Cuisenaire Rods

Cuisenaire rods are math-learning aids that provide students a hands-on way of learning fundamental math concepts, from the four arithmetical operations to working with fractions. They were originally designed by Georges Cuisenaire, a Belgian educator in the 1920s, and later popularized by Caleb Gattegno in the 1950s. Just as written musical notes allow one to visualize and perform music, Cuisenaire rods make mathematics visible by using wooden rods of varying lengths and colors. According to a primary school in Thuin, Belgium, students who were taught in this manner were said to have "learned mathematics faster than most other students in the world." 

Week 4, Day 2

Today we learned how to form augmented matrices and solve for their variables through two methods: Gaussian Elimination with Back Substitution and Gauss-Jordan elimination. When performimg Gaussian Elimination, you should consider the following steps: First, write the augmented matrix of the system of linear equations. Second, use elementary row operations to rewrite the augmented matrix in row-echelon form. Finally, write the system of linear equations corresponding to the matrix in row-echelon form, and use back-substitution to find the solution. Ultimately, this method requires that the main diagonal of the matrix consists of one's and everything below it are zeros. The Gauss-Jordan Elimination method, however, requires that the main diagonal of the matrix consists of one's and everything around it are zeros. The steps to performing Gauss-Jordon are similar to that of Gaussian Elimination. You simply continue using elementary row operations to achieve the appropriate form until each variable has been solved for without the use of back substitution. 

In the above problem, the student didn't quite understand the process of Gauss-Jordan Elimination. While he has achieved row-echelon form, the student forgot that the operations he used apply to the solutions on the right side of the dotted line as well. When he multiplies -2 by row one and adds it to row 2, he must remember to do so for not only 2 and 3 in the second row, but also the  5. The operation would yield a solution of -8 + 5 = 3. The same goes for the first row in which he multiplies row 2 by -1 and adds it to row 1. This operation would yield a solution of -5 + 4 = -1. The final solutions would be -1 for the first variable and 3 for the second variable. 

Problem 1:

Problem 2:

Problem 3:

I used the Gauss-Jordan Elimination method in problems 1 and 2, and the Gaussian Elimination method in problem 3. 

Magic in Numbers

There are various ways to manipulate numbers, and see true magic in their properties.

Here are a few tricks you can enjoy:

Trick 1: Number Below 10

Step 1: Think of a number below 10. 

Step 2: Double the number you have thought of.

Step 3: Add 6 to your result.

Step 4: Divide the answer in half.

Step 5: Subtract your original number from this quotient. 

Step 6: Your answer should always be 3.

For example, I chose the number 8.

8 x 2 = 16

6 + 16 = 22

22 / 2 = 11

11 - 8 = 3

Trick 2: Any Number

Step 1: Think of any number. It can be either positive or negative.

Step 2: Subtract 1 from the number you thought of.

Step 3: Multiply your result by 3.

Step 4: Add 12 to your result.

Step 5: Divide this answer by 3.

Step 6: Add 5 to this answer.

Step 7: Subtract your original number from this answer.

Step 8: Your answer should always be 8.

For example, I chose the number 20.

20 - 1 = 19

19 x 3 = 57

12 + 57 = 69

69 / 3 = 23

23 + 5 = 28

28 - 20 = 8

Trick 3: Same 3 Digit Number

Step 1: Think of any 3 digit number containing the same 3 digits. 

Step 2: Add up each of the digits in this number.

Step 3: Divide your original number with the sum of its 3 digits.

Step 4: Your answer should always be 37.

For example, I chose the number 111.

1 + 1 + 1 = 3

111 / 3 = 37

All in all, these simple math tricks should demonstrate that numbers can produce constant results, just as we know God's love is constant and does not waver. :) 

Chapter 7: Systems of Equations and Inequalities Review

Section 7.1: Solving Systems of Equations

In this lesson, we learned how to solve for variables through substitution. The first step is to isolate the variable. Once you have determined its value, substitute it for the same variable in the second equation. By basic computation, you should be able to identify the value of each variable. If the solutiom is, for example, 2 = 8, then write "no solution." If the solution is, for example, 3 = 3, then write "all real solutions." The following is an example of how to perform substitution: 

x + y = 4, x + y = -2

1) x = 4 - y

2) (4 - y) - y = -2

4 - 2y = -2

-2y = -6

y = 3

3) x + 3 = 4

x = 1

Section 7.2: Systems of Linear Equations in 2 Variables

In this lesson, we learned how to solve for variables through elimination. First, obtain coefficients that differ only in sign. Second, add equations to eliminate a variable. Then back substitute to solve for the second equation. Finally, check your solution! There are three possible types of solutions: intersecting lines (one solution), parallel lines (no solution), and the same line (infinitely many solutions). The following is an example of how to perform elimination:

5x + 3y = 9 

2x - 4y = 14

Step 1: 

4(5x + 3y = 9) --> 20x + 12y = 36

3(2x - 4y = 14) --> 6x - 12y = 42

Step 2:

20x + 12y = 36

  6x - 12y = 42

+ -----------------

26x = 78

x = 3

Step 3:

2(3) - 4y = 14

6 - 4y = 14

-4y = 8

y = -2

Step 4:

2(3) - 4(-2) = 14

6 + 8 = 14

14 = 14

(3, -2)

Section 7.3: Multivariable Linear Systems

The following are methods to solve 3 equations with 3 unknown variables:

Using Back-Substitution

x - 2y + 3z = 9 

y + 3z = 5

z = 2

1) Substitute 2 for z in the second equation.

y + 3(2) = 5

y = -1

2) Substitute -1 for y and 2 for z in the first equation.

x - 2(-1) + 3(2) = 9 

x = 1

3) Write your solution as an ordered pair.

(1, -1, 2)

Using Elimination 

x - 2y + 3z = 9

-x + 3y + 0z = -4

2x - 5y + 5z = 17

First, eliminate all the x's from the first column. 

1) Add the first equation to the second equation, producing a new second equation.

x - 2y + 3z = 9

-x + 3y + 0z = -4

+ -----------------

y + 3z = 5

2) Add -2 times the first equation to the third equation, producing a new third equation.

-2(x - 2y + 3z = 9)

2x - 5y + 5z = 17

+ -----------------

-y - z = -1

Second, eliminate all the y's from the second column.

3) Add the second equation to the third equation, producing a new third equation.

x - 2y + 3z = 9

       y + 3z = 5

             2z = 4

Finally, solve for z in the third equation.

4) Multiply the third equation by 1/2, producing a new third equation.

x - 2y + 3z = 9

       y + 3z = 5

               z = 2

5) Write your solution as an ordered pair.

(1, -1, 2)

Nonsquare systems occur when there are 3 unknown variables and only 2 equations.

x - 2y + z = 2

2x - y - z = -3

1) Add -2 times the first equation to the second equation, producing a new second equation.

x - 2y + z = 2

     3y - 3z = -3

2) Multiply the second equation by 1/3, producing a new second equation.

x -2y + z = 2

      y - z = -1

3) Solving for y in terms of z, you get y = z - 1, and back-substitution into equation 1 yields

x - 2(z - 1) + z = 2

x - 2z + 2 + z = 2

x = z

4) Finally, by letting z = a, you have the solution written as an ordered pair

(a, a - 1, a)

Section 7.4: Partial Fractions

In this lesson, we learned how to perform partial fraction decomposition, which involves writing a rational expression as the sum of two or more rational expressions. For instance, (x+7) / (x^2-x-6) can be written as the sum of two fractions with linear denominators. That is, (x+7) / (x^2-x-6) = (2 / x-3) + (-1 / x+2). Each fraction on the right side of the equation is a partial fraction, and together they make up the partial fraction decomposition of the left side. The following are steps to carry out partial fraction decomposition: 

1. Multiply by LCD

2. Distribute

3. Collect like terms

4. Factor out the variable

5. Equate the coefficients

6. Solve the system of equations

7. Write the answer as a partial fraction

When the denominator of a partial fraction is, for example, (x+1)^2, it is called a repeated factor.

When the denominator of a partial fraction is, for example, (x^2+1), it is called a quadratic factor. 

Here are a few examples:

Example 1: Distinct Linear Factors

5/(x^2+x-6)--> 5/(x+3)(x-2) = A/(x+3) + B/(x-2)

1. 5 = A(x-2) + B(x+3)

2. 5 = Ax - 2A + Bx + 3B

3. 5 = Ax + Bx - 2A + 3B 

4. 5 = (A + B)x - 2A + 3B 

5. 0 = A + B, 5 = -2A + 3B

6. 5 = 5B--> B = 1, A = -1

7. -1/(x+3) + 1/(x-2)

Example 2:Repeated Linear Factors

2x - 3/(x-1)^2 

In this case, a factor (x-1) repeats, and thus the steps to solve it are as follows: 

1. Multiply by LCD

2. Distribute

3. Collect like terms

4. Equate the coefficients

5. Solve the system of equations

6. Write the answer as a partial fraction

2x - 3/(x-1)^2 = A/(x-1) + B/(x-1)^2 Every exponent gets a factor!

1. 2x-3 = A(x-1) + B

2. 2x-3 = Ax - A + B

3. 2x-3 = Ax + (-A + B)

4. A = 2, -3 = -A + B

5. -3 + 2 = B = -1

6. 2/(x-1) - 1/(x-1)^2

Example 3: Distinct Linear Factors

4x^2 + 2x - 1 / x^2(x +1) = A/x + B/x^2 + C/x+1

1. 4x^2 + 2x - 1 = Ax(x+1) + B(x+1) + Cx^2

2. 4x^2 + 2x - 1 = Ax^2 + Ax + Bx + B + Cx^2

3. 4x^2 + 2x - 1 = Ax^2 + Cx^2 + Ax + Bx + B

4. 4x^2 + 2x - 1 = (A + C)x^2 + (A + B)x + B

5. A + C = 4, A + B = 2, B = -1

6. A = 3, B = -1, C = 1

7. 3/x -1/x^2 + 1/x+1

Example 4: Distinct Linear and Quadratic Factors

x^2 - 1 / x(x^2 + 1) 

In this case, the denominator contains an irreducible quadratic factor, so the set up is as follows:

x^2 - 1 / x(x^2 + 1) = A/x + Bx + C / x^2 + 1

1. x^2 - 1 = A(x^2 + 1) + (Bx + C)x

2. x^2 - 1 = Ax^2 + A + Bx^2 + Cx

3. x^2 - 1 = Ax^2 + Bx^2 + Cx + A

4. x^2 - 1 = (A + B)x^2 + Cx + A

5. A + B = 1, C = 0, A = -1

6. A = -1, B = 2, C = 0

7. -1/x + 2x/ x^2 + 1

When the exponent of the numerator is greater than or equal to the exponent of the denominator, use long division. Once you have found the quotient, take the remainder and put it over the original denominator, using it as your new partial fraction. 

Section 7.5: Systems of Inequalities

In this lesson, we learned how to solve and graph systems of inequalities. The graph of the equation will normally separate the plane into two or more regions. In each region, one of the following must be true: a) All points in the region are solutions of the inequality. b) No point in the region is a solution of the inequality. Thus, you can determine whether the points in an entire region satisfy the inequality by simply testing one point in the region. The following are steps to graph an inequality in 2 variables:

1. Replace the inequality sign with an equal sign, and sketch the graph of the resulting equation. 

(Use a dashed line for < or > and a solid line for <(=) or >(=)). 

2. Test one point in each of the regions formed by the graph in Step 1. If the point satisfies the inequality, shade the entire region todenote that every point in the region satisfies the inequality.

3. A solution of a system of inequalities in x and y is a point (x , y) that satisfies each inequality.

4. For a system of inequalities, it is helpful to find the vertices of the solution region.

Here are some examples:

Example 1:

Sketch the graph of y >(=) x^2 - 1

The graph, as shown above, of the corresponding equation y = x^2 - 1 is a parabola. By testing the point (0,0) above the parabola and the point (0,-2) below the parabola, you can see that the points that satisfy the inequality are those lying above (or on) the parabola. This inequality is also nonlinear.

Example 2:

Sketch the graphs of x >(=) -2 and y <(=) 3.

a) The graph of the corresponding equation x = -2 is a vertical line. The points that satisfy the inequality are those lying to the right of this line. b) The graph of the corresponding equation y = 3 is a horizontal line. The points that satisfy the inequality are those lying below (or on) this line.

Example 3:

Sketch the region containing all points that satisfy the system. 

x^2 - y <(=) 1

-x + y <(=) 1

As shown in the graph above, the points that satisfy the inequality x^2 - y <(=) 1 are the points lying above (or on) the parabola given by y = x^2 - 1. The points that satisfy the inequality -x + y <(=) are rhe points lying below (or on) the line given by y = x + 1. To find the points of the intersection of the parabola and line, solve the system of equations. Using the method of substitution, you will find the solutions to be (-1,0) and (2,3). 

Lesson 7.6: Linear Programming

In this lesson, we learned how to perform linear programming. A 2D linear programming problem consists of a linear objective function and a system of linear inequalities called constraints. The objective function gives the quantity that is to be maximized or minimized, and the constraints determine the set of feasible solutions.

The following are steps to solving a linear programming problem:

1) Sketch the region corresponding to the system of constraints.

2) Find the vertices of the region.

3) Test the objective function, z = ax + by, at each of the vertices and select the values of the variables that optimize the objective function. For a bounded region, both a maximum and minimum will exist. For an unbound region, if an optimized solution exists, it will occur at a vertex. 

Week 2 What's Your Age?

Here is a little math-related trick you can use to calculate your age. 

1. Multiply the first number of the age by 5. 

2. Add 3 to the result of that calculation.

3. Double the answer from Step 2.

4. Add the second digit of the number with the result.

5. Subtract 6 from that calculation.

For example, I am 16 years old.

1. 1 x 5 = 5

2. 5 + 3 = 8

3. 8 x 2 = 16

4. 16 + 6 = 22

5. 22 - 6 = 16

And there you go, 16 years old.

This is just another way of demonstrating one of math's many unique properties. :) 

Week 2, Day 2

Today we learned how to solve and graph systems of inequalities. The graph of the equation will normally separate the plane into two or more regions. In each region, one of the following must be true: a) All points in the region are solutions of the inequality. b) No point in the region is a solution of the inequality. Thus, you can determine whether the points in an entire region satisfy the inequality by simply testing one point in the region. The following are steps to graph an inequality in 2 variables:

1. Replace the inequality sign with an equal sign, and sketch the graph of the resulting equation. 

(Use a dashed line for < or > and a solid line for <(=) or >(=)). 

2. Test one point in each of the regions formed by the graph in Step 1. If the point satisfies the inequality, shade the entire region todenote that every point in the region satisfies the inequality.

3. A solution of a system of inequalities in x and y is a point (x , y) that satisfies each inequality.

4. For a system of inequalities, it is helpful to find the vertices of the solution region.

Here are some examples:

Example 1:

Sketch the graph of y >(=) x^2 - 1

The graph, as shown above, of the corresponding equation y = x^2 - 1 is a parabola. By testing the point (0,0) above the parabola and the point (0,-2) below the parabola, you can see that the points that satisfy the inequality are those lying above (or on) the parabola. This inequality is also nonlinear.

Example 2:

Sketch the graphs of x >(=) -2 and y <(=) 3.

a) The graph of the corresponding equation x = -2 is a vertical line. The points that satisfy the inequality are those lying to the right of this line. b) The graph of the corresponding equation y = 3 is a horizontal line. The points that satisfy the inequality are those lying below (or on) this line.

Example 3:

Sketch the region containing all points that satisfy the system. 

x^2 - y <(=) 1

-x + y <(=) 1

As shown in the graph above, the points that satisfy the inequality x^2 - y <(=) 1 are the points lying above (or on) the parabola given by y = x^2 - 1. The points that satisfy the inequality -x + y <(=) are rhe points lying below (or on) the line given by y = x + 1. To find the points of the intersection of the parabola and line, solve the system of equations. Using the method of substitution, you will find the solutions to be (-1,0) and (2,3). 

Week 2, Day 1

Last week, we learned how to perform partial fraction decomposition, which involves writing a rational expression as the sum of two or more rational expressions. For instance, (x+7) / (x^2-x-6) can be written as the sum of two fractions with linear denominators. That is, (x+7) / (x^2-x-6) = (2 / x-3) + (-1 / x+2). Each fraction on the right side of the equation is a partial fraction, and together they make up the partial fraction decomposition of the left side. The following are steps to carry out partial fraction decomposition: 

1. Multiply by LCD

2. Distribute

3. Collect like terms

4. Factor out the variable

5. Equate the coefficients

6. Solve the system of equations

7. Write the answer as a partial fraction

When the denominator of a partial fraction is, for example, (x+1)^2, it is called a repeated factor.

When the denominator of a partial fraction is, for example, (x^2+1), it is called a quadratic factor. 

Here are a few examples:

Example 1:

5/(x^2+x-6)--> 5/(x+3)(x-2) = A/(x+3) + B/(x-2)

1. 5 = A(x-2) + B(x+3)

2. 5 = Ax - 2A + Bx + 3B

3. 5 = Ax + Bx - 2A + 3B 

4. 5 = (A + B)x - 2A + 3B 

5. 0 = A + B, 5 = -2A + 3B

6. 5 = 5B--> B = 1, A = -1

7. -1/(x+3) + 1/(x-2)

Example 2:

2x - 3/(x-1)^2 

In this case, a factor (x-1) repeats, and thus the steps to solve it are as follows: 

1. Multiply by LCD

2. Distribute

3. Collect like terms

4. Equate the coefficients

5. Solve the system of equations

6. Write the answer as a partial fraction

2x - 3/(x-1)^2 = A/(x-1) + B/(x-1)^2 Every exponent gets a factor!

1. 2x-3 = A(x-1) + B

2. 2x-3 = Ax - A + B

3. 2x-3 = Ax + (-A + B)

4. A = 2, -3 = -A + B

5. -3 + 2 = B = -1

6. 2/(x-1) - 1/(x-1)^2

Example 3:

4x^2 + 2x - 1 / x^2(x +1) = A/x + B/x^2 + C/x+1

1. 4x^2 + 2x - 1 = Ax(x+1) + B(x+1) + Cx^2

2. 4x^2 + 2x - 1 = Ax^2 + Ax + Bx + B + Cx^2

3. 4x^2 + 2x - 1 = Ax^2 + Cx^2 + Ax + Bx + B

4. 4x^2 + 2x - 1 = (A + C)x^2 + (A + B)x + B

5. A + C = 4, A + B = 2, B = -1

6. A = 3, B = -1, C = 1

7. 3/x -1/x^2 + 1/x+1

Example 4:

x^2 - 1 / x(x^2 + 1) 

In this case, the denominator contains an irreducible quadratic factor, so the set up is as follows:

x^2 - 1 / x(x^2 + 1) = A/x + Bx + C / x^2 + 1

1. x^2 - 1 = A(x^2 + 1) + (Bx + C)x

2. x^2 - 1 = Ax^2 + A + Bx^2 + Cx

3. x^2 - 1 = Ax^2 + Bx^2 + Cx + A

4. x^2 - 1 = (A + B)x^2 + Cx + A

5. A + B = 1, C = 0, A = -1

6. A = -1, B = 2, C = 0

7. -1/x + 2x / x^2 + 1

Week 1 7.3 Example #6

Today I learned how to expand our use of elimination to solve for not one, not two, but three variables. This can be referred to as a non-square system. The following link is a video demonstrating how to solve a system of equations with three variables: 

http://www.educreations.com/lesson/view/hma-period-5-square-and-nonsquare-systems-6/15521951/?s=7a1VCG&ref=app

Week 1 Interesting Math Facts

1. In a group of 23 people, at least two have the same birthday with the probability greater than 1/2

2. Among all shapes with the same perimeter, a circle has the largest area

3. Among all shapes with the same area, a circle has the shortest perimeter

4. There are curves that fill a plane without holes

5. 12 + 3 - 4 + 5 + 67 + 8 + 9 = 100. 

There is at least one other representation of 100 with 9 digits in the correct numerical order

6. It is possible to cut a pie into 8 pieces with only three movements

7. The number 5 is pronounced 'ha' in Thai, and .555 is used as slang for 'ha, ha, ha'

8. Notches on animal bones prove that humans have been doing math since 30,000 BC

9. Dyscalculia means difficulty in learning arithmetic, such as understanding how to operate numbers

10. Googol (origin of the 'Google' brand) is the term used for a number 1 followed by 100 zeros

Week 1, Day 2

Today I learned how to solve for variables through elimination. The steps to perform this operation is as follows: First, obtain coefficients that differ only in sign. Second, add equations to eliminate a variable. Then back substitute to solve for the second equation. Finally, check your solution! Once you have determined the value of each variable, write your solution as an ordered pair. There are a number of possibilities in regard to solutions, three of which are intersecting lines (one solution), parallel lines (no solution), and the same line (infinitely many solutions). The following are example problems demonstrating how to solve for variable through elimination:

Example 1:

5x + 3y = 9 

2x - 4y = 14

Step 1:

 4(5x + 3y = 9) --> 20x + 12y = 36

 3(2x - 4y = 14) --> 6x - 12y = 42

Step 2: 

 20x + 12y = 36

   6x - 12y = 42

+ -----------------

26x = 78

x = 3

Step 3: 

2(3) - 4y = 14

6 - 4y = 14

-4y = 8

y = -2

Step 4:

2(3) - 4(-2) = 14

6 + 8 = 14

14 = 14

(3, -2) 

Example 2:

An airplane flying into a headwind travels the 2000-mile flying distance between two cities in 4 hours and 24 minutes. On the return flight, the same distance is traveled in 4 hours. Find the airspeed of the plane and the speed of the wind, assuming that both remain constant.

r(1) = speed of the plane

r(2) = speed of the wind

r(1) - r(2) = speed of the plane against the wind

r(1) + r(2) = speed of the plane with the wind

Use the formula: Distance = (rate)(time)

2000 = (r(1) - r(2)) (4 + 24/60) --> 5000 = 11r(1) - 11r(2)

2000 = (r(1) + r(2)) (4) --> 500 = r(1) + r(2)

By elimination, the solution is:

r(1) = 5250/11 = 477.27 miles per hour (speed of the plane)

r(2) = 250/11 = 22.73 miles per hour (speed of the wind)

Week 1, Day 1

Today I learned how to solve for variables through substitution. The first step is to isolate the variable. Once you have determined its value, subsitute it for the same variable in the second equation. By basic computation, you should be able to identify the value of each variable. If the solution is, for example, 2 = 8, then there is no solution because the statement is mathematically false. If the solution is, for example, 3 = 3, then all solutions are real numbers. Factoring may also come into play if, for instance, you get the equation x^2 - 2 = 0, in which case you would remove the x and get x (x - 2) = 0. Setting each quantity equal to zero, you would conclude that there are two solutions, 0 and 2. The following are example problems demonstrating how to solve for variables through substitution: 

Example 1: 

x + y = 4, x - y = -2

1) x = 4 - y 

2) (4 - y) - y = -2

4 - 2y = -2

-2y = -6

y = 3

3) x + 3 = 4

x = 1

Example 2:

x^2 - x - y = 1

-x + y = -1

1) y = -1 + x

2) x^2 - x - (-1 + x) = 1

x^2 - x + 1 - x = 1

x^2 - 2x + 1 = 0 

x^2 - 2x = 0 

x (x - 2) = 0 

x = 0, 2

3) If x = 0, then 

-0 + y = -1

y = -1

If x = 2, then

2 + y = -1

y = -3

y = -1, -3

I also learned two equations dealing with the concept of "breaking-even."

1) Total cost = cost per unit x # of units sold + initial cost

2) Total revenue = price per unit x # of units sold

3) Break-even if total cost = total revenue