course MTH 163
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Question: `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.
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Your solution:
2a + 3b + c = 128
60a + 5b + c = 90
200a + 10b + c = 0
200a + 10b + c = 0
-2a - 3b - c =-128
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198a + 7b =-128
60a + 5b + c = 90
-2a - 3b - c =-128
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58a + 2b = -38
198a + 7b = -128
58a + 2b = -38
Confidence rating: -[3]-
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Self-critique (if necessary): OK
Self-critique Rating: OK
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Question: `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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Your solution:
Common factor b of 14.
( 58a + 2b = -38) * 7
406a + 14b = -266
(198a + 7b = -128) * 2
396a + 14b = -256
406a + 14b = -266
-396a - 14b = 256
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10a = -10
a = -1
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Self-critique (if necessary): OK
Self-critique Rating: OK
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Question: `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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Your solution:
58(-1) + 2b = -38
-58 + 2b = -38
2b = 20
b = 10
Confidence rating: -[3]-
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Self-critique (if necessary): OK
Self-critique Rating: OK
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Question: `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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Your solution:
2a + 3b + c = 128
2(-1) + 3(10) + c = 128
-2 + 30 + c = 128
c = 100
200a + 10b + c = 0
200(-1) + 10(10) + (100) = 0
-200 + 100 + 100 = 0
0 = 0
Confidence rating: -[3]-
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Self-critique (if necessary): OK
Self-critique Rating: OK
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Question: `q005. The graph you sketched in a previous assignment contained the given points (1, -2), (3, 5) and (7, 8).
We are going to use simultaneous equations to obtain the equation of that parabola.
*A graph has a parabolic shape if its the equation of the graph is quadratic.
*The equation of a graph is quadratic if it has the form y = a x^2 + b x + c.
*y = a x^2 + b x + c is said to be a quadratic function of x.
To find the precise quadratic function that fits our points, we need only determine the values of a, b and c.
*As we will discover, if we know the coordinates of three points on the graph of a quadratic function, we can use simultaneous equations to find the values of a, b and c.
The first step is to obtain an equation using the first known point.
*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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Your solution:
y = ax^2 + bx + c
-2 = a(1)^2 + b(1) + c
-2 = a + b + c
Confidence rating: -[3]-
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Self-critique (if necessary): OK
Self-critique Rating: OK
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Question: `q006. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as in the preceding question, then what two equations do we get if we substitute the x and y values corresponding to the point (3, 5), then the point (7, 8) into the form y = a x^2 + b x + c? (each point will give us one equation)
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Your solution:
y = ax^2 + bx + c
5 = a(3)^2 + b(3) + c
5 = 9a + 3b + c
y = ax^2 + bx + c
8 = a(7)^2 + b(7) + c
8 = 49a + 7b + c
Confidence rating:
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Given Solution:
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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Self-critique (if necessary): OK, but I think that there is a simple error in your given answer, you had:
8 = a * 7^2 + b * 7 + c, or
49 a + 7 b + c = 7
Where did the 7 come from? Is it
really an 8?
It's really an 8, and I know about the error but choose to leave it in there to see who's paying attention.
As expected, you are.
Self-critique Rating: OK
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Question: `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 we obtain three equations with unknowns a, b and c. You have already done this.
Write down the system of equations we got when we substituted 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.
Solve the system to find the values of a, b and c.
*What is the solution of this system?
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Your solution:
8 = 49a + 7b + c
5 = 9a + 3b + c
-2 = a + b + c
49a + 7b + c = 8
-9a - 3b - c = -5
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40a + 4b = 3
49a + 7b + c = 8
-a - b - c = 2
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48a + 6b = 10
(48a + 6b = 10) * 2
96a + 12b = 20
(40a + 4b = 3) * 3
120a + 12b = 9
120a + 12b = 9
-96a - 12b = -20
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24a = -11
a = -11/24 or -0.458333
40a + 4b = 3
40(-11/24) + 4b = 3
b = 16/3 or 5.33333
(-11/24) + (16/3) + c = -2
c = -55/8 or -6.875
Confidence rating: -[3]-
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Self-critique (if necessary): OK
Self-critique Rating: OK
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Question: `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, in order to obtain a specific quadratic function.
*What is your function?
*What y values do you get when you substitute x = 1, 3, 5 and 7 into this function?
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Your solution:
y = -11/24x^2 + 16/3x - 55/8
y = -11/24(1)^2 + 16/3(1) - 55/8
y = -2
y = -11/24(3)^2 + 16/3(3) - 55/8
y = 5
y = -11/24(5)^2 + 16/3(5) - 55/8
y = 25/3 or 8.3333
y = -11/24(7)^2 + 16/3(7) - 55/8
y = 8
Confidence rating: -[3]-
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&#This looks good. See my notes. Let me know if you have any questions. &#