course Mth 158 `ǾS}Et_assignment #010
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17:53:35 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 --> 5y + 6 = -18 - y (5y+6)-6=(-18-y)-6 5y=-24-y (5y)+y=(-24-y)+y 6y=-24 (6y/6)=(-24/6) y= - 4 confidence assessment: 3
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17:55:09 ** 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 --> yes...I notice that it was wrote as you done soo... self critique assessment: 3
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18:14:11 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 --> (2x+)/3+16=3x 3[(2x+1)/3+16]=[3x]3 6x+3+48=9x 6x+51=9x (6x+51)-51=(9x)-51 6x=9x-51 (6x)-9x=(9x-51)-9x -3x=-51 (-3x/-3)=(-51/-3) x=17 But when checked both sides do not equal out. NO SOLUTION! confidence assessment: 2
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18:16:08 ** 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 --> oh..Ii did it wrong...sorry! I see what I did, too. I multiplied the 2x+1 by 3, too when I should have cancelled out the 3. self critique assessment: 2
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18:26:21 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 --> ok??? x can not = -3 (x+2)(x+3)=(x+3)^2 x^2+5x+6=x^2+9 (x^2+5x+6)-9=(x^2+9)-9 x^2+5x-3=x^2 (x^2+5x-3)-x^2=(x^2)-x^2 5x-3=0 (5x-3)+3=0+3 5x=3 5x/5=3/5 x=3/5 confidence assessment: 2
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18:32:05 ** 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 --> So I know when to multiply,subtract and add and when to divide. Why can't I get these right??? hmmmmmm.... Question??? How do you get x^2 - x - 6 from (x+2)(x+3)?
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18:41:00 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 --> x/(x^2-9)+4/(x+3)=3/(x^2-9) factor out x^2-9 which is (x-3)(x+3) multiply both sides by (x-3)(x+3) (x-3)(x+3)[x/(x-3)(x+3)+4/(x-3)=[3/(x-3)(x+3)](x-3)(x+3) x+4(x-3)=3 x+4x-12=3 5x-12=3 (5x-12)+12=3+12 5x=15 5x/5=15/5 x=3 confidence assessment: 3
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18:41:22 ** 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 self critique assessment: 2
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18:42:41 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 --> ???? confidence assessment: 0
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18:47:53 ** 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 maybe I have been looking at this screen tooo long! I understand. self critique assessment: 2
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18:53:42 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 --> x=(b-1)/-a 1-ax=b (1-ax)-1=b-1 -ax=b-1 -ax/-a=(b-1)/-a x=(b-1)/-a confidence assessment: 1
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18:54:54 ** 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 so I could further went with this expression and gotten ride of the negative. But I was close. self critique assessment: 2
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19:00:17 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 --> t=(v-v0)/g confidence assessment: 1
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19:00:59 ** 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 --> yeah! I got this one. but didn't show my work, sorry. self critique assessment: 2
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