#$&* course Mth158 10/9/12 around 8:50 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y Subtract 6 from both sides, giving us 5y = 12 - y Add y to both sides, 5y + y = 12 or 6y = 12 divide both sides by 6 y = 2 INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation in the above note is 5y + 6 = -18 - y. The solution to this equation is found by practically the same steps but you end up with y = -4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: 1.1.38 \ 44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2x+1) / 3 + 16 = 3x First you must multiply each side by 3 to get the fraction out. (2x+1) + 48 = 9x Then at the 48 + 1 to get 49. 2x + 49 = 9x Then subtract the 2x from both sides. 49 = 7x Then divide both sides by 7. 7 = x confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: (2x + 1) / 3 + 16 = 3x First, multiply both sides of the equation by 3 2x +1 + 48 =9x or 2x + 49 = 9x subtract 2x from both sides to get 49 = 7x Divide both sides by 7 to get x = 7. STUDENT QUESTION I was wondering at the end since it ended up 49 = 7x and you divide by 7 and say x = 7…would you have to make it a -7 if you move it to the opposite side of the equation? INSTRUCTOR RESPONSE It's not a matter of 'moving things around', but a matter of adding or subtracting the same quantity on both sides, or multiplying or dividing both sides by the same quantity. In this case both sides are divided by 7, which doesn't involve any negative signs. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * was 1.1.44 \ 36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x+2)(x-3) = (x+3)^2 The first thing that we must do is get the parentheses from this problem out. You can use the distributive property or FOIL. x^2 -x - 6 = x^2 +6x +9 Then you subtract X^2 from both sides. -x - 6 = 6x + 9 Then I added 6 to both sides. -x = 6x + 15 Then I subtracted 6x from both sides. -7x = 15 Then I divided both sides by -7. x = -15/7 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2 First, we use the distributive property to remove the parenthesis and get x^2 - x - 6 = x^2 + 6x + 9 subtract x^2 from both sides, -x - 6 = 6x + 9 Subtract 9 from both sides - x - 6 - 9 = 6x or -x - 15 = 6x add x to both sides -15 = 7x Divide both sides by 7 x = -15/7 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) First I factored the x^2-9 which is (x-3)(x+3) Then I multiplied each term by (x-3)(x+3) This caused cancellation, leaving me with x + 4(x-3) = 3 x + 4x -12 = 3 5x -12 = 3 +12 +12 5x = 15 5x / 5 = 15 / 5 x = 3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x / (x^2 -9) + 4 / (x+3) = 3 / (x^2 -9), first factor x^2 - 9 to get x / ( (x-3)(x+3) ) + 4 / (x+3) = 3 / ( (x-3)(x+3) ). Multiply both sides by the common denominator ( (x-3)(x+3) ): ( (x-3)(x+3) ) * x / ( (x-3)(x+3) ) + ( (x-3)(x+3) ) * 4 / (x+3) = ( (x-3)(x+3) ) * 3 / ( (x-3)(x+3) ). Simplify: x + 4(x-3) = 3. Apply the Distributive Law, rearrange and solve: x + 4x - 12 = 3 5x = 15 x = 3. If there is a solution to the original equation it is x = 3. However x = 3 results in denominator 0 when substituted into the original equation, and division by 0 is undefined. So there is no solution to the equation. STUDENT COMMENT x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) Since you have like terms (x^2 - 9) on both sides, they cancel each other out INSTRUCTOR RESPONSE If something 'cancels' by multiplication or division, it has to 'cancel' from all terms. (x^2 - 9) is not a multiplicative or divisive factor of the term 4 / (x + 3) so that factor does not 'cancel'. You can multiply or divide both sides by the same quantity, or add and subtract the same quantity from both sides. Anything called 'cancellation' that doesn't result from these operations is invalid. Because 'cancellation' errors are so common among students at this level, my solutions never mention anything called 'cancellation'. If you multiply both sides of the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) by (x^2 - 9), you get ( x / (x^2-9) + 4 / (x+3) ) * (x^2 - 9) = 3 / (x^2-9) * (x^2 - 9) so that x / (x^2-9) * (x^2 - 9) + 4 / (x+3) * (x^2 - 9) = 3 / (x^2-9) * (x^2 - 9). The (x^2 - 9) does then 'cancel' from two of the three terms, but not from the third. You get x + 4 / (x+3) * (x^2 - 9) = 3. You're still stuck with an x^2 - 9 factor on one of the terms, and a denominator x - 3. However this equation does represent progress. If you factor x^2 - 9 into (x-3)(x+3), things quickly simplify. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 2 Well I do see that I worked the problem correctly, but I honestly had not thought about there not being a true solution. I do see that if you substitute the 3 that you do get 0. It makes sense now. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.1.58 (was 1.1.54). Explain, step by step, how you solved the equation (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7) First we must multiply both sides by the LCM (10w-7)(5w+7) This changes the problem to: (8w+5)(5w+7) = (4w-3)(10w-7) Work this problem using distributive law 40w^2 + 56w + 25w + 35 = 40w^2 -28w -30w +21 40w^2 + 81w + 35 = 40w^2 - 58w +21 Then you must subtract 40w^2 from both sides. 81w + 35 = -58w + 21 Add 58w to both sides. 139w + 35 = 21 Subtract 35 from both sides. 139w = -14 Divide both sides by 139. w = -14 / 139 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * GOOD STUDENT SOLUTION: 1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7) After cancellation the left side reads: (5w+7)(8w + 5) After cancellation the right side reads: (10w - 7)(4w - 3) multiply the factors on each side using the DISTRIBUTIVE LAW Left side becomes: (40w^2) + 81w + 35 Right side becomes: (40w^2) - 58w + 21 3) subtract 40w^2 from both sides add 58w to both sides subtract 35 from both sides Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139) STUDENT QUESTION: (5w+7)(8w+5) = (10w-7)(4w-3) work what you can 40w^2 + 35 = 40w^2 +21 take away 40w^2 from both sides didnt understand this one..; INSTRUCTOR RESPONSE: It doesn't look like you used the distributive law to multiply those binomials. (5w+7)(8w+5) = 5w ( 8w + 5) + 7 ( 8w + 5)= 40 w^2 + 25 w + 56 w + 35 = 40 w^2 + 81 w + 35. (10w-7)(4w-3) = 10 w ( 4 w - 3) - 7 ( 4 w - 3) = 40 w^2 - 30 w - 28 w + 21 = 40 w^2 - 58 w + 21. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 - a x = b, a <> 0 The point of this one is to get x all by itself. To do this you must start adding or subtracting. The first thing I did was subtract a 1 from both sides. -ax = b - 1 Then divide each side by -a x = (b - 1) / a confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Start with 1 -ax = b, a <> 0. Adding -1 to both sides we get 1 - ax - 1 = b - 1, which we simplify to get -ax = b - 1. Divide both sides by -a, which gives you x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get x = (-b + 1) / a or x = (1 - b) / a. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to multiply the right hand side by -1/-1 which made my answer wrong. I see now. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * extra problem (was 1.1.72). Explain, step by step, how you solved the equation x^3 + 6 x^2 - 7 x = 0 using factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^3 + 6 x^2 - 7 x = 0 The first thing to do with this problem is to factor out the x. x(x^2 + 6x - 7) = 0 Then you have to factor the trinomial. x(x+7)(x-1) = 0 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^3 + 6 x^2 - 7 x = 0 factor x out of the left-hand side: x(x^2 + 6x - 7) = 0. Factor the trinomial: x ( x+7) ( x - 1) = 0. Then x = 0 or x + 7 = 0 or x - 1 = 0 so x = 0 or x = -7 or x = 1. ** STUDENT QUESTION I don’t understand this part of the equation x = 0 or x + 7 = 0 or x - 1 = 0 so x = 0 or x = -7 or x = 1 ??? where do you get all of this from??? INSTRUCTOR RESPONSE x ( x+7) ( x - 1) = 0 says that three different quantities, multiplied together, give you zero. Now if three quantities multiplied together give you zero, what is the one thing you know for sure? You know for sure that one of them is zero, because if you multiply three quantities that aren't zero you don't get zero (more specifically if you multiply three numbers, none of which are zero, you don't get zero). The three quantities are x, x + 7 and x - 1. The only way you can get zero by multiplying these quantities is if one of them is zero. So if x ( x+7) ( x - 1) = 0 you know that x = 0, or x + 7 = 0, or x - 1 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Well I did get somewhat confused with this one. I thought that the problem was completed. When really I needed to go a step further. I feel that I understand this now. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 1.1.90 (was 1.2.18). The final exam counts as two tests. You have scores of 86, 80, 84, 90. What score do you need on the final in order to end up with a B average, which requires an average score of 80, and an A average, which requires a score of 90? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would need to score a 70 on the final exam in order to end up with a B average and you would need to score a 100 on the final exam in order to end up with a A average. I came up with my answer by substituting different numbers and adding them together and dividing by 6 to get my average. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * This can be solved by trial and error but the only acceptable method for this course, in which we are learning to solve problems by means of equations, is by an equation. Let x be the score you make on the exam. The average of the four tests is easy to find: 4-test average = ( 86 + 80 + 84 + 90 ) / 4 = 340 / 4 = 85. The final grade can be thought of as being made up of 3 parts, 1 part being the test average and 2 parts being the exam grade. We would therefore have final average = (1 * test average + 2 * exam grade) / 3. This gives us the equation final ave = (85 + 2 * x) / 3. If the ave score is to be 80 then we solve (85 + 2 * x) / 3 = 80. Multiplying both sides by 3 we get 85 + 2x = 240. Subtracting 85 from both sides we have 2 x = 240 - 85 = 155 so that x = 155 / 2 = 77.5. We can solve (340 + x) / 5 = 90 in a similar manner. We obtain x = 92.5. Alternative solution: If we add 1/3 of the test average to 2/3 of the final exam grade we get the final average. So (using the fact that the test ave is 85%, as calculated above) our equation would be 1/3 * 85 + 2/3 * x = final ave. For final ave = 80 we get 1/3 * 85 + 2/3 * x = 80. Multiplying both sides by 3 we have 85 + 2 * x = 240. The rest of the solution goes as before and we end up with x = 77.5. Solving 1/3 * 85 + 2/3 * x = 90 we get x = 92.5, as before. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Well my solution was a lot more simple. I must have mis understood the example. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 1.1.82 (was 1.1.90). Explain, step by step, how you solved the equation v = -g t + v0 for t. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v = -g t + v0 for t First you need to get t by itself. To do this you have to subtract v0 from both sides. v - v0 = -gt Then you must divide both sides by -g. (v-v0)/ -g = t Then simplify and get: t=(-v + v0) / g confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * NOTE: v0 stands for v with subscript 0; the whole expression v0 stands for the name of a variable. It doesn't mean v * 0. Starting with v = -g t + v0, add -v0 to both sides to get v - v0 = -gt. Divide both sides by -g to get (v - v0) / (-g) = t so that t = -(v - v0) / g = (-v + v0) / g. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 2 * Add comments on any surprises or insights you experienced as a result of this assignment. I learned quite a bit working through all of these problems. A brand new depth for Linear Equations. But I will practice some more, because some were confusing. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ________________________________________ #$&*
#$&* course Mth158 10/9/12 around 8:50 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y Subtract 6 from both sides, giving us 5y = 12 - y Add y to both sides, 5y + y = 12 or 6y = 12 divide both sides by 6 y = 2 INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation in the above note is 5y + 6 = -18 - y. The solution to this equation is found by practically the same steps but you end up with y = -4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: 1.1.38 \ 44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2x+1) / 3 + 16 = 3x First you must multiply each side by 3 to get the fraction out. (2x+1) + 48 = 9x Then at the 48 + 1 to get 49. 2x + 49 = 9x Then subtract the 2x from both sides. 49 = 7x Then divide both sides by 7. 7 = x confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: (2x + 1) / 3 + 16 = 3x First, multiply both sides of the equation by 3 2x +1 + 48 =9x or 2x + 49 = 9x subtract 2x from both sides to get 49 = 7x Divide both sides by 7 to get x = 7. STUDENT QUESTION I was wondering at the end since it ended up 49 = 7x and you divide by 7 and say x = 7…would you have to make it a -7 if you move it to the opposite side of the equation? INSTRUCTOR RESPONSE It's not a matter of 'moving things around', but a matter of adding or subtracting the same quantity on both sides, or multiplying or dividing both sides by the same quantity. In this case both sides are divided by 7, which doesn't involve any negative signs. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * was 1.1.44 \ 36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x+2)(x-3) = (x+3)^2 The first thing that we must do is get the parentheses from this problem out. You can use the distributive property or FOIL. x^2 -x - 6 = x^2 +6x +9 Then you subtract X^2 from both sides. -x - 6 = 6x + 9 Then I added 6 to both sides. -x = 6x + 15 Then I subtracted 6x from both sides. -7x = 15 Then I divided both sides by -7. x = -15/7 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2 First, we use the distributive property to remove the parenthesis and get x^2 - x - 6 = x^2 + 6x + 9 subtract x^2 from both sides, -x - 6 = 6x + 9 Subtract 9 from both sides - x - 6 - 9 = 6x or -x - 15 = 6x add x to both sides -15 = 7x Divide both sides by 7 x = -15/7 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) First I factored the x^2-9 which is (x-3)(x+3) Then I multiplied each term by (x-3)(x+3) This caused cancellation, leaving me with x + 4(x-3) = 3 x + 4x -12 = 3 5x -12 = 3 +12 +12 5x = 15 5x / 5 = 15 / 5 x = 3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x / (x^2 -9) + 4 / (x+3) = 3 / (x^2 -9), first factor x^2 - 9 to get x / ( (x-3)(x+3) ) + 4 / (x+3) = 3 / ( (x-3)(x+3) ). Multiply both sides by the common denominator ( (x-3)(x+3) ): ( (x-3)(x+3) ) * x / ( (x-3)(x+3) ) + ( (x-3)(x+3) ) * 4 / (x+3) = ( (x-3)(x+3) ) * 3 / ( (x-3)(x+3) ). Simplify: x + 4(x-3) = 3. Apply the Distributive Law, rearrange and solve: x + 4x - 12 = 3 5x = 15 x = 3. If there is a solution to the original equation it is x = 3. However x = 3 results in denominator 0 when substituted into the original equation, and division by 0 is undefined. So there is no solution to the equation. STUDENT COMMENT x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) Since you have like terms (x^2 – 9) on both sides, they cancel each other out INSTRUCTOR RESPONSE If something 'cancels' by multiplication or division, it has to 'cancel' from all terms. (x^2 - 9) is not a multiplicative or divisive factor of the term 4 / (x + 3) so that factor does not 'cancel'. You can multiply or divide both sides by the same quantity, or add and subtract the same quantity from both sides. Anything called 'cancellation' that doesn't result from these operations is invalid. Because 'cancellation' errors are so common among students at this level, my solutions never mention anything called 'cancellation'. If you multiply both sides of the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) by (x^2 - 9), you get ( x / (x^2-9) + 4 / (x+3) ) * (x^2 - 9) = 3 / (x^2-9) * (x^2 - 9) so that x / (x^2-9) * (x^2 - 9) + 4 / (x+3) * (x^2 - 9) = 3 / (x^2-9) * (x^2 - 9). The (x^2 - 9) does then 'cancel' from two of the three terms, but not from the third. You get x + 4 / (x+3) * (x^2 - 9) = 3. You're still stuck with an x^2 - 9 factor on one of the terms, and a denominator x - 3. However this equation does represent progress. If you factor x^2 - 9 into (x-3)(x+3), things quickly simplify. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 2 Well I do see that I worked the problem correctly, but I honestly had not thought about there not being a true solution. I do see that if you substitute the 3 that you do get 0. It makes sense now. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.1.58 (was 1.1.54). Explain, step by step, how you solved the equation (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7) First we must multiply both sides by the LCM (10w-7)(5w+7) This changes the problem to: (8w+5)(5w+7) = (4w-3)(10w-7) Work this problem using distributive law 40w^2 + 56w + 25w + 35 = 40w^2 -28w -30w +21 40w^2 + 81w + 35 = 40w^2 - 58w +21 Then you must subtract 40w^2 from both sides. 81w + 35 = -58w + 21 Add 58w to both sides. 139w + 35 = 21 Subtract 35 from both sides. 139w = -14 Divide both sides by 139. w = -14 / 139 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * GOOD STUDENT SOLUTION: 1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7) After cancellation the left side reads: (5w+7)(8w + 5) After cancellation the right side reads: (10w - 7)(4w - 3) multiply the factors on each side using the DISTRIBUTIVE LAW Left side becomes: (40w^2) + 81w + 35 Right side becomes: (40w^2) - 58w + 21 3) subtract 40w^2 from both sides add 58w to both sides subtract 35 from both sides Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139) STUDENT QUESTION: (5w+7)(8w+5) = (10w-7)(4w-3) work what you can 40w^2 + 35 = 40w^2 +21 take away 40w^2 from both sides didnt understand this one..; INSTRUCTOR RESPONSE: It doesn't look like you used the distributive law to multiply those binomials. (5w+7)(8w+5) = 5w ( 8w + 5) + 7 ( 8w + 5)= 40 w^2 + 25 w + 56 w + 35 = 40 w^2 + 81 w + 35. (10w-7)(4w-3) = 10 w ( 4 w - 3) - 7 ( 4 w - 3) = 40 w^2 - 30 w - 28 w + 21 = 40 w^2 - 58 w + 21. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 - a x = b, a <> 0 The point of this one is to get x all by itself. To do this you must start adding or subtracting. The first thing I did was subtract a 1 from both sides. -ax = b - 1 Then divide each side by -a x = (b - 1) / a confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Start with 1 -ax = b, a <> 0. Adding -1 to both sides we get 1 - ax - 1 = b - 1, which we simplify to get -ax = b - 1. Divide both sides by -a, which gives you x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get x = (-b + 1) / a or x = (1 - b) / a. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to multiply the right hand side by -1/-1 which made my answer wrong. I see now. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * extra problem (was 1.1.72). Explain, step by step, how you solved the equation x^3 + 6 x^2 - 7 x = 0 using factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^3 + 6 x^2 - 7 x = 0 The first thing to do with this problem is to factor out the x. x(x^2 + 6x - 7) = 0 Then you have to factor the trinomial. x(x+7)(x-1) = 0 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^3 + 6 x^2 - 7 x = 0 factor x out of the left-hand side: x(x^2 + 6x - 7) = 0. Factor the trinomial: x ( x+7) ( x - 1) = 0. Then x = 0 or x + 7 = 0 or x - 1 = 0 so x = 0 or x = -7 or x = 1. ** STUDENT QUESTION I don’t understand this part of the equation x = 0 or x + 7 = 0 or x - 1 = 0 so x = 0 or x = -7 or x = 1 ??? where do you get all of this from??? INSTRUCTOR RESPONSE x ( x+7) ( x - 1) = 0 says that three different quantities, multiplied together, give you zero. Now if three quantities multiplied together give you zero, what is the one thing you know for sure? You know for sure that one of them is zero, because if you multiply three quantities that aren't zero you don't get zero (more specifically if you multiply three numbers, none of which are zero, you don't get zero). The three quantities are x, x + 7 and x - 1. The only way you can get zero by multiplying these quantities is if one of them is zero. So if x ( x+7) ( x - 1) = 0 you know that x = 0, or x + 7 = 0, or x - 1 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Well I did get somewhat confused with this one. I thought that the problem was completed. When really I needed to go a step further. I feel that I understand this now. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 1.1.90 (was 1.2.18). The final exam counts as two tests. You have scores of 86, 80, 84, 90. What score do you need on the final in order to end up with a B average, which requires an average score of 80, and an A average, which requires a score of 90? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would need to score a 70 on the final exam in order to end up with a B average and you would need to score a 100 on the final exam in order to end up with a A average. I came up with my answer by substituting different numbers and adding them together and dividing by 6 to get my average. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * This can be solved by trial and error but the only acceptable method for this course, in which we are learning to solve problems by means of equations, is by an equation. Let x be the score you make on the exam. The average of the four tests is easy to find: 4-test average = ( 86 + 80 + 84 + 90 ) / 4 = 340 / 4 = 85. The final grade can be thought of as being made up of 3 parts, 1 part being the test average and 2 parts being the exam grade. We would therefore have final average = (1 * test average + 2 * exam grade) / 3. This gives us the equation final ave = (85 + 2 * x) / 3. If the ave score is to be 80 then we solve (85 + 2 * x) / 3 = 80. Multiplying both sides by 3 we get 85 + 2x = 240. Subtracting 85 from both sides we have 2 x = 240 - 85 = 155 so that x = 155 / 2 = 77.5. We can solve (340 + x) / 5 = 90 in a similar manner. We obtain x = 92.5. Alternative solution: If we add 1/3 of the test average to 2/3 of the final exam grade we get the final average. So (using the fact that the test ave is 85%, as calculated above) our equation would be 1/3 * 85 + 2/3 * x = final ave. For final ave = 80 we get 1/3 * 85 + 2/3 * x = 80. Multiplying both sides by 3 we have 85 + 2 * x = 240. The rest of the solution goes as before and we end up with x = 77.5. Solving 1/3 * 85 + 2/3 * x = 90 we get x = 92.5, as before. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Well my solution was a lot more simple. I must have mis understood the example. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 1.1.82 (was 1.1.90). Explain, step by step, how you solved the equation v = -g t + v0 for t. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v = -g t + v0 for t First you need to get t by itself. To do this you have to subtract v0 from both sides. v - v0 = -gt Then you must divide both sides by -g. (v-v0)/ -g = t Then simplify and get: t=(-v + v0) / g confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * NOTE: v0 stands for v with subscript 0; the whole expression v0 stands for the name of a variable. It doesn't mean v * 0. Starting with v = -g t + v0, add -v0 to both sides to get v - v0 = -gt. Divide both sides by -g to get (v - v0) / (-g) = t so that t = -(v - v0) / g = (-v + v0) / g. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 2 * Add comments on any surprises or insights you experienced as a result of this assignment. I learned quite a bit working through all of these problems. A brand new depth for Linear Equations. But I will practice some more, because some were confusing.