Linear Equations, Part 1: Basics

With finals coming up I have been doing a lot of linear equation problems with my students, so I thought it appropriate to do a post about linear equations. One notable characteristic about linear equations is that they have constant slopes. This leads to a few simple equations and techniques that will allow the student to solve any linear problem.

First, let's talk about the three forms of linear equations:

*Standard Form: Ax + By = C
*Slope-Intercept Form: y = mx + b, where b is the y-intercept
*Point-Slope Form: y - y1 = m(x - x1), where (x1,y1) is a known coordinate.

In these equations m represents the slope. If there is no information given about the line except for two points (x1,y1) and (x2,y2), then the slope can be found using the slope formula:

  m = (y2-y1)/(x2-x1), also known as "rise over run"

We can then put the slope and one of the known points into the Point-Slope form. Solving for y will yield the Slope-Intercept form. If necessary we could rearrange the Slope-Intercept form into the Standard form.

Example:


Given the two points (2,6) and (6,8) find the equation of the line that passes through these points,
a) in slope-intercept form
b) in standard form
c) state the coordinate of the y-intercept

Solution

a) First we have to find the slope. Using the slope formula we have:
           m = (8-6)/(6-2) = 2/4 = 1/2  *always simplify fractions when possible
   
    Now that we know the slope, we substitute it and one of the points into the point-slope form:
          y - 6 = (1/2)(x - 2)  *either known point will work. I chose (2,6) because of the smaller numbers
    
     Solve for y
          y - 6 = (1/2)x - 1  *distributive property
          y = (1/2)x + 5       *add 6 to both sides to isolate y
     
     The slope-intercept form is y = (1/2)x + 5

b) All we have to do here is rearrange the slope-intercept form, solving for the constant
         y = (1/2)x + 5   *start with slope-intercept form
         -(1/2)x + y = 5  *subtract (1/2)x from both sides to isolate the constant.
         2[-(1/2)x + y] = 5(2)   *multiply both sides by 2 to clear the fraction
         -x + 2y = 10   *Standard form

c) Refer to the slope-intercept form, y = (1/2)x + 5
          In this instance b = 5, so the coordinate of the y-intercept is (0,5)  *x is always zero at the y-intercept

So we have gone from knowing nothing but two points on a particular line and was able to determine the slope, all three forms of the equation, and identify the y-intercept. All of this is possible because linear equations have constant slopes.

In Part 2, I will analyze the above example graphically and give some other graphical examples.

-mike-

Limit Basics

Today's tutoring session with Chris was mostly about series and sequences, but a key concept in understanding them is the idea of a limit. Limits are also fundamental to differential and integral calculus. What is a limit? A limit describes the behavior of a sequence or a function as the variable approaches some number, usually infinity or zero. Lets look at some examples:

1) What is the limit of the sequence {1, 1/2, 1/3, 1/4,...,1/n} as n approaches infinity?

We can see that as n gets larger the number gets smaller. So as n goes out towards infinity the number becomes infinitesimally smaller coming closer and closer to zero, and so we say that the limit is zero. This sequence converges to zero.

2) What is the limit of the sequence {1, 4, 9, 16, 25,...,n^2} as n approaches infinity?

We can see that as n gets larger so does the number--exponentially. So as n goes out towards infinity the number becomes infinitely larger, and so we say that the limit is infinity. The limit diverges.

3) What is the limit of the sequence {cos(π), cos(2π), cos(3π),...,cos(nπ)} as n approaches infinity?

From trigonometry we know that this is equivalent to {1, -1, 1, -1,...}, and so this sequence oscillates forever between 1 and -1. This limit does not exist and is considered divergent.

To denote a limit symbolically we write lim n→∞ which is read as "the limit as n approaches infinity." (It is important to note that on paper "n→∞" is written underneath "lim" however we cannot do that on a computer).

These are simple examples. What if the sequence is explicitly defined by a rational function? There are three simple rules that will allow such a limit to be found quickly. Lets take a rational function defined by (a^n+c)/(b^m+d)

*If n=m, then the limit is the ratio of the leading coefficients.
*If n>m, then the limit diverges to infinity
*If n<m, then the limit converges to zero.

 For example:

1) lim n→∞ (3n+1)/(n-2)  = 3/1 = 3       
2) lim n→∞ (n^3)/(n^2+10)  = infinity and is divergent      
3) lim n→∞ (3-n)/(n^2+5) = 0

Limits are fundamental to calculus. Both derivatives (rates of change) and integrals (areas under curves) are driven by limits. This shall be discussed further in another post.

-mike-

Problem Solving

Today's lesson with Brian was mostly preparing for his EOC (End Of Course) exam. Word problems often trip people up, but not because they are hard; because they have lots of words. The secret to quickly solving a word problem is to identify the math: What are the operations? What is given? What is unknown? What is the problem asking?

Example:

You burn 20 calories per minute running and 10 calories per minute swimming. Your total workout is 40 minutes of running and swimming where r is the total minutes spent running. How many calories do you burn in your 40 minute workout if you spend 20 minutes running?

The first issue here is that there are two unknowns: time spent running, r, and time spent swimming. However, since we know the total time of the workout we can put time spent swimming in terms of time spent running. Time spent swimming is the total time minus the time spent running: So time spent swimming is (40-r).

Lets build an equation: y = 20r + 10(40-r). Here we have the number of minutes spent swimming in terms of the difference between the total work out and time spent running. Each coefficient is the number of calories burned per minute for running and swimming respectively.

Now lets substitute the given information into the equation. From the last line of the word problem we know that r = 20. So we have: y = 20(20) + 10(40-20).

Easy to evaluate: y = 400 + 10(20) = 400 + 200 = 600.

So you would burn 600 calories on your 40 minute workout.

-mike-

Welcome

I love math. It is a language just like any other tongue. Not only does understanding math help problem solving skills, but it also helps to cultivate intuition and creativity. I am passionate about teaching math to everyone from grade school to college, from young kids to adults decades out of school.

I have been tutoring for 3 years both privately and as a math tutor at Valencia College. I recently opened up my own private studio out of which I conduct tutoring when not at the college. This blog will serve several purposes:

*To act as interaction with my students outside of our tutoring session. Students can ask questions, pick up some extra problems, or communicate with me about math in general.

* To be a good read for anyone interested in math

*To advertise my services

*To talk in general about math; I will be talking about various techniques and posting instructional videos for the benefit of anyone who reads my blog.

If you like math, don't like math, want to learn math, like a challenge, or are just interested in numbers, then please subscribe and tell your friends to check out the blog as well.

-mike-