#$&* course Mth 158 9/27 10 010. `* 10
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y Subtract 6 from both sides, giving us 5y = 12 - y Add y to both sides, 5y + y = 12 or 6y = 12 divide both sides by 6 y = 2 INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation in the above note is 5y + 6 = -18 - y. The solution to this equation is found by practically the same steps but you end up with y = -4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: 1.1.38 \ 44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2x+1) / 3 + 16 = 3x multiply both sides by 3 2x+1+48=9x subtract 2x from both sides 49=7x divide both side by 7 7=x confidence rating #$&*: 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:ok ********************************************* 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 distribute x^2 -x -6= x^2 +6x+9 add 6 and subtract 6x on both sides x^2 -7x= x^2 +15 subtract x^2 from both sides -7x=15 divide by -7 on both sides x= -(15/7) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2 First, we use the distributive property to remove the parenthesis and get x^2 - x - 6 = x^2 + 6x + 9 subtract x^2 from both sides, -x - 6 = 6x + 9 Subtract 9 from both sides - x - 6 - 9 = 6x or -x - 15 = 6x add x to both sides -15 = 7x Divide both sides by 7 x = -15/7 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9)/ YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) factor it where needed to find common denominator x/[(x-3)(x+3)] +4/(x+3)= 3/[(x-3)(x+3)] LCD=(x-3)(x+3) so multiply both sides by that and get x + 4(x-3)=3 distribute x+4x-12=3 add 12 to both sides 5x=15 divide both sides by 5 x=3 confidence rating #$&*: 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): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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) find LCM and multiply both sides by it [(8w + 5)*(5w + 7)(10w - 7)] / (10w - 7) = [(4w - 3)*(5w + 7)(10w - 7)] / (5w + 7) cancel where needed and get (8w + 5)*(5w + 7)= (4w - 3)(10w - 7) distribute until you get 40w^2+81w+35=40w^2-58w+21 subtract 40^2 from both sides 81w+35=-58w+21 add 58w to both sides and subtract 35 from both sides to get 139w=-14 divide 139 on both sides w=-(14/139) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * GOOD STUDENT SOLUTION: 1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7) After cancellation the left side reads: (5w+7)(8w + 5) After cancellation the right side reads: (10w - 7)(4w - 3) multiply the factors on each side using the DISTRIBUTIVE LAW Left side becomes: (40w^2) + 81w + 35 Right side becomes: (40w^2) - 58w + 21 3) subtract 40w^2 from both sides add 58w to both sides subtract 35 from both sides Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139) STUDENT QUESTION: (5w+7)(8w+5) = (10w-7)(4w-3) work what you can 40w^2 + 35 = 40w^2 +21 take away 40w^2 from both sides didnt understand this one..; INSTRUCTOR RESPONSE: It doesn't look like you used the distributive law to multiply those binomials. (5w+7)(8w+5) = 5w ( 8w + 5) + 7 ( 8w + 5)= 40 w^2 + 25 w + 56 w + 35 = 40 w^2 + 81 w + 35. (10w-7)(4w-3) = 10 w ( 4 w - 3) - 7 ( 4 w - 3) = 40 w^2 - 30 w - 28 w + 21 = 40 w^2 - 58 w + 21. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 - a x = b, a <> 0 subtract 1 from both sides -ax=b-1 divide -a from both sides x=(b-1)/-a confidence rating #$&*: 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): ???do you need to multiply by the -1???? ------------------------------------------------ Self-critique Rating: ok
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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):ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * 1.1.90 (was 1.2.18). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) scores 86, 80, 84, 90, scores to ave B (80) and A (90). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: final exam =2/3 of final average scores= 86,80, 84, 90 final exam=x so [(86+80+84+90)/4 +2x]/3= final average (340/4 + 2x)/3=fa (85+2x)/3=fa substitute 80 for fa so we have (85+2x)/3= 80 and solve for x which is 85+2x=240 after multiplying both sides by 3. Subtract 85 from both sides to get 2x=155 and divide both sides by x to get x=77.5 , at least to get an 80. Go through the same process, except substitute 90 for fa so (85+2x)/3=90 and you find that x has to = at least 92.5 for Mike to get an A for his final average. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * This can be solved by trial and error but the only acceptable method for this course, in which we are learning to solve problems by means of equations, is by an equation. Let x be the score you make on the exam. The average of the four tests is easy to find: 4-test average = ( 86 + 80 + 84 + 90 ) / 4 = 340 / 4 = 85. The final grade can be thought of as being made up of 3 parts, 1 part being the test average and 2 parts being the exam grade. We would therefore have final average = (1 * test average + 2 * exam grade) / 3. This gives us the equation final ave = (85 + 2 * x) / 3. If the ave score is to be 80 then we solve (85 + 2 * x) / 3 = 80. Multiplying both sides by 3 we get 85 + 2x = 240. Subtracting 85 from both sides we have 2 x = 240 - 85 = 155 so that x = 155 / 2 = 77.5. We can solve (340 + x) / 5 = 90 in a similar manner. We obtain x = 92.5. Alternative solution: If we add 1/3 of the test average to 2/3 of the final exam grade we get the final average. So (using the fact that the test ave is 85%, as calculated above) our equation would be 1/3 * 85 + 2/3 * x = final ave. For final ave = 80 we get 1/3 * 85 + 2/3 * x = 80. Multiplying both sides by 3 we have 85 + 2 * x = 240. The rest of the solution goes as before and we end up with x = 77.5. Solving 1/3 * 85 + 2/3 * x = 90 we get x = 92.5, as before. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.1.82 (was 1.1.90). Explain, step by step, how you solved the equation v = -g t + v0 for t. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v = -g t + v0 subtract v0 from both sides v-v0= -gt divide -g on both sides (v-v0)/-g=t or -(v-v0)/g=t confidence rating #$&*: 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): ok ------------------------------------------------ Self-critique Rating: ok * 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: #*&!