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

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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

000.  As with other documents you have seen so far, you will copy this document into a text editor to complete it before submitting it.  However in a text editor the graphs won't show.  So you might have to look back at this document to see the graphs. 

We begin with the idea of constructing graphs from just a few points.  In this exercise we will be concerned with parabolas.  The first topic in this course will be functions with parabolic graphs.  This exercise will then continue with the process of solving a system of two simultaneous linear equations.  You will very soon see how the two topics are connected.

Below are three dots, representing three points, with a certain symmetry

The symmetry condition is this:

• Two of the points are at the same 'height' on the page, and the third is at equal distances from these two points.

Any time we have three dots with this condition of symmetry, we can easily sketch a parabola through them:

The three points below have the same condition of symmetry, but the distance of the third dot from the other two is greater.

If we sketch the parabola corresponding to these points, it will be narrower that the previous parabola:

The three points below satisfy the same symmetry condition.  The third point is still equidistant from the other two, but in this case the equidistant point lies 'above' the other two.

The resulting parabola opens downward:

In the figure below copies of these three parabolas are shown at different locations in the xy plane.  The three 'basic points' of one of these parabolas are shown. 

The coordinates of the three 'basic points' shown in the figure above are (1, 3), (2, 1) and (3, 3).

The vertex of a parabola is its highest or lowest point, corresponding to the 'equidistant point' in our three-basic-point scheme.

We will define the 'three basic point' scheme for parabolas as follows:

• One of the points is the vertex.
• The other two points line along the same horizontal line (that is, they have the same y coordinate) and they are 2 units apart (as measured on the scale of the x axis).

Note that there are seven questions in this assignment.

Question `q001:  What is the vertex of each of the other two parabolas depicted above?

Your Solution:

Confidence Rating:

Given Solution:  The vertex of the 'blue' parabola is at the point (2, -1), the 'lowest' point on the parabola.

The vertex of the 'purple' parabola is at the point (-2, 3), the 'highest' point on the parabola.

Self-critique:

Self-critique rating:

Question `q002:  What are the coordinates of the other two 'basic points' of each parabola?

Your Solution:

Confidence Rating:

Given Solution:  For the 'blue' parabola, the points (1, 0) and (3, 0) are two units apart, and lie on the same horizontal line.  The horizontal line is the x axis.

For the 'purple' parabola, the points (-3, 1) and (-1, 1) are two units apart.  These points lie on the horizontal line where y = 2. 

Self-critique:

Self-critique rating:

Question `q003:  For the first parabola, the one whose vertex is (2, 1), how far would we have to move to the right or the left, starting from the vertex, in order to be directly above or below another of its 'basic points'?  How far would we then have to move in the vertical direction to reach that point?

Your Solution:

Confidence Rating:

Given Solution:  For the parabola with vertex (2, 1), if we move 1 unit to the right or left we will be at the point (3, 1) or (1, 1), putting us directly below one of the other two basic points.  If we then move 2 units upward, we will be at the point (3, 3) or (1, 3).

So if we move 1 unit to the right or left, we need to move 2 units upward to get to another basic point.

Self-critique:

Self-critique rating:

Question `q004:  For each of the other two parabolas, how far would we have to move to the right or the left, starting from the vertex, in order to be directly above or below another of its 'basic points'?  How far would we then have to move in the vertical direction to reach that point?

Your Solution:

Confidence Rating:

Given Solution:  For either of the other two parabolas, if we move 1 unit to the right or left we will be directly above or below one of its basic points (above in the case of the 'third' parabola, whose vertex is (-2, 3), below in the case of the 'second' parabola, whose vertex is (2, -1).)

To get to the basic points of the 'third' parabola we will need to move 2 units downward. 

To get to the basic points of the 'second' parabola we will need to move 1 unit upward. 

Self-critique:

Self-critique rating:

Question `q005.  Solve the following system of simultaneous linear equations: 

3a + 3b = 9

6a + 5b = 16.

Your solution: 

Confidence rating:

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. 

STUDENT QUESTION

I got my answer in a very different way than the solution given. I have been trying to remember things from the classes I
took a long time ago and came up with this answer. Is it alright to use this method?

INSTRUCTOR RESPONSE

Here is a synopsis of your solution:

I'll first solve the first equation for a:

 

3a+3b=9 so

a+b=3 so

a=3-b.

 

Now I'll substitute this expression for a into the second equation

 

6 a + 5 b = 16  

 

Replacing a with 3 - b:

6(3-b)+5b=16
18-6b+5b=16
-b=-2
b=2

a = 3 - b so a=3 - 2 = 1

 

Substituting a = 1 and b = 2 into the two equations we get

3(1)+3(2)=9 so 9 = 9
6(1)+5(2)=16 so 16 = 16.

 

The solution checks with the two equations.

You have an excellent solution.

The method you have used is performed correctly and is equally valid with the method used in the solutions. It is called the 'substitution method'. For these first few problems in this course the substitution method and the elimination method are equally efficient.

However the elimination method is also important, and since elimination works better on most of the problems we'll be encountering in the near future, it is the method I use in the given solutions. 

You can use either method, as long as you know both. However you might find the given solutions easier to understand if you use the elimination method.

Self-critique (if necessary):

Self-critique Rating:

Question:  `q006.  Solve the following system of simultaneous linear equations using the method of elimination:

4a + 5b = 18

6a + 9b = 30.

Your solution: 

Confidence rating:

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.

Self-critique (if necessary):

Self-critique Rating:

Question:  `q007.  If y = 5x + 8, then for what value of x will we have y = 13?

Your solution: 

Confidence rating:

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.

Self-critique (if necessary):

Self-critique Rating: