course Mth 163 `ΌMЁassignment #002
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12:57:57 `q001. Note that this assignment has 8 questions Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables: 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. NOTE SOLN IS -1, 10, 100.
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RESPONSE --> We start be eliminating the easiest variable, which would be c since it is just 1. So we will subtract the first equation from the third equation to get a new equation. (200a + 10b + c) - (2a + 3b + c) = 198a + 7b 0 - 128 = -128, so our first new equation is 198a + 7b = 128. We subtract the first equation from the second equation to get our other new equation: (60a + 5b + c) - (2a + 3b + c) = 58a +2b 90 - 128 = -38 The second new equation is 58a + 2b = -38 confidence assessment: 2
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13:01:19 The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results. Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b. The right-hand side will be the difference 90 - 128 = -38, so the second equation will become 'new' 2d equation: 58 a + 2 b = -38. The 'new' third equation by a similar calculation will be 'new' third equation: 198 a + 7 b = -128. You might have obtain this system, or one equivalent to it, using a slightly different sequence of calculations.
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RESPONSE --> I got both of the ""new"" equations the same way, by subtracting one of the ""old"" equations from another. self critique assessment: 2
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13:07:20 `q002. Solve the two equations 58 a + 2 b = -38 198 a + 7 b = -128 , which can be obtained from the system in the preceding problem, by eliminating the easiest variable.
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RESPONSE --> To eliminate the next variable we have to multiply both equations by the other variables number. So b is the next easiest to get rid of: -7 * (58a + 2b = -38) 2 * (198a + 7b = -128) So we multiply that out: -406a - 14b = 266 396a + 14b = -256 Now we subtract these two newer equations: -10a = 10 a = -1 confidence assessment: 2
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13:15:19 Neither variable is as easy to eliminate as in the last problem, but the coefficients of b a significantly smaller than those of a. So we will eliminate b. To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the program multiplications: -7 * ( 58 a + 2 b) = -7 * -38 2 * ( 198 a + 7 b ) = 2 * (-128). Doing the arithmetic we obtain -406 a - 14 b = 266 396 a + 14 b = -256. Adding the two equations we obtain -10 a = 10, so we have a = -1.
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RESPONSE --> self critique assessment: 3
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13:21:08 `q003. Having obtained a = -1, use either of the equations 58 a + 2 b = -38 198 a + 7 b = -128 to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation.
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RESPONSE --> So we will substitute -1 into the equation for a to find the value for b 58 (-1) + 2b = -38 -58 + 2b = -38, add 58 to both sides of the equation 58 + (-58) + 2b = -38 + 58 2b = 20 b = 10 Now we can substitute the value for b into the other equation to validate the value for a. 198a + 7(10) = -128 198a + 70 = -128, subtract 70 on both sides 198a + 70 - 70 = -128 - 70 198a = -198 a = -1 confidence assessment: 2
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13:23:18 You might have completed this step in your solution to the preceding problem. Substituting a = -1 into the first equation we have 58 * -1 + 2 b = -38, so 2 b = 20 and b = 10.
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RESPONSE --> That's exactly how I solved for the b value. I did use to b value in the other equation to validate that a = -1 self critique assessment: 2
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13:26:37 `q004. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations.
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RESPONSE --> Now that we've found a and b values, we are able to find the value for c. So we substitute those values into the equation: 2(-1) + 3(10) + c = 128 -2 + 30 + c = 128 c = 100 If we substitute those values into the other equations we will find that they are the correct values for the variables. confidence assessment: 2
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13:28:09 Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which was some simple arithmetic gives us c = 100. Substituting these values into the second equation we obtain 60 * -1 + 5 * 10 + 100 = 90, or -60 + 50 + 100 = 90, or 90 = 90. We could also substitute the values into the third equation, and will begin obtain an identity. This would completely validate our solution.
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RESPONSE --> I did use the values in the other equations to validate the values. self critique assessment: 2
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13:33:18 `q005. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case for the graph you sketched in the preceding assignment, then what equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c?
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RESPONSE --> We just choose a set of points to substitute into the equation, I'll use the (1, -2) -2 = a * 1^2 + b(1) + c -2 = a + b + c confidence assessment: 2
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13:34:29 We substitute y = -2 and x = 1 to obtain the equation -2 = a * 1^2 + b * 1 + c, or a + b + c = -2.
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RESPONSE --> The a x^2 made me think a little but since 1^2 is just 1, it will not make a difference in this case. self critique assessment: 2
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13:38:50 `q006. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then what equations do we get if we substitute the x and y values corresponding to the point (3, 5), then (7, 8) into the form y = a x^2 + b x + c?
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RESPONSE --> So now we'll use the other points to substitute in the equation: (3, 5) 5 = a * 3^2 + b*3 + c 5 = 9a + 3b + c (7, 8) 8 = a * 7^2 + b*7 + c 8 = 49a + 7b + c confidence assessment: 2
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13:39:54 Using the second point we substitute y = 5 and x = 3 to obtain the equation 5 = a * 3^2 + b * 3 + c, or 9 a + 3 b + c = 5. Using the third point we substitute y = 8 and x = 7 to obtain the equation 8 = a * 7^2 + b * 7 + c, or 49 a + 7 b + c = 7.
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RESPONSE --> Those are the same equations I arrived at from substituting the values in. self critique assessment: 3
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13:55:09 `q007. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then what system of equations do we get if we substitute the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c? What is the solution of this system?
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RESPONSE --> For the three points we get the following equations: (1, -2) - - - a + b + c = -2 (3, 5) - - - 9a + 3b + c = 5 (7, 8) - - - 49a + 7b + c = 8 Now if we solve for two 'new' equations we get: (49a + 7b + c = 8) - (a + b + c = -2) = 48a + 6b = 10 (9a + 3b + c = 5) - ( a + b + c = -2) = 8a + 2b = 7 We can solve for a by multiplying the second equation by 3: (48a + 6b = 10) - (24a + 6b = 21) = 24a = -11 We find that a = -0.4583 If we substitute that value in we can solve for b, which b = 5.333. Now we can use those two values to solve for c, which c = -6.875. confidence assessment: 2
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13:56:31 The system consists of the three equations obtained in the last problem: a + b + c = -2 9 a + 3 b + c = 5 49 a + 7 b + c = 8. This system is solved in the same manner as in the preceding exercise. However the solutions don't come out to be whole numbers. The solution of this system, in decimal form, is approximately a = - 0.45833, b = 5.33333 and c = - 6.875. If you obtained a different system, you should show the system of two equations you obtained when you eliminated c, then indicate what multiple of each equation you put together to eliminate either a or b.
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RESPONSE --> I did the multiplication to find all three values. self critique assessment: 3
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14:12:26 `q008. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c. What function do you get? What do you get when you substitute x = 1, 3, 5 and 7 into this function?
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RESPONSE --> If we substitute the a, b, and c values into y = ax^2 + bx + c we would get: y = (-0.4583)x^2 + 5.3333 x -6.875 If we substitute the values 1, 3, 5, and 7 into this function for x, we would get the following: (I am not going to put all the multiplication out because its a little lengthy, but I did it on paper) x = 1; y = -0.4583 + 5.3333 -6.875 y = -2 x = 3; y = (-0.4583)*3^2 + 5.3333(3) - 6.875 y = 5 x = 5; y = (-0.4583)*5^2 + 5.3333(5) - 6.875 y = 8.334 x = 7; y = (-0.4583)*7^2 + 5.3333(7) - 6.875 y = 8 confidence assessment: 2
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14:15:47 Substituting the values of a, b and c into the given form we obtain the equation y = - 0.45833 x^2 + 5.33333 x - 6.875. When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2. When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5. When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333. When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8. Thus the y values we obtain for our x values gives us the points (1, -2), (3, 5) and (7, 8) we used to obtain the formula, plus the point (5, 8.33333).
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RESPONSE --> I did get the points (1, -2), (3, 5), (5, 8.334), and (7, 8). self critique assessment: 3
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