course Mth 158 \…Ò`ͤȱÈ\Áª²“íñÑÊùGß[Òassignment #010
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15:04:11 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 --> First I added y to both sides 6y + 6 = -18. then subtracted 6 from both sides 6y = -24, then divided by 6 and got y= - 4. confidence assessment: 2
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15:04:44 ** 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 --> I got it right. self critique assessment: 3
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15:10:26 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 --> I first added 16 to both sides (2x +1)/ 3 = 3x - 16. Then to get rid of the fraction I multiplied both sides by 3 and got 2x + 1 = 3(3x - 16).Next I distributed the 3, that gave me 2x + 1 = 9x -48. Then I subtracted 2x from both sides. 1 = 7x - 48, next I added 48 to both sides 49 = 7x, divided by 7 and found x = 7 confidence assessment: 3
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15:11:14 ** 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 did the steps in a diferent order, but still arrived at the correct solution. self critique assessment: 3
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15:21:25 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 --> First, I multiplied the two binomials together. I distributed x to the second set and got x^2 - 3x. Then distributed 2 and got 2x - 6. Then I put the two together and got x^2 - 3x + 2x - 6, combined like terms to find x^2 -x - 6.Then I sqared ( x + 3) and got x^2 + 3x + 3x+ 9, combined like terms and got x^2+6x + 9 = x^2 - x - 6. Next I added 6 to both sides and got x^2 + 6x + 15 = x^2 - x. Then I added x to both sides and got x^2 + 7x + 15 = x^2. Next I subtracted x^2 from both sides 7x + 15 = 0. Then I subtracted 15 from both sides 7x = -15 divided by 7 and found x to be -15/ 7. confidence assessment: 1
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15:22:04 ** 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 --> That what I did. self critique assessment: 3
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15:27:04 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 factored (x^2 - 9) into (x + 3)(x - 3). To get a common denominator I multiplied 4 / (x + 3) by ( x- 3) to get 4x- 12/ ( x+3)(x - 3). Then I used the equation x + 4x - 12 = 3 to find x. I added x+ 4x = 5x. 5x - 12 = 3, added 12 to both sides 5x = 15 divided by 5 and x = 3. confidence assessment: 2
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15:28:15 ** 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 --> I found x= 3, but I didn't think about it making the denominator 0. I know that I denominator of 0 is undefined. self critique assessment: 2
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15:33:58 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 set both numerators equal to one another. 8w + 5 = 4w - 3. Subtracted 4w from both sides and got 4w+5= -3, then subtracted 5 from both sides and got 4w = -8, divided by 4 and got w= -2. When I set the denominators equal to each other 10w - 7 = 5w + 7 and subtracted 5w from both sides I got 5w - 7 = 7, added 7 to both sides 5w = 14, divided by 5 and got w= 5/14. So w= - 2 and 5/14 confidence assessment: 2
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15:36:13 ** 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 --> I did that one wrong I didn't use LCM I understand how to get LCM and how to use it in this problem. I can see how you used the distributive property to find the equation and how you found the solution. self critique assessment: 2
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15:38:38 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 --> I first subtracted 1 from both sides -ax = b -1. then divided by -a to get x = (b - 1) / -a confidence assessment: 1
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15:40:28 ** 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 --> I didn't multiply the right side by -1. Why did you? self critique assessment: 2
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15:43:07 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 --> v0 = 0 because any number* 0 = 0, so I re-wrote the problem as v = gt then divied by g to get v/g = t. confidence assessment: 1
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15:44:14 ** 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 --> I thought that v0 meant v* 0. I understans that it is a subscript. I understand how you solved the problem for the correct answer. self critique assessment: 2
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15:45:16 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> It suprised me that I remember how to solve basic equations, its been awhile since I had done that. confidence assessment: 3
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course Mth 158 \…Ò`ͤȱÈ\Áª²“íñÑÊùGß[Òassignment #010
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15:04:11 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 --> First I added y to both sides 6y + 6 = -18. then subtracted 6 from both sides 6y = -24, then divided by 6 and got y= - 4. confidence assessment: 2
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15:04:44 ** 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 --> I got it right. self critique assessment: 3
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15:10:26 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 --> I first added 16 to both sides (2x +1)/ 3 = 3x - 16. Then to get rid of the fraction I multiplied both sides by 3 and got 2x + 1 = 3(3x - 16).Next I distributed the 3, that gave me 2x + 1 = 9x -48. Then I subtracted 2x from both sides. 1 = 7x - 48, next I added 48 to both sides 49 = 7x, divided by 7 and found x = 7 confidence assessment: 3
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15:11:14 ** 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 did the steps in a diferent order, but still arrived at the correct solution. self critique assessment: 3
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15:21:25 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 --> First, I multiplied the two binomials together. I distributed x to the second set and got x^2 - 3x. Then distributed 2 and got 2x - 6. Then I put the two together and got x^2 - 3x + 2x - 6, combined like terms to find x^2 -x - 6.Then I sqared ( x + 3) and got x^2 + 3x + 3x+ 9, combined like terms and got x^2+6x + 9 = x^2 - x - 6. Next I added 6 to both sides and got x^2 + 6x + 15 = x^2 - x. Then I added x to both sides and got x^2 + 7x + 15 = x^2. Next I subtracted x^2 from both sides 7x + 15 = 0. Then I subtracted 15 from both sides 7x = -15 divided by 7 and found x to be -15/ 7. confidence assessment: 1
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15:22:04 ** 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 --> That what I did. self critique assessment: 3
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15:27:04 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 factored (x^2 - 9) into (x + 3)(x - 3). To get a common denominator I multiplied 4 / (x + 3) by ( x- 3) to get 4x- 12/ ( x+3)(x - 3). Then I used the equation x + 4x - 12 = 3 to find x. I added x+ 4x = 5x. 5x - 12 = 3, added 12 to both sides 5x = 15 divided by 5 and x = 3. confidence assessment: 2
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15:28:15 ** 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 --> I found x= 3, but I didn't think about it making the denominator 0. I know that I denominator of 0 is undefined. self critique assessment: 2
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15:33:58 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 set both numerators equal to one another. 8w + 5 = 4w - 3. Subtracted 4w from both sides and got 4w+5= -3, then subtracted 5 from both sides and got 4w = -8, divided by 4 and got w= -2. When I set the denominators equal to each other 10w - 7 = 5w + 7 and subtracted 5w from both sides I got 5w - 7 = 7, added 7 to both sides 5w = 14, divided by 5 and got w= 5/14. So w= - 2 and 5/14 confidence assessment: 2
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15:36:13 ** 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 --> I did that one wrong I didn't use LCM I understand how to get LCM and how to use it in this problem. I can see how you used the distributive property to find the equation and how you found the solution. self critique assessment: 2
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15:38:38 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 --> I first subtracted 1 from both sides -ax = b -1. then divided by -a to get x = (b - 1) / -a confidence assessment: 1
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15:40:28 ** 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 --> I didn't multiply the right side by -1. Why did you? self critique assessment: 2
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15:43:07 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 --> v0 = 0 because any number* 0 = 0, so I re-wrote the problem as v = gt then divied by g to get v/g = t. confidence assessment: 1
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15:44:14 ** 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 --> I thought that v0 meant v* 0. I understans that it is a subscript. I understand how you solved the problem for the correct answer. self critique assessment: 2
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15:45:16 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> It suprised me that I remember how to solve basic equations, its been awhile since I had done that. confidence assessment: 3
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