x - x^2 - 1 + x^4 / (x^3)(x^2 + 1) indicates that you first divide x^4 by x^3, then multiply the result by x^2 + 1. You finally add this result to x^3 + x - x^2 - 1. I do not believe this is what you intended. I believe every step in your solution was correct as you wrote it on paper, and that what you intended to say was correct. However to put it correctly into typewriter notation, you need to group the denominator throughout, and in the last step you need to group your numerator. Using brackets to emphasize the grouping of numerator and denominator: [ x^3 + x - x^2 - 1 + x^4 ] / [(x^3)(x^2 + 1)] We would then put the numerator into order of decreasing exponents, so arrive at an easily-recognized standard form: [ x^4 + x^3 - x^2 + x - 1 ] / [(x^3)(x^2 + 1)] STUDENT QUESTION Why doesn’t the x^3 cancel out??? x^4+x^3-x^2+x-1/x^3(x^2+1)
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Given Solution: x-3, x^3+3x and x^3-9x factor into x-3, x(x^2+3) and x(x^2-9) then into (x-3) , x(x^2+3) , x(x-3)(x+3). The factors x-3, x, x^2 + 3 and x + 3 'cover' all the factors of the three polynomials, and all are needed to do so. The LCM is therefore: • LCM = x(x-3)(x+3)(x^2+3) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.7.64 (was R.7.60). Show how you found and simplified the difference 3x / (x-1) - (x - 4) / (x^2 - 2x + 1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3x/(x-1)(x-4) / (x-1)(x-1) After factoring, the numerator and denominator both have a common denominator of (x-1) confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. Since the first denominator (x - 1) is already a factor of the second, our common denominator is (x - 1)^2. To express the given expression in terms of the common denominator we then multiply the first expression by (x-1) / (x-1) to get 3x(x-1)/(x-1)^2 - (x-4)/(x-1)^2, which gives us (3x^2-3x-x+4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2. STUDENT QUESTION: I don’t see where the 3x(x-1) comes in. When I tried working it on paper, I started with 3x / (x -1) - (x-4) / (x^2 - 2x + 1) and reduced to3x / (x-1) - (x-4) / (x-1)^2But then I tried to solve it by(3x) - (x-4) / x-1 which isn’t right INSTRUCTOR RESPONSE: 3x / (x-1) - (x-4) / (x-1)^2 consists of two fractions with different denominators. You can't do the subtraction until you have a common denominator. The denominator of the first is (x - 1); the denominator of the second is (x - 1)^2. If you multiply the first denominator by (x - 1) you get the second denominator. So to express the first fraction with the same denominator as the second you multiply its numerator and denominator both by ( x - 1 ). You get (3x / (x - 1) ) * (x - 1) / (x - 1) = 3x ( x - 1) / (x - 1)^2. So the subtraction becomes 3x ( x - 1) / ( x - 1) ^ 2 - (x - 4) / (x - 1)^2. The denominator of the combined fraction is (x - 1)^2, while the numerator is (3 x * (x - 1) - (x - 4) ); this simplifies to 3 x^2 - 3 x - x + 4, whic gives us (3x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: QUESTION FROM STUDENT: On the practice test I'm having problems with problem #5 I don't know where to start or how to set it up. I'm probably missing something simple and will probably feel stupid by seeing the solution. Could you help with this problem. A retailer is offering 35% off the purchase price of any pair of shoes during its annual charity sale. The sale price of the shoes pictured in the advertisement is $44.85. Find the original price of the shoes by solving the equation p-.35p = 44.85 for p. INSTRUCTOR RESPONSE: It's very easy to grab onto the wrong idea on a problem and then have trouble shaking it, or to just fail to look at it the right way. Nothing stupid about it, just human nature. See if the following makes sense. If not let me know. p - .35 p = 44.85. Since p - .35 p = 1 p - .35 p = (1 - .35) p = .65 p we have .65 p = 44.85. Multiplying both sides by 1/.65 we get p = 44.85 / .65 = etc. (you can do the division on your calculator); you'll get something near $67). " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: QUESTION FROM STUDENT: On the practice test I'm having problems with problem #5 I don't know where to start or how to set it up. I'm probably missing something simple and will probably feel stupid by seeing the solution. Could you help with this problem. A retailer is offering 35% off the purchase price of any pair of shoes during its annual charity sale. The sale price of the shoes pictured in the advertisement is $44.85. Find the original price of the shoes by solving the equation p-.35p = 44.85 for p. INSTRUCTOR RESPONSE: It's very easy to grab onto the wrong idea on a problem and then have trouble shaking it, or to just fail to look at it the right way. Nothing stupid about it, just human nature. See if the following makes sense. If not let me know. p - .35 p = 44.85. Since p - .35 p = 1 p - .35 p = (1 - .35) p = .65 p we have .65 p = 44.85. Multiplying both sides by 1/.65 we get p = 44.85 / .65 = etc. (you can do the division on your calculator); you'll get something near $67). " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!