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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
.
Your solution, attempt at solution. If you are unable to attempt a solution, give
a phrase-by-phrase interpretation of the problem along with a statement of what
you do or do not understand about it.
This response should be given, based on the work you did in completing
the assignment, before you look at the given solution.
002.
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.
Your solution:
Confidence Assessment:
Given Solution:
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
The right-hand side will be the
difference 90 - 128 = -38, so the second equation will become
The 'new' third equation by a similar calculation will be
You might well have obtained this system, or one equivalent to it,
using a slightly different sequence of calculations. (As one example
you might have subtracted the second from the first, and the third from the
second).
Self-critique (if necessary):
Self-critique Rating:
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.
Your solution:
Confidence Assessment:
Given Solution:
Neither variable is as easy to eliminate as in the last
problem, but the coefficients of b are significantly smaller than those of
a. So here we choose eliminate b. It
would also have been OK to choose to eliminate a.
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 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.
Self-critique (if necessary):
Self-critique Rating:
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.
Your solution:
Confidence Assessment:
Given Solution:
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.
Self-critique (if necessary):
Self-critique Rating:
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.
Your solution:
Confidence Assessment:
Given Solution:
Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which we
easily solve to get c = 100.
Substituting these values into the second equation, in order
to check our solution, 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 again obtain an identity. This
would completely validate our solution.
Self-critique (if necessary):
Self-critique Rating:
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.
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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.
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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.
Your solution:
Confidence Assessment:
Given Solution:
We substitute y = -2 and x = 1 to obtain the equation
-2 = a * 1^2 + b * 1 + c, or
a + b + c = -2.
Self-critique (if necessary):
Self-critique Rating:
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)
Your solution:
Confidence Assessment:
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.
Self-critique (if necessary):
Self-critique Rating:
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.
Your solution:
Confidence Assessment:
Given Solution:
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 in this case 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 solution, you should show your
solution. Start by indicating 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.
Self-critique (if necessary):
Self-critique Rating:
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.
Your solution:
Confidence Assessment:
Given Solution:
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.
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When we substitute 1 into the equation we obtain y = -.45833
* 1^2 + 5.33333 * 1 - 6.875 = -2.
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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 yield the points
(1, -2), (3, 5) and (7, 8). These are the points we used to obtain the formula.
We also get the additional point (5,
8.33333).
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