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course Phy 231 8/30 4:15 I have done the assignments that were due Tuesday. I use Windows 7 and I expect to use a laptop, calculator, and a smartPhone..............................................
Given Solution: The system 3a + 3b = 9 6a + 5b = 16 can be solved by adding an appropriate multiple of one equation in order to eliminate one of the variables. Since the coefficient of a in the second equation (the coefficient of a in the second equation is 6)) is double that in the first (the coefficient of a in the first equation is 3), we can multiply the first equation by -2 in order to make the coefficients of a equal and opposite: -2 * [ 3a + 3b ] = -2 [ 9 ] 6a + 5b = 16 gives us -6a - 6 b = -18 6a + 5b = 16 . Adding the two equations together we obtain -b = -2, or just b = 2. Substituting b = 2 into the first equation we obtain 3 a + 3(2) = 9, or 3 a + 6 = 9 so that 3 a = 3 and a = 1. Our solution is therefore a = 1, b = 2. We used the first equation in our last step, so we verigy this solution is by substituting these values into the second equation, where we get 6 * 1 + 5 * 2 = 6 + 10 = 16. STUDENT QUESTION I got my answer in a very different way than the solution given. I have been trying to remember things from the classes I took a long time ago and came up with this answer. Is it alright to use this method? INSTRUCTOR RESPONSE Here is a synopsis of your solution: I'll first solve the first equation for a: 3a+3b=9 so a+b=3 so a=3-b. Now I'll substitute this expression for a into the second equation 6 a + 5 b = 16 Replacing a with 3 - b: 6(3-b)+5b=16 18-6b+5b=16 -b=-2 b=2 a = 3 - b so a=3 - 2 = 1 Substituting a = 1 and b = 2 into the two equations we get 3(1)+3(2)=9 so 9 = 9 6(1)+5(2)=16 so 16 = 16. The solution checks with the two equations. You have an excellent solution. The method you have used is performed correctly and is equally valid with the method used in the solutions. It is called the 'substitution method'. For these first few problems in this course the substitution method and the elimination method are equally efficient. However the elimination method is also important, and since elimination works better on most of the problems we'll be encountering in the near future, it is the method I use in the given solutions. You can use either method, as long as you know both. However you might find the given solutions easier to understand if you use the elimination method. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Substitution 3a + 3b = 9 6a + 5b = 16 3b = 9 - 3a b = 3 - a 6a + 5(3 - a) = 16 6a + 15 - 5a = 16 a + 15 = 16 a = 1 6(1) + 5b = 16 6 + 5b = 16 5b = 10 b = 2 ------------------------------------------------ Self-critique Rating: 3
********************************************* Question: `q006. Solve the following system of simultaneous linear equations using the method of elimination: 4a + 5b = 18 6a + 9b = 30. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To get the same coefficient in front of both a's with one a having an opposite sign, you can multiply the top equation by 6 and the bottom by -4. 6(4a + 5b = 18) 24a + 30b = 108 -4(6a + 9b = 30) + (-24a - 36b = -120) After adding like variables, you eliminate variable a 0a - 6b = -12 -6b = -12 b = 2 Plug value for b into equation 4a + 5(2) = 18 4a + 10 = 18 4a = 8 a = 2 a = 2, b = 2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: In the system 4a + 5b = 18 6a + 9b = 30 we see that the coefficients of b are relatively prime; they therefore have a least common multiple equal to 5 * 9. The coefficients 4 and 6 of a have a least common multiple of 12. We have a choice of which variable to eliminate. We could 'match' the b by multiplying the first equation by 9 and the second by -5, or we could match the coefficients of a by multiplying the first equation by 3 and the second by -2. Either choice would work. The numbers required to 'match' the coefficients of a are smaller, but the numbers required to 'match the coefficients of b would otherwise work equally well. Choosing to 'match' the coefficient of a, we obtain 3 * [4a + 5b ] = 3 * 18 -2 * [ 6a + 9b ] = -2 * 30, so the system becomes 12 a + 15 b = 54 -12 a - 18 b = -60. Adding the equations we get -3 b = -6, so b = 2. Substituting this value of b into the first equation we obtain 4 a + 5 * 2 = 18, or 4 a + 10 = 18, which we easily solve to obtain a = 2. Substituting this value of a into the second equation we obtain 6 * 2 + 9 * 2 = 30, which verifies our solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It didn't occur to me that the least common multiple of 6 and 4 is 12. However, you can safely multiply 6 by 4 to eliminate one of the variables and still get the correct values for a and b ------------------------------------------------ Self-critique Rating: 3
" Self-critique (if necessary): ------------------------------------------------ Self-critique rating:#$&* course Phy 231 8/30 4:15 I have done the assignments that were due Tuesday. I use Windows 7 and I expect to use a laptop, calculator, and a smartPhone.
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Given Solution: The system 3a + 3b = 9 6a + 5b = 16 can be solved by adding an appropriate multiple of one equation in order to eliminate one of the variables. Since the coefficient of a in the second equation (the coefficient of a in the second equation is 6)) is double that in the first (the coefficient of a in the first equation is 3), we can multiply the first equation by -2 in order to make the coefficients of a equal and opposite: -2 * [ 3a + 3b ] = -2 [ 9 ] 6a + 5b = 16 gives us -6a - 6 b = -18 6a + 5b = 16 . Adding the two equations together we obtain -b = -2, or just b = 2. Substituting b = 2 into the first equation we obtain 3 a + 3(2) = 9, or 3 a + 6 = 9 so that 3 a = 3 and a = 1. Our solution is therefore a = 1, b = 2. We used the first equation in our last step, so we verigy this solution is by substituting these values into the second equation, where we get 6 * 1 + 5 * 2 = 6 + 10 = 16. STUDENT QUESTION I got my answer in a very different way than the solution given. I have been trying to remember things from the classes I took a long time ago and came up with this answer. Is it alright to use this method? INSTRUCTOR RESPONSE Here is a synopsis of your solution: I'll first solve the first equation for a: 3a+3b=9 so a+b=3 so a=3-b. Now I'll substitute this expression for a into the second equation 6 a + 5 b = 16 Replacing a with 3 - b: 6(3-b)+5b=16 18-6b+5b=16 -b=-2 b=2 a = 3 - b so a=3 - 2 = 1 Substituting a = 1 and b = 2 into the two equations we get 3(1)+3(2)=9 so 9 = 9 6(1)+5(2)=16 so 16 = 16. The solution checks with the two equations. You have an excellent solution. The method you have used is performed correctly and is equally valid with the method used in the solutions. It is called the 'substitution method'. For these first few problems in this course the substitution method and the elimination method are equally efficient. However the elimination method is also important, and since elimination works better on most of the problems we'll be encountering in the near future, it is the method I use in the given solutions. You can use either method, as long as you know both. However you might find the given solutions easier to understand if you use the elimination method. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Substitution 3a + 3b = 9 6a + 5b = 16 3b = 9 - 3a b = 3 - a 6a + 5(3 - a) = 16 6a + 15 - 5a = 16 a + 15 = 16 a = 1 6(1) + 5b = 16 6 + 5b = 16 5b = 10 b = 2 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q006. Solve the following system of simultaneous linear equations using the method of elimination: 4a + 5b = 18 6a + 9b = 30. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To get the same coefficient in front of both a's with one a having an opposite sign, you can multiply the top equation by 6 and the bottom by -4. 6(4a + 5b = 18) 24a + 30b = 108 -4(6a + 9b = 30) + (-24a - 36b = -120) After adding like variables, you eliminate variable a 0a - 6b = -12 -6b = -12 b = 2 Plug value for b into equation 4a + 5(2) = 18 4a + 10 = 18 4a = 8 a = 2 a = 2, b = 2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In the system 4a + 5b = 18 6a + 9b = 30 we see that the coefficients of b are relatively prime; they therefore have a least common multiple equal to 5 * 9. The coefficients 4 and 6 of a have a least common multiple of 12. We have a choice of which variable to eliminate. We could 'match' the b by multiplying the first equation by 9 and the second by -5, or we could match the coefficients of a by multiplying the first equation by 3 and the second by -2. Either choice would work. The numbers required to 'match' the coefficients of a are smaller, but the numbers required to 'match the coefficients of b would otherwise work equally well. Choosing to 'match' the coefficient of a, we obtain 3 * [4a + 5b ] = 3 * 18 -2 * [ 6a + 9b ] = -2 * 30, so the system becomes 12 a + 15 b = 54 -12 a - 18 b = -60. Adding the equations we get -3 b = -6, so b = 2. Substituting this value of b into the first equation we obtain 4 a + 5 * 2 = 18, or 4 a + 10 = 18, which we easily solve to obtain a = 2. Substituting this value of a into the second equation we obtain 6 * 2 + 9 * 2 = 30, which verifies our solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It didn't occur to me that the least common multiple of 6 and 4 is 12. However, you can safely multiply 6 by 4 to eliminate one of the variables and still get the correct values for a and b ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: