Systems of Linear Equations

I'm back! Sorry for the tardiness, but college students never have much extra time on their hands.

Anyway, today's post will be a mathematical review; coming directly from a recent tutoring session in which I helped a student with systems of linear equations. Systems of linear equations can be solved by using either the substitution method, the addition method, or the graphical method. Sometimes more than one method is required, especially for systems of three equations, each with the same three variables. Personally, I prefer the addition and substitution methods, but the graphical method is fine, too.

Say we are given the following system of linear equations:

This is very simple and straightforward example. We can solve this easily using the addition method, which would simplify to 3x = 6, which would further simplify to x = 2. Using this result, we can solve for y: (2) - y = 1 à 2 - 1 = y à y = 1. Or 2(2) + y = 5 à 4 + y = 5 à y = 1. Our answers, therefore, are x = 2 and y = 1, or an ordered pair of (2,1).

If we decided to use the substitution method, on the other hand, we would begin by solving for a variable in either of the equations. I'll solve for y in equation 2: x - y = 1 à y = x - 1. Then we substitute this in equation 1 for the variable y: 2x + (x - 1) = 5 à 3x - 1 = 5 à 3x = 6 à x = 2. The rest of the answer is the same as that already used for the addition method.

The same method applies if we have a system of three equations, each with the same three variables. There will just be more steps involved. We will solve the following example:

I will start with the substitution method, solving for z in the third equation. z = 3x - y - 10. Next, we will substitute this value for z in our other equations, before using the addition method. If we do use the addition method first, as it is tempting to do, we will still end up with a value for y. So that will give us 2x + 2y - (3x - y - 10) = 4, which simplifies to -x + 3y + 10 = 4 or -x + 3y = -6. In our second equation, we get x - 2y + 4(3x - y - 10) = 5, which, after simplifying, results in x - 2y + 12x - 4y - 40 = 5. After further simplification, we find 13x - 6y = 45.

In order to ensure that we can cancel one of our variables, I will multiply the entirety of the first equation by 2. Our result being -2x + 6y = -12. And now, we can go ahead and use the addition method:

Which gives us the conclusive result that x = 3. Taking that answer, we can substitute that value for x in either of the most recently derived equations. We will use that derived from the original equation two. 13(3) - 6y = 45 à 39 - 6y = 45 à -6y = 6 à y = -1. And finally, we can use those values in the equation solved for z in the first step to find a value for z. z = 3(3) - (-1) - 10 à z = 9 + 1 - 10 à z = 0. Our ordered triplet, then, is (3,-1,0).

I hope this finds those that may need it, and that it helps them to understand better how to find the intersection point, or solution, to a system of linear equations. You can contact me through my ANONYMOUS contact form: Anonymous Contact Form—it does have a place for a name and email, BUT those are NOT REQUIRED—the one caveat is that you might not be able to get a response.

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