Query 0

#$&*

course Mth 163

9/6 10

002.

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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:

The easiest variable to eliminate is c.

The first and second equation elimination is

(60a + 5b + c) - (2a + 3b + c) = 90 -128

58a + 2b = -38

The first and third equation is

(200a + 10b + c) - (2a + 3b + c ) = 0 - 128

198a + 7b = -128

The two remaining equations are 198a +7b = -128 and 58a + 2b = -138

I'm not sure if you need me to go any farther in this problem.

confidence rating #$&*: 2

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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

• 2d eqn - 1st eqn left-hand side: (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 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).

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Self-critique (if necessary):

OK

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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:

58a + 2b = -38

198a + 7b = -128

(-7) (58a + 2b) = -38 (-7) = -406a -14b = 266

(2) (198a + 7b) = -128 (2) = 396a + 14b = -256

-406a - 14b = 266

396a + 14b = -256

= -10a = 10

a = -1

58 (-1) + 2b = -38

-58 + 2b = -38

2b = 20

b = 10

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

Do I not need to solve for b in this question??? If so, did I arrive at the right answer???

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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:

a = -1

58 (-1) + 2b = -38

-58 + 2b = -38

2b = 20

b = 10

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

I answered my own question from the last comment :)

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Self-critique Rating:

@&

Very good.

*@

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:

2(-1) + 3(10) + c = 128

-2 + 30 + c = 128

28 + c = 128

c = 100

60 ( -1) + 5 (10) + 100 = 90

-60 + 50 + 100 = 90

90 = 90

confidence rating #$&*:

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3

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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.

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Self-critique (if necessary):

OK

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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:

-2 = a (1)^2 + b(1) + c

-2 = a + b + c

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

OK

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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:

(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 rating #$&*:

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3

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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

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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:

A + b + c = -2

9a + 3b + c = 5

49a + 7b + c = 8

9a + 3b + c = 5

-a - b - c = 2

8a + 2b = 7

49a + 7b + c = 8

-a - b - c = 2

48a + 6b = 10

8a + 2b = 7

48a + 6b = 10

3(8a + 2b) = 7(3)

24a + 6b = 21

24a + 6b = 21

-48a - 6b = -10

24a = 11

a = .45833

48 (-.45833) + 6b = 10

-21.99984 + 6b = 10

6b = 31.99984

B = 5.3333

9(-.45833) + 3(5.3333) + c = 5

-4.12497 + 15.9999 + c = 5

11.87493 + c = 5

C = -6.8749

A = -.45833

B = 5.3333

C = -6.8749

confidence rating #$&*: 1

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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.

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Self-critique (if necessary):

Shew…..this was rough. Finally got it, and got it correct.

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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 = (-.45833)x^2 + (5.3333)x - 6.8749

Y = (-.45833)(1)^2 + (5.3333) (1) - 6.8749

Y = -1.99 or -2

Y = (-.45833)(3)^2 + (5.3333)(3) - 6.8749

Y = 5

Y = (-.45833)(5)^2 + (5.333)(5) - 6.8749

Y = 8.33

Y = (-.45833)(7)^2 + (5.333)(7) - 6.8749

Y = 8

confidence rating #$&*:

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3

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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.

• 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 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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Self-critique (if necessary):

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Self-critique rating:

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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 = (-.45833)x^2 + (5.3333)x - 6.8749

Y = (-.45833)(1)^2 + (5.3333) (1) - 6.8749

Y = -1.99 or -2

Y = (-.45833)(3)^2 + (5.3333)(3) - 6.8749

Y = 5

Y = (-.45833)(5)^2 + (5.333)(5) - 6.8749

Y = 8.33

Y = (-.45833)(7)^2 + (5.333)(7) - 6.8749

Y = 8

confidence rating #$&*:

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3

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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.

• 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 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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Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Very good responses. Let me know if you have questions. &#