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course Mth 163

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:

To eliminate the variable in the first 2 equations we are going to have to eliminate the letter c in both of the equations by subtracting.

(60a + 5b + c) -(2a + 3b + c) = 58a + 2b

Then subtract 90 - 128 = -38

So the second equation becomes 58a + 2b = -38

To get the third equation we will subtract the third equation from the first.

(200a + 10b + c) - (2a + 3b + c) = 198a + 7b

Then subtract 0 - 128 = -128

So the third equation becomes 198a + 7b = -128

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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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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:

To solve the equation we have to eliminate one of the variables. To eliminate the letter b we have to multiply the first equation by -7 and the second equation by 2.

-7* 58a + -7* 2b = -7* -38

-406a -14b = 266

2* 198a + 2* 7b = 2* -128

396a +14b = -256

Now we have to add the two equations together.

(-406a - 14b = 266)

(396a + 14b = -256)=

-10a = 10 ; to get a by itself we have to divide both sides by -10

a = -1

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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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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:

To get b we have to plug in a= -1 into the first equation.

58a + 2b = -38

58(-1) + 2b = -38

-58 + 2b = -38; add -58 to both sides

2b = 20 ; divide 2 on both sides

b = 10

to make sure the two numbers we got are correct we have to use the other equation by plugging in the numbers.

198(-1) + 7(10) = -128

-198 + 70 =-128

-128 = -128

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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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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:

To get the answer for c we have to plug in the numbers a= -1 and b = 10 into the first equation.

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

-2 + 30 + c = 128 ; to get c by itself we have add 2 to both sides and subtract 3 from both sides.

c= 100

To check this we have to plug in all three numbers into the second equation

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

-60 + 50 + 100 = 90

90=90

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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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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:

If we substitute the points (1,-2), y = -2 and x = 1, we get the equation

-2 = a (1)^2 + b(1) + c which simplifies to

a + b + c = -2

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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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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:

The first set of points ( 3, 5) will give us the equation

5 = a(3)^2 + b(3) + c or

9a + 3b + c = 5

The second set of points ( 7, 8) gives us the equation

8 = a(7)^2 + b(7) + c

49a + 7b + c = 8

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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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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

First we have to subtract the first equation by the second equation to eliminate the letter c.

(9a + 3b + c) -(a + b + c)

8a + 2b

Then we subtract 5 - (-2) = 7

8a + 2b = 7

To get the third equation we have to subtract the third equation from the first.

(49a + 7b + c) - (a + b + c) = 48a + 6b

then subtract 8 - (-2) =10

SO therefore we get 48a + 6b = 10

Now we have to take the second equation (multiply it by -6) and take the third equation and add it to the second equation to eliminate the letter b.

-6*8a + -6*2b = -6 * 7

-48a -12b = -42

(-48a -12b = -42)

(48a + 6b = 10)

-6b = -32 ; divide by -6

b = 5.33333

Now plug in b=5.33333 into the second equation.

8a + 2* 5.33333 = 7

8a + 10.66666 = 7 ; subtract 10.66666 from both sides

8a = -3.66666 ; divide 8 on both sides

a= -0.4583325

Now to get c we plug in a and b into the first equation.

a + b + c = -2

-0.4583325 + 5.33333 + c = -2 ; add a and b together then subtract from both sides

c= -6.8749975

So therefore we have

a= -0.4583325

b= 5.33333

c= -6.8749975

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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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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:

When we plug in the numbers this is the equation we get:

y= -0.45833x^2 + 5.33333x + -6.875

Once we plug in x=1 we get

Y= -0.45833(1)^2 + 5.33333 + -6.875

Y= -2

When we plug in x=3 we get

Y=-0.45833(3)^2 + 5.33333(3) + -6.875

Y= -4.12497 +15.99999 + -6.875

Y=5.0

When we plug in x=5 we get

Y= -0.45833(5)^2 + 5.33333(5) + -6.875

Y=-11.45825 +26.66665 +-6.875

Y=8.3334

When we plug in x=7 we get

Y=-0.45833(7)^2 + 5.33333(7) + -6.875

Y=-22.45817 +37.33331 +-6.875

Y=8.00014

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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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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:

When we plug in the numbers this is the equation we get:

y= -0.45833x^2 + 5.33333x + -6.875

Once we plug in x=1 we get

Y= -0.45833(1)^2 + 5.33333 + -6.875

Y= -2

When we plug in x=3 we get

Y=-0.45833(3)^2 + 5.33333(3) + -6.875

Y= -4.12497 +15.99999 + -6.875

Y=5.0

When we plug in x=5 we get

Y= -0.45833(5)^2 + 5.33333(5) + -6.875

Y=-11.45825 +26.66665 +-6.875

Y=8.3334

When we plug in x=7 we get

Y=-0.45833(7)^2 + 5.33333(7) + -6.875

Y=-22.45817 +37.33331 +-6.875

Y=8.00014

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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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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:

When we plug in the numbers this is the equation we get:

y= -0.45833x^2 + 5.33333x + -6.875

Once we plug in x=1 we get

Y= -0.45833(1)^2 + 5.33333 + -6.875

Y= -2

When we plug in x=3 we get

Y=-0.45833(3)^2 + 5.33333(3) + -6.875

Y= -4.12497 +15.99999 + -6.875

Y=5.0

When we plug in x=5 we get

Y= -0.45833(5)^2 + 5.33333(5) + -6.875

Y=-11.45825 +26.66665 +-6.875

Y=8.3334

When we plug in x=7 we get

Y=-0.45833(7)^2 + 5.33333(7) + -6.875

Y=-22.45817 +37.33331 +-6.875

Y=8.00014

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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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