The idea behind the addition / elimination method is to multiple one or more equations by a constant so when they are added together, one of the variables eliminates. Then you have one equation with one variable and you can solve for that variable.

- Choose a variable to eliminate. Usually the variable that can be eliminated by multiplying by smaller numbers is the better choice.
- Multiply one or both equations by a constant so that the least common multiple of the coefficients on the variable to be eliminated is obtained. Care should be taken so that one coefficient becomes negative and the other is positive.
- Add the two equations together so the variable is eliminated.
- Solve the resulting equation for the remaining variable.
- Back-substitute that value into the one of the two original equations to find the remaining variable.
- Check your answer into the other equation.

As an alternative to step 5, and this is extremely helpful when the answer is a fraction or decimal value and not pleasant to work with (I know, I really should stop talking bad about my friends), you can go through the elimination process again with the other variable, and then you don't have to work with the fractions until the check process.

You may have either no solution, one unique solution, or many solutions when solving a 2x2 system of linear equations.

- Intersecting lines
- Consistent System (consistent means there is a solution, there is no contradiction).
- Independent System (the value for y doesn't depend on what x is).

Write your answer as the set containing the ordered pair {(x,y)} (replace the x and y with the actual values).

This would be analogous to a conditional system from section 2.1, certain values of x and y make it true, while others don't.

Consider the system of linear equations 3x + 2y = 17 and 2x - y = 2. The solution is x = 3 and y = 4, so you write the answer as { ( 3, 4 ) }.

- Parallel lines
- Inconsistent System (contradiction)
- Independent / Dependent doesn't apply since there is no solution.

Write your answer as "no solution", the symbol for the empty or null set, Ø, or the empty set { }, but do not write it as the set containing the null set {Ø}. Be careful if you're a person who slashes their zeros. Make sure I can tell a zero apart from the null set.

This case occurs when both variables eliminate, and you are left with a false statement.

This would be analogous to a contradiction from section 2.1.

Consider the system of linear equations 3x - 2y = 3 and -3x + 2y = 2. When you add them together, you get 0 = 5, which is a contradiction. Therefore, the answer is "no solution", { }, or Ø

- Coincident lines (same line)
- Consistent System (identity)
- Dependent System (the value for y will depend on what x is, it's not always the same value).

Write your answer as one of the two equations. Do NOT say "many solutions" or "all real numbers". All real numbers will not work. First of all, it the solutions are ordered pairs, not individual x or y coordinates. Second, not every point works, only those on the line work.

You may also write your answer in parametric form. This will be the preferred method for higher ordered systems, so you might as well learn it now.

This case occurs when both variables eliminate, and you are left with a true statement.

This would be analogous to an identity from section 2.1.

Consider the system of linear equations 3x - 2y = 3 and -3x + 2y = -3. When you add them together, you get 0 = 0, which is always true. Therefore, any values of x and y that satisfy the equation are solutions. You would write your solution as

- 3x - 2y = 3
- { ( x, y ) | 3x - 2y = 3 }
- x = t, y = 3/2 t - 3/2 (this is known as parametric form and is covered in section 8.3)

Ever wonder why most all of the problems in an algebra textbook come out with nice integer answers? Is it because life is that way? Certainly not, it is because the problems are contrived to have nice answers.

How do you get a made up problem to have nice answers? The answer is that you don't. You start with the answer and work backwards.

Let's say that we want the answer to be (3,-2).

Make up anything for the left side of the equation, say 2x-5y. Then, plug in x=3 and y=-2, so 2(3) - 5(-2) = 6 + 10 = 16. The first equation is 2x-5y=16.

Now, repeat the process again with a different left side, say 3x+2y. Well, 3(3) + 2(-2) = 9 - 4 = 5, so 3x+2y=5.

Your system of linear equations is 2x-5y=16 and 3x+2y=5.

You can make up anything for the left hand side, just make sure the right hand side is what you get when you plug those values in. You will also need to have two equations if there are two variables, one equation if there is only one variable, three equations if there are three variables, etc.

Up until this point, when we have found the linear regression model, we have just used the functions on the calculator to obtain the results, and it has been fairly easy and painless. Now we learn that the calculator is actually solving a system of linear equations to obtain the model.

The capital Greek letter sigma stands for sum. Normally, there is an index with a starting point (k=1) written below the sigma and an ending point (n, meaning k=n) written above the sigma. Then, each variable will have a subscript to let you know that it is a function or sequence that depends on the value of the index, k.

Confused? Well, maybe you should be. The book doesn't discuss sequences and series until chapter 7, and here you are in chapter 5 and they expect you to know how to do it.

I'm going to use a short hand notation to keep things simpler and easier to remember. Instead of writing it the way the book does, I'll just use a sigma symbol and then what I want to total.

Remember that sigma means sum, so sigma x means add up all the x's.

In statistics, we like to simplify things, and get a little sloppy, and drop all the index stuff and just know that it applies to all the points. In the notation above, I will use the form on the right. ∑x just means add up all the x values. Not too bad when you look at it that way.

Consider the linear model y = ax + b. The values for a and b can be found by solving this system of linear equations.

b∑1 + a∑x = ∑y

b∑x + a∑x^{2} = ∑xy

Notice each term in the second equation has one more x in it than the corresponding term in the first equation. This pattern will repeat when we do quadratic regression (see problems 105-108 in section 5.3 or Appendix B.3 for an explanation), or cubic, or quartic, or ...

You will add up each variable in the summation for each different point. The first summation is the sum of 1. So, if you add 1 for each point, you will simply have the number of points. The other values are the sum of the x's, the sum of the y's, the sum of the squares of the x's, and the sum of the products of the x's and y's.

Writing these ordered pairs in a columnar table, and then adding columns
for the x^{2} and xy will help. After you solve the system of linear equations, substitute
the values for a and b into the equation y = ax + b to get the model.

Find the equation of the linear model that best fits the points (2,3), (5,2), (6,1), and (8,-1).

Set up a table with columns for x, y, x^{2},
and xy.

x | y | x^{2} |
xy |
---|---|---|---|

2 | 3 | 4 | 6 |

5 | 2 | 25 | 10 |

6 | 1 | 36 | 6 |

8 | -1 | 64 | -8 |

21 | 5 | 129 | 14 |

The numbers in the bottom row represent the sums that go into the system. Since there are 4 points, the ∑1 = 4.

The system of linear equations to solve is 4b + 21a = 5 and 21b + 129a = 14. When you solve that, you get a = -49/75 and b = 117/25.

When you stick those back into the model, you get the y = -49/75 x + 117/25.