Don't forget my book if you are having trouble in algebra.

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Suppose we wish to solve a system of two equations in two variables by the elimination method or substitution method.  Which method is easiest to use depends on the way the equations are written and personal preference.

In the system     x = 2y + 5       it's easiest to subsitute 2y + 5 for x in the second equation.
                         2x + 3y = 10


But in the system    x + 6y = 12   it's easiest to eliminate one of the two variables first.
                               5x + 6y = 20   

A rule of thumb that I use to determine which method to use is if either equation in the system is already solved for one variable in terms of the other (see first system above), use the substitution method.


Solving the first sytem using substitution, we get

2(2y+ 5) + 3y = 10

4y + 10 + 3y = 10

7y + 10 = 10

7y = 0

  y = 0

Substitute 0 for y in the first equation to get x = 2(0) + 5 = 5

Therefore, the solution is x = 5, y = 0.

Solving the second system, it's easiest to eliminate the y variable by subtracting the second equation from the first. This gives us

-4x = -8

x = 2

Substituting 2 for x in the first equation of the system gives us

2 + 6y = 12

6y = 10

y = 5/3

Therefore, the solution is  x = 2, y = 5/3

When constructing a box and whisker, we need to know the minimum value of the set of data, the maximum value, the first quartile (the middle value of the lower half of the data), the median (the middle value of the set of data when arranged from lowest to highest), the third quartile (the middle value of the upper half of the data) and the maximum value. An example of a box and whisker is shown with these five values identified.  Notice the box drawn enclosing
the first quartile, the median and the third quartile. 
 

Today I was working with a student calculating probabilities. One problem involved a square inscribed in a circle. What is the probability that a randomly placed dot will fall within the square?  To solve such a problem, we simply find the area of the square and divide it by the area of the circle.  Remember that the area of a square is given by A = s^2, there s is the length of the side.  The area of circle is given by A = Pi(r^2).  You can use the approximation 22/7 for Pi or 3.14 to make calculations easier.

Remember that the absolute value of any quantity is positive. To solve an equation involving absolute value, remember to solve the positive case and the negative case. 

For example |x + 5| = 10,  the positive case is x + 5 = 10 and the negative case is x + 5 = -10. Therefore, the solutions are x = 5 or x = -15.

The graph of the function f(x) = |ax + b| where a and b are real numbers will always fall in the first and second quadrants since absolute value is always positive.

The graph of the function f(x) = c|ax + b| where a and b are real numbers and c is negative will be in the third and fourth quadrants.

We are always told that division by 0 is undefined, in the case of 1/0 or n/0, where n is any number except for 0.  If we examine the fraction 1/n and choose smaller and smaller numbers for n approaching 0, what happens?

For n = 1,  1/1 = 1
For n = .1,  1/.1 = 10
For n = .0001, 1/.0001 = 10,000
For n = .0000001, 1/.0000001 = 10,000,000

Notice as n approaches 0, 1/n approaches infinity.

Can we say then that 1/0 is infinity since we never actually use 0 in the denominator of the fraction 1/n?

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