course Mth 158 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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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):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* Question: 1.1.38 \ 44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to get rid of the denominator, multiply both sides and all terms by 3 to give you 2x+49 = 9x Then subtract 2x from similar terms on both sides of the eqution. 49=7x Divide by 7 on both sides to get x by itself. 7=x confidence rating #$&* 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):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* Question: * was 1.1.44 \ 36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, distribute all terms in the equation, giving you x^2-x-6 =x^2+6x+9 Subtract x^2 from both equations to cancel x^2, leaving you with -x-6=6x+9 Add –x to the second equation to get all x values together,giving you -6=7x+9 Subtract 9 from -6, giving you -15=7x Divide 7 out of x in order to get x by itself. X=-15/7 confidence rating #$&* 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* Question: * 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9)/ YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would want to first factor all terms to their lowest factor possible, so x^2-9 is a difference of squares and can be factored as follows, x/(x+3)(x-3)+4/(x+3)=3/(x+3)(x-3) Then, find a common denominator for the all terms, which is (x-3)(x+3). Multiply these terms by the terms in the entire equation and simplify, giving you X+4(x-3)=3 Distribute the 4 to the terms in the parenthesis and combine like terms. x + 4x - 12 = 3 5x = 15 Divide 5 from both equations to get x by itself. x = 3. confidence rating #$&* 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):0
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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. * 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: First, add -1 to both sides 1-ax-1= b–1 Then simplify to get -ax = b–1 Divide both sides by -a x = (b - 1) / (-a) Then multiply both sides by -1 to get rid of the negative in the denominator x = (1 - b) / a confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Start with 1 -ax = b, a <> 0. Adding -1 to both sides we get 1 - ax - 1 = b - 1, which we simplify to get -ax = b - 1. Divide both sides by -a, which gives you x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get x = (-b + 1) / a or x = (1 - b) / a. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Instead of multiplying by -1, I should have been more clear and stated that the exact term I was multiplying was -1/-1, although it gives me the same answer. ------------------------------------------------ Self-critique rating #$&*1 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This equation is in quadratic form, Therefore, factor out likle terms, in this case, X x(x^2 + 6x - 7) = 0 Then you are left with a trinomial, which can be factored to find values for x x ( x+7) ( x - 1) = 0 Then set each equation equal to zero and solve for X. x = 0 or x = -7 or x =1 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^3 + 6 x^2 - 7 x = 0 factor x out of the left-hand side: x(x^2 + 6x - 7) = 0. Factor the trinomial: x ( x+7) ( x - 1) = 0. Then x = 0 or x + 7 = 0 or x - 1 = 0 so x = 0 or x = -7 or x = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* 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: This equation is in quadratic form, Therefore, factor out likle terms, in this case, X x(x^2 + 6x - 7) = 0 Then you are left with a trinomial, which can be factored to find values for x x ( x+7) ( x - 1) = 0 Then set each equation equal to zero and solve for X. x = 0 or x = -7 or x =1 confidence rating #$&* 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 did not understand the initial question. Now that I understand that I am trying to find an equation to give me the final average grade. First give designate X as my score on the final. Then find the average of all the tests by adding all scores and dividing by 4. ( 86 + 80 + 84 + 90 ) / 4 = 340 / 4 = 85 The equation for the final average is as follows. final ave = (85 + 2 * x) / 3 • final ave = (85 + 2 * x) / 3. If the ave score is to be 80 then we solve (85 + 2 * x) / 3 = 80. Multiply both sides by 3 85 + 2x = 240 Subtract 85 from both sides 2 x = 240 - 85 = 155 so that x = 155 / 2 = 77.5 Then solve (340 + x) / 5 = 90 x = 92.5 ------------------------------------------------ Self-critique rating #$&*1 ********************************************* 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: With the equation v = -g t + v0 for t, first divide the second equation by t to get it on the other side. V/T=-g+v0 Then multiply each side by V to get rid of the V in the numerator in the equation on the left. T=(-g+v0)*v T=-vg+v^20
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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): If I start with v = -g t + v0, I would need to add -v0 to both sides to get v - v0 = -gt Then I would divide both sides by -g to get (v - v0) / (-g) = t Then, set the equation equal to zero to get t = -(v - v0) / g = (-v + v0) / g ------------------------------------------------ Self-critique rating #$&*1 * Add comments on any surprises or insights you experienced as a result of this assignment. Some of the questions were a little confusting and hard to understand at first, which made me doubt my answers. "