Assignment 1

course Mth 163

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.

id: qa pc1

001.

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Question: `q001. Note that this assignment has 10 questions

Solve the following system of simultaneous linear equations:

3a + 3b = 9

6a + 5b = 16.

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Your solution:

3a + 3b = 9

6a + 5b = 16

First you have to find a coefficient of a in the next problem to match

-2 makes it to where it will work with 6 in the next problem

-2 ( 3a + 3b) = -2 (9)

Thus,

-6a – 6b = - 18

6a + 5b = 16

Adding them together gives us

B= 2

and

3 a + 3 (2) = 9

A=1

Therefore,

A=1, B= 2

Confidence Assessment:

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Given Solution:

The system

3a + 3b = 9

6a + 5b = 16

can be solved by adding an appropriate multiple of one equation in order to eliminate one of the variables.

Since the coefficient of a in the second equation (the coefficient of a in the second equation is 6)) is double that in the first (the coefficient of a in the first equation is 3), we can multiply the first equation by -2 in order to make the coefficients of a equal and opposite:

-2 * [ 3a + 3b ] = -2 [ 9 ]

6a + 5b = 16

gives us

-6a - 6 b = -18

6a + 5b = 16

. Adding the two equations together we obtain

-b = -2, or just b = 2.

Substituting b = 2 into the first equation we obtain

3 a + 3(2) = 9, or

3 a + 6 = 9 so that

3 a = 3 and

a = 1.

Our solution is therefore a = 1, b = 2.

We used the first equation in our last step, so we verigy this solution is by substituting these values into the second equation, where we get 6 * 1 + 5 * 2 = 6 + 10 = 16.

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Self-critique (if necessary):

Self-critique Rating:

2

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Question: `q002. Solve the following system of simultaneous linear equations using the method of elimination:

4a + 5b = 18

6a + 9b = 30.

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Your solution:

4a + 5b = 18

6a + 9b = 30

First we have to see what they have in common. 4 and 6 both go into 12 so we have to multiply them to get 12.

3 (4a + 5b ) = 3 (18)

-2 (6a +9b) = -2 (30)

Then

12 a + 15 b = 54

-12 a – 18 b = 60

Add them together

-3b = -6

B= 2

Confidence Assessment:

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Given Solution:

In the system

4a + 5b = 18

6a + 9b = 30

we see that the coefficients of b are relatively prime; they therefore have a least common multiple equal to 5 * 9. The coefficients 4 and 6 of a have a least common multiple of 12.

• We have a choice of which variable to eliminate. We could 'match' the b by multiplying the first equation by 9 and the second by -5, or we could match the coefficients of a by multiplying the first equation by 3 and the second by -2.

• Either choice would work. The numbers required to 'match' the coefficients of a are smaller, but the numbers required to 'match the coefficients of b would otherwise work equally well.

Choosing to 'match' the coefficient of a, we obtain

3 * [4a + 5b ] = 3 * 18

-2 * [ 6a + 9b ] = -2 * 30,

so the system becomes

12 a + 15 b = 54

-12 a - 18 b = -60.

Adding the equations we get

-3 b = -6, so

b = 2.

Substituting this value of b into the first equation we obtain

4 a + 5 * 2 = 18, or

4 a + 10 = 18,

which we easily solve to obtain

a = 2.

Substituting this value of a into the second equation we obtain

6 * 2 + 9 * 2 = 30,

which verifies our solution.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q003. If y = 5x + 8, then for what value of x will we have y = 13?

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Your solution:

To find that y = 13 you have to plug in y

So the equation would read

13= 5x + 8

Then subtract 8 on both sides to get x alone.

Then divide by x

13 – 8 = 5

5 = 5x

5/5x = 1

X= 1

Confidence Assessment:

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Given Solution:

We first substitute y = 13 into the equation y = 5 x + 8 to obtain

13 = 5 x + 8.

Subtracting 8 from both equations and reversing the equality we obtain

5 x = 5,

which we easily solve to obtain

x = 1.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q004. Sketch a set of coordinate axes representing y vs. x, with y on the vertical axis and x on the horizontal axis.

• Plot the points (1, -2), (3, 5) and (7, 8).

• Sketch a smooth curve passing through these three points.

On your curve, what are the y coordinates corresponding to x coordinates 1, 3, 5 and 7? Estimate these coordinates as accurately as you can from your graph.

Retain your sketch for use in future questions.

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Your solution:

With the x coordinates 1,3, and 7 the y coordinates are -2, 3, and 8

Confidence Assessment:

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Given Solution:

The x coordinates 1, 3 and 7 match the x coordinates of the three given points, the y coordinates will be the y coordinates -2, 3 and 8, respectively, of those points.

• At x = 5 the precise value of x, for a perfect parabola, would be 8 1/3, or about 8.333.

• You are unlikely to have drawn a perfect parabola and your estimate will almost certainly differ from this value. Any estimate with .5 or so of this value would be a good estimate.

Drawn with complete accuracy a parabola through these points will peak between x = 3 in and x = 7.

• The peak of the actual parabola will occur close to x = 6. Any estimate between x = 5 and x = 7 is pretty good, and even an estimate between x = 7 and x = 8 is not unreasonable.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q005. Using your sketch from the preceding exercise, estimate the x coordinates corresponding to y coordinates 1, 3, 5 and 7. Also estimate the x values at which y is 0.

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Your solution:

Y= 1, 3, 5, 7

X= 1, 2 ,3 ,4

When y is 0 x = 9, 5, 7

Confidence Assessment:

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Given Solution:

The easiest way to estimate your points would be to make horizontal lines on your graph at y = 1, 3, 5 and 7. You would easily locate the points were these lines intersect your graph, then estimate the x coordinates of these points.

For the actual parabola passing through the given points, y will be 1 when x = 1.7 (and also, if your graph extended that far, near x = 10).

• y = 3 near x = 2.3 (and near x = 9.3).

• y = 5 at the given point (3, 5), where x = 3.

• y = 7 near x = 4 (and also near x = 7.7).

Any set of estimates in the vicinity of these values is good.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q006. Suppose the graph you used in the preceding two exercises represents the profit y on an item, with profit given in cents, when the selling price is x, with selling price in dollars. According to your graph:

• What would be the profit if the item is sold for 4 dollars?

• What selling price would result in a profit of 7 cents?

• Why is this graph not a realistic model of profit vs. selling price?

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Your solution:

X= 4 when put on the graph followed up the graph it is closest to 7

If the profit was 7 cents then the selling price would be 7.7

This isn’t a realistic model because it doesn’t show how profit and selling prices go together, it doesn’t match up.

Confidence Assessment:

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Given Solution:

To find the profit for a selling price of x = 4 dollars, we would look at the x = 4 point on the graph.

• This point is easily located by sketching a vertical line through x = 4. Projecting over to the y-axis from this point, you should have obtained an x value somewhere around 7, representing 7 cents.

The profit is the y value, so to obtain the selling price x corresponding to a profit of y = 7 we sketch the horizontal line at y = 7, which as in a preceding exercise will give us x values of about 4 (y = 7 would also occur at x = 7.7, approx., if the entire parabola was drawn).

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q007. On another set of coordinate axes, plot the points (-3, 4) and (5, -2). Sketch a straight line through these points.

We will proceed step-by-step obtain an approximate equation for this line:

First substitute the x and y coordinates of the first point into the form y = m x + b.

• What equation do you obtain when you make this substitution?

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Your solution:

( -3 , 4) x= -3 y = 4

4= -3m +b

Confidence Assessment:

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Given Solution:

Substituting x = -3 and y = 4 into the form y = m x + b, we obtain the equation

• 4 = -3 m + b.

We can reverse the right- and left-hand sides to get

• -3 m + b = 4.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q008. Substitute the coordinates of the point (5, -2) into the form y = m x + b. What equation do you get?

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Your solution:

X= 5 y= -2

-2= m 5 + b

Confidence Assessment:

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Given Solution:

Substituting x = 5 and y = -2 into the form y = m x + b, we obtain the equation

• -2 = 5 m + b.

Reversing the sides we have

• 5 m + b = -2

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q009. You have obtained the equations -3 m + b = 4 and 5 m + b = -2. Use the method of elimination to solve these simultaneous equations for m and b.

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Your solution:

Just subtract the first equation from the second

-3 m + b = 4

- 5m + b = -2

-8 m = 6

M= -3/4

Substitute b

(-3/4) -3 + b= 4

B= 7/4

Confidence Assessment:

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Given Solution:

Starting with the system

-3 m + b = 4

5 m + b = -2

we can easily eliminate b by subtracting the equations. If we subtract the first equation from the second we obtain

-8 m = 6, with solution

m = -3/4.

Substituting this value into the first equation we obtain

(-3/4) * -3 + b = 4, which we easily solve to obtain

b = 7/4.

To check our solution we substitute m = -3/4 and b = 7/4 into the second equation, obtaining

5 ( -3/4) + 7/4 = -2, which gives us -8/4 = -2 or -2 = -2.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q010. Substitute your solutions b = 7/4 and m = -3/4 into the original form y = m x + b.

• What equation do you obtain?

• What is the significance of this equation?

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Your solution:

Y= -3/4 + 7/4

Equation of a straight line through points (-3, 4) and (5, -2)

Confidence Assessment:

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Given Solution:

Substituting b = 7/4 and m = -3/4 into the form y = m x + b, we obtain the equation

y = -3/4 x + 7/4.

This is the equation of the straight line through the given points (-3, 4) and (5, -2).

&#Good work. Let me know if you have questions. &#