#$&* course Math 158 to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y Subtract 6 from both sides, giving us 5y = 12 - y Add y to both sides, 5y + y = 12 or 6y = 12 divide both sides by 6 y = 2 INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation in the above note is 5y + 6 = -18 - y. The solution to this equation is found by practically the same steps but you end up with y = -4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: 1.1.38 \ 44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You need to multiply each side by 3 to get rid of the denominator. It gives you 2x + 1+48 = 9x. You collect like terms and get 2x +49 = 9x. You would subtract 2x from both sides. You have 49= 7x. You will divide by 7 on both sides and your answer is x = 7. confidence rating #$&*: 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: You will use the distributive property. You will get x^2 - 3x + 2x - 6 = x^2 + 6x + 9. You need to collect like terms on the left: x^2 - x - 6 = x^2 + 6x + 9. You need to subtract x^2 from both sides which cancels out. You will have -x-6= 6x + 9. You will need to subtract 6x from each side. You will have -7x -6 = 9. You will add six to both sides. You get -7x = 15. You will divide by -7 on both sides. You get x = -15/7. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2 First, we use the distributive property to remove the parenthesis and get x^2 - x - 6 = x^2 + 6x + 9 subtract x^2 from both sides, -x - 6 = 6x + 9 Subtract 9 from both sides - x - 6 - 9 = 6x or -x - 15 = 6x add x to both sides -15 = 7x Divide both sides by 7 x = -15/7 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You factor your 1st and 3rd denominators. You get x/(x-3)(x+3) + 4/(x+3) = 3/(x+3)(x-3). You multiply both sides by the common denominator: x*(x-3)(x+3) and 4*(x-3)(x+3) = 3(x-3)(x+3). This gives you: x +4x - 12 =3. You collect like terms and get 5x-12 =3. You add 12 to both sides. You have 5x = 15. You divide both sides by 5 and your answer is x =3. confidence rating #$&*: 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: I would cross multiply and distribute to get: 40w^2 + 81w+35 = 40w^2 - 58w+21. You subtract 40w^2 from both sides. This gives you 0w^2 which means there is no solution. You cannot have 0 in front of a variable. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * GOOD STUDENT SOLUTION: 1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7) After cancellation the left side reads: (5w+7)(8w + 5) After cancellation the right side reads: (10w - 7)(4w - 3) multiply the factors on each side using the DISTRIBUTIVE LAW Left side becomes: (40w^2) + 81w + 35 Right side becomes: (40w^2) - 58w + 21 3) subtract 40w^2 from both sides add 58w to both sides subtract 35 from both sides Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139) STUDENT QUESTION: (5w+7)(8w+5) = (10w-7)(4w-3) work what you can 40w^2 + 35 = 40w^2 +21 take away 40w^2 from both sides didnt understand this one..; INSTRUCTOR RESPONSE: It doesn't look like you used the distributive law to multiply those binomials. (5w+7)(8w+5) = 5w ( 8w + 5) + 7 ( 8w + 5)= 40 w^2 + 25 w + 56 w + 35 = 40 w^2 + 81 w + 35. (10w-7)(4w-3) = 10 w ( 4 w - 3) - 7 ( 4 w - 3) = 40 w^2 - 30 w - 28 w + 21 = 40 w^2 - 58 w + 21. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would subtract 1 from both sides giving you -ax = b-1. You need to divide by -a. You get b-1/-a =x. confidence rating #$&*: 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: You would factor to get (x+7)(x-1). You will then solve x+7 = 0. You will subtract 7 from both sides. You get x = -7. You solve x -1= 0. You add 1 to both sides. You get x=1. Your answers are: x=0, x= -7, and x = 1. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^3 + 6 x^2 - 7 x = 0 factor x out of the left-hand side: x(x^2 + 6x - 7) = 0. Factor the trinomial: x ( x+7) ( x - 1) = 0. Then x = 0 or x + 7 = 0 or x - 1 = 0 so x = 0 or x = -7 or x = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.1.90 (was 1.2.18). The final exam counts as two tests. You have scores of 86, 80, 84, 90. What score do you need on the final in order to end up with a B average, which requires an average score of 80, and an A average, which requires a score of 90? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Problem 1: You need to average the test grades. You get 340/4 = 85. You work out. 85 +2x/3 = 80. The exam is 2 parts of the grade and the exam 1 part. You need to multiply 3 and 80 to get 240. Your problem is 85 + 2x = 240. You subtract 85 from both sides. You get 2x = 155. Divide by 2 on both sides. x= 77.5 to earn 80%. Problem 2: You need to use the test average from above: 85. Set up your problem again: 85 + 2x/3 = 90. You would cross multiply to get 85 + 2x = 270. You would subtract 85 from both sides. You get 2x = 185. You divide by 2 and get x = 92.5 to have a grade of 90%. confidence rating #$&*:: 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: You would divide by g to both sides. You would have v/-g = t +v0. You would subtract v0. Your answer is v -v0/-g =t confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: * Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: * Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!