course Mth 158 010. `* 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): yes must look at the equation carefully 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 ********************************************* Your solution: Take the 16 to the other side and multiply by 3 giving 2x+1 = (3x-16)3 so 2x+1 = 9x -48 then subtract 1 from -48 and subtract 9x from 2x giving: -7x = -49 and then divide by -7 giving x = 7 Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Must watch out for negative numbers and making sure you subtract and add properly 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 ********************************************* Your solution: Get rid of parenthesis so x^2 - x - 6 = x^2 + 6x + 9 The cancel out the x^2 from both sides leaving -x - 6 = 6x + 9 then Subtract 9 and then x giving -15 = 7x Then divide by 7 so x = -15/7 Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): You have to apply the rules we learned earlier like the distributive law 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)/ ********************************************* Your solution: Factor x^2 – 9 and multiply by the now denominator (x-3)(x+3) and simplify giving x + 4(x-3) = 3 then solve x + 4x - 12 = 3 and then 5x = 15 x = 3. Confidence Assessment: it seemed complicated but once you split it up and factor and see all the common factors it is simpler ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Self-critique (if necessary I didn’t think it all the way through to check if x = 0 it makes the denominator 0 which can not happen so there is no solution. This means that you should always go back and double check because you never know! 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) ********************************************* Your solution: Not sure…but to make common denominator in order to get rid of parenthesis and then solve… Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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) Ok. After seeing the solution it makes more sense on how to approach the problem. * 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0. ********************************************* Your solution: Add -1 to get -ax = b – 1 divide by -a to get x = (1 - b) / a after you multiply by -1 to get rid of the negatives Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): You have to remember what you are solving for when there are other letter in the equation 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. ********************************************* Your solution: Take a common x out giving x(x^2 + 6x - 7) = 0 then factor x ( x+7) ( x - 1) = 0 and then solve meaning x=0, x = -7, and x = 1 Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): with the expression equaling zero you have to solve each with 0 like above question Rating: ********************************************* 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). ********************************************* Your solution: To find the average of the test scores add them up and divide by 4 so (86+80+84+90)/4 And for the average of B and A you must add them together and divide by 2 so (80+90)/2 Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I didn’t do this completely and properly but I can see what the question meant 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. ********************************************* Your solution: Subtract –v0 and divide by g and you get t = -(v-v0) / g Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.