course MTH 158
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20:42:11 query 1.1.20 (was 1.1.12). Explain, step by step, how you solved the equation 5y + 6 = -18 - y
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RESPONSE --> I first add the 18 to the 5y+6 side, you do this because it was neg. on the opposite side. this gives you 5y+24=-y. My personal preference is to subtract the 24 back to the other side and add the y to the 5y side. You now have 5y+y=-24 or 6y=-24 and your final answer is y= -4
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20:42:45 ** 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 as I note it is 5y + 6 = -18 - y. If that's the correct equation the solution is found by practically the same steps but you end up with y = -4. **
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RESPONSE --> ok
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20:56:03 query 1.1.44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x
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RESPONSE --> first multiply by 3 to get 6x+3/9+48. Add the denominator to get 6x+3/57=x now cancel out the 3 because it divides into the 57. Now you have 6x+1/19=x . Next muliply by 6x to get 1/19=6x^2 x^2=6/19, then x=sqrt6/19
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20:57:51 ** 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. 49 = 7x Divide both sides by 7 7 = x **
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RESPONSE --> I understand now
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21:04:01 query was 1.1.36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2
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RESPONSE --> use the distributive property to get x^2-x-6=x^2+9 now add 6 x^2-x=x^2+15 substract the x^2 and now you have -x=15, multiply by the neg to get x=-15
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21:04:38 ** 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 -15/7 = x **
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RESPONSE --> I had the x+3^2 set up as x+3 and x-3
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21:11:22 query 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)/
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RESPONSE --> I canceled out like terms after breaking down the x^2-9 s an got x/4=3/x=3 youi now have x^2+3x=12 now divide by 3 to get x^2/3=4 now multiply by the 3 to get x^2=12 and its x=sqrt12
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21:11:35 ** 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. Simplify 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. When you multiplied both sides by x-3, if x = 3 you were multiplying by zero, which invalidated your solution **
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RESPONSE --> ok
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21:14:25 query 1.1.58 (was 1.1.54). Explain, step by step, how you solved the equation (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7)
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RESPONSE --> i cross multiplied and got 40w^2+35=40w2+21 you subtract the 40w^2 and now you have 35=21 its not solvable
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21:14:56 ** 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) DER**
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RESPONSE --> ok
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21:18:49 query 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0.
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RESPONSE --> first subtract 1 to get b-1=-ax now divide by a to have b-1/a=-x change the sign to have -b+1/a=x
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21:19:09 ** 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. **
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RESPONSE --> ok
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21:22:15 query 1.1.72. Explain, step by step, how you solved the equation x^3 + 6 x^2 - 7 x = 0 using factoring.
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RESPONSE --> x^2+6x^2-7x=0 add 7x to get 7x^2=7x now divide by 7x to get x=0
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21:22:22 ** 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. **
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RESPONSE --> ok
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21:29:41 query 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).
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RESPONSE --> I first got his average by adding all 4 grades together and dividing by 4. This gives me 64. i set up each equation as 65+x=80. For this one i substract the 65 from the 50 to get x=15 then i multiplied this by 4. So he needs to make a 60 to get a 80. Then 64+x=90 to get x=26 then i mulitplied thsi by 4 to get 104. So unless there is extra credit there is no way he can make a A
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21:29:57 ** 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. **
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RESPONSE --> ok
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21:31:42 query 1.1.82 (was 1.1.90). Explain, step by step, how you solved the equation v = g t + v0 for t.
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RESPONSE --> first substract v0 to get v-v0=-gt next divide by t to get v-v0/t=-g
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21:31:59 ** 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 }or t = (v - v0) / g. **
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RESPONSE --> sorry i solved for g, stupid mistake!
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21:32:07 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> none
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21:32:13 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> none
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If you self-critique your work thoroughly, which will give me the information I need to know whether (and if so you) to assist you, you should succeed in this course.