SOLVING A SYSTEM OF (TWO) LINEAR EQUATIONS - A Guide of Hints, Strategies and Simple Explanations - Algebra in Words

Algebra in Words: A Guide of Hints, Strategies and Simple Explanations (2014)

SOLVING A SYSTEM OF (TWO) LINEAR EQUATIONS

Before continuing, there are some important things to be aware of.

A system of equations means: lines on the same graph that may intersect.

What does it mean to solve a system of two linear equations? To solve means to find the point of intersection, which is literally in the form (x, y)… so you’re essentially finding an x and corresponding y. The (x,y)-point is the solution (when there is a solution, which there won’t always be). For more on this, see: Interpreting the "Solutions", including: No Solution, in the next few pages.

You always need as many equations as you have unknown variables to solve for. Here, there are 2 equations and 2 unknown variables. (If you have 3 unknown variables, you would need 3 equations with those variables, etc.).

What Does “Solving In Terms Of” Mean?

When rearranging equations, you will often hear it explained as “Solve for one variable in terms of the other variable.” More specifically, you might hear it as “Solve for y in terms of x.” It is important that you understand the context of the words “in terms of.” In the beginning of algebra when you learn to solve simple algebraic equations, you are used to solving one equation with one variable, and finding the numeric value of that variable; that is the endpoint. From that point, students often get accustomed to: solving, and getting a number for an answer, and being done.

But sometimes, especially for a multi-step problem (as a system of two linear equations is), you solve for (isolate) a variable, but you don’t get a number-answer (at least right away). But this is ok. Some students who expect to solve and get a number (instantly) think they made a mistake when they don’t get a number. This is not unusual. This is where one variable is solved in terms of another variable, meaning neither variable goes away, and you don’t solve to get a number… but you are still rearranging the equation to isolate a variable of interest, and whatever variable(s) still remains is shown on the other side with the other terms. To solve for something in terms of a variable can be translated and broken down like this:

“Solve for” = isolate

“Something” = whatever you’re isolating. It could be a variable. It could also be a number or an entire term. Whatever it is, get it to one side, and whatever remains goes on the other side.

“In terms of” = the variable(s) or term(s) which go on the opposite side of the entity being solved for.

Note: This type of equation rearrangement is not only used for solving systems of equations. You will often also see a small chapter in your textbooks dedicated just to multi-variable equation rearrangement. The equations used are often associated with geometry, trigonometry, statistics, economics, physics and chemistry. Exercises in solving multi-variable equations in terms of other variables prepare you for actual application of those equations in their related fields.

The Three Ways to Solve

There are three general ways to solve a system of two linear equations (two lines in a two dimensional space, also known as a plane):

1. Graph & Check

2. The Substitution Method

3. The Addition/Elimination Method

All three methods will yield the same outcome. I will discuss the best time to use each method, especially when it is better to use the Substitution Method vs. the Addition/Elimination Method.

How many solutions should you expect? There will either be:

·   one solution (made up of one x and one y),

·   no solution, or

·   infinite solutions

These are the only possible outcomes. For instance, there can’t be two solutions. This explanation is continued in: Interpreting the Solutions, in which the situations and associated vocabulary for these solutions are explained in more detail. But the next three sections give a closer look at each of the three solving methods.

Graph & Check

  Sometimes this method is broken up separately into Check or Graph, and sometimes they must (or can) be done together.

  Often the Check-only method is introduced first. In this case, an ordered pair (an x, y point) will be given to you. To check, you must substitute the x-value in for x and the y-value in for y into both equations, then simplify. Your goal is to determine if the point is or is not a solution. In order for the point to be a solution, each equation will reduce to a number that equals itself, such as 3 = 3. But if only one (or neither) simplifies to a number that equals itself, then the proposed point is not a solution. You will know when the point is not a solution because an equation will simplify to a number that appears to equal a different number, which reveals itself to be a blatant inequality, and therefore not a solution.

  In another related instance, the point itself isn’t given to you, but a pre-drawn graph is. In this case, you are expected to read the point of intersection, and then check that point as explained in the previous paragraph. Often times, a pre-drawn graph will be made to have an obvious point, and by obvious I mean “integers,” not some obscure decimal numbers.

  Or you may have to do every step: create the graph, interpret the point of intersection and check it, as previously described. You could be given two sets of points (two or three sets of points for each of the two lines), which you will have to plot and graph. You might also just be given two equations and be instructed to “find the solution,” in which case you must:

·   Determine three points for each graph (so a total of 6 points),

·   Plot the points and sketch the two lines,

·   Make a judgment as to the point of intersection, then

·   Check the point by substituting the values into each of the two equations, simplifying, and interpreting the outcome.

A few reminders:

·   Use the procedure for: How to Graph a Linear Equation.

·   Upon making a table of points to plot, take notice of any x-y points that are the same in both tables. If you find a matching pair, that is your solution. You should probably still graph, then check to confirm.

·   Neatness is essential when graphing. Try to use graph paper. If you don’t have any, consider using a straight edge. But more importantly, try your best to draw each unit on your x and y axes with equal lengths. This will make your point of intersection more accurate, and will give you a better ability to find the correct point, which should then allow it to successfully check.

·   Also, even when you are not required to Graph & Check, you still can if you want to double check your results from the Substitution Method or the Addition/Elimination Method.

The Substitution Method

The substitution method is started in one of two ways.

One way is by taking one equation and solving it for one variable. When doing this, aim for the variable that will be most easily isolated. A good way to identify the best variable to isolate is by finding a term with either a small coefficient, or a term with a coefficient that the other terms in the equation will be easily divisible by. For instance, if one of the terms is “2y” and the other two terms are even (perhaps they are 6x and -8), then solving for y is a good choice because the other terms are (easily and noticeably) divisible by 2.

Sometimes a term already has no coefficient (meaning it has an unwritten coefficient of 1). In this case, it’s a good idea to employ the Substitution Method because part of the work is done for you (that part being to make its coefficient 1). All you have to do then is isolate that variable. There are many systems of equations where this is the case. Often times, the writers of the math problems set you up to notice this variable. Sometimes, the variable is even already isolated, so all you have to do is realize that, then proceed to substitute what it equals in for that variable in the other equation.

Once you have solved for one of the variables in terms of the other variable, you must take that equality and substitute it into the variable for which you just isolated, in the other equation. For instance, if you just solved for y in one equation, then you must take what y equals and substitute that in for y in the other equation. When you do this, you should take notice of three things:

1.  All the variables in the equation you just substituted into will be the same. It is only when they are the same that allows you to solve for the numeric value of that variable.

2. Once you do the substitution, you will be solving for the numeric value of the other variable. For instance, if you originally solve for y in terms of x and then substitute in for y in the other equation, you will then solve for the numeric value of x. Also,

3. Be careful not to make the common mistake of substituting into the equation you just solved for in the first step. If you do, then once you simplify, you will end up with a number that equals itself, and this may leave you confused and wondering where to go from there. So remember to substitute into the other equation.

You are now one step away from completing this problem, but students sometimes get confused at this final step.

The last step is to take the numeric value of the variable you just solved for and substitute that back in for that variable into either of the original equations, then solve for the other variable. For instance, if you just found the value of x to be -5, substitute -5 back in for x in one of the original equations, then solve for the value of y. I have two comments about this:

·   Students are often confused by: Which of the original equations should I substitute my value back into? And the answer is: either. Sometimes the choice seems to confuse students, so here’s how you can choose. You can either

just randomly pick one, or

·   Choose the equation which appears to simplify easier. The one that will be easier to simplify may be the one with smaller coefficients or the one that does not contain fractions. If both look like similar difficulty, just choose one randomly. The answer will come out the same.

Sometimes students forget the final step, perhaps because, up to this point, you are used to a one-number or one-variable answer. Don’t forget this step. Remember, the solution is a point (an x and a y).

The most common mistake made by students using this method is getting confused about what to substitute. So in summary, you solve for one variable in terms of the other… leaving you with one variable isolated on one side of the equal sign, and the other two terms on the other side. You then substitute in for the variable you just solved for into the other equation. The “stuff” you substitute into the other equation will replace the variable with the two terms from the other side of the first equation. You must then distribute and simplify in order to solve for your first value.

The Addition/Elimination Method

Some books call this “The Addition Method” and some books call this “The Elimination Method.” I call it a hybrid of both, because you start by adding the two equations (after any necessary conversions), which eliminates one variable, making it a new, one-variable equation that can be solved (for the other variable). Remember, to perform this method, terms of the same variable in each equation must be opposites so they cancel out to zero when (the equations are) added together.

But my focus is to tell you when it is advantageous to use this method. Here are some clues to look for:

·   You notice two terms of the same variable in each equation which are already opposites [meaning same term (variable and coefficient) but opposite signs]. These are already set up to cancel each other out to zero once the equations are added. All you have to do is add the equations, then proceed to the next step.

·   You notice that one term is one multiple away from making it the opposite of a term of the same variable in the other equation. For instance, if one term in one equation is 3x, and the other equation has a -9x, then 3x can become +9x by multiplying it by 3 (and don’t forget to multiply that factor through by the other terms in that equation). Or you have the term -5x in one equation and -5x in the other. You must multiply one of the equations through by -1, to make a -5x become +5x.

·   Or, whenever neither equation is given with a variable with (an unwritten) coefficient of 1… mainly, the opposite of what is explained for The Substitution Method. In other words, all variables havecoefficients (other than 1).

Common Mistakes:

·   A very common mistake students make is adding the two equations without checking and converting one equation (to manipulate one variable into the opposite of a term from the other equation). If you do not properly set the equations up to have opposite terms, then adding the equations will just give you a third equation, still having two variables. Students often get stuck here, and rightfully so, because this is a dead-end; there’s nothing you can do with it.

·   Another common mistake students make is forgetting to add the constants when the equations are added. The constants are the numbers with no variables attached. Don’t forget to add them, as they are just as much a part of the problem as the terms with variables.

Examples for Choosing the Method

In this section there are two “systems of linear equations” given: System A and System B. Each equation is already simplified and put into standard form for this type of problem. I want you to examine each set, and using the clues explained in the previous two sections, determine which method (Substitution or Addition/Elimination) would be best to use in each case. The answers and explanations will be given on the following page. The “solutions” to the system will also be given in case you want to do the problem for practice.

System A:

Equation 1:

Equation 2:

System B:

Equation 3:

Equation 4:

System A would best be solved using the Substitution Method because the “y” in Equation 2 already has an (unwritten) coefficient of 1. The next step would be to isolate the y by adding to both sides, giving you: . Then substitute the in for y in Equation 1. The solution to System A is (-2, -1).

System B would best be solved using the Addition/Elimination Method. There are two ways to approach this. First, take notice of the “14x” and the “-7x”. If the “-7x” is multiplied by “2,” it will become “-14x” which is the opposite of “14x”. You would need to multiply each term in Equation 4 by 2 to properly convert “-7x” into “-14x”. Then, when you add the two equations, the “x” terms will cancel out to zero, allowing you to then simplify and solve for y (and then x). The solution to System B is (2, -14).

Or, you could solve System B another way, by multiplying Equation 3 by “2” and multiplying Equation 4 by “3”. This would make

“3y” would become “6y,”

“-2y” would become “-6y,” and

The y-terms cancel out to zero because 6y – 6y = 0.

As I mentioned before, some students take a mistaken approach to this first by adding the two equations together without multiplying through the necessary term(s) to manipulate terms of one variable to cancel. If mistakenly added, you would then get (what I will call)

Equation 5: y + 14x = 8

… Notice how both variables still remain in the equation? This leaves you at a dead end, because you can’t successfully use this equation to solve for either variable.

Interpreting the “Solutions”

One Solution - Consistent

When the two lines cross, this is called a consistent system. In this case, there is one solution to be found, which is the point of intersection. The lines can be a combination of diagonal, horizontal and/or vertical lines. Keep in mind, all sets of perpendicular lines have one solution and make a consistent system.

No Solution - Inconsistent, Parallel

The two lines don’t cross… because they are parallel; parallel lines by definition never touch. This is an inconsistent system. There are 3 ways you can tell that lines are parallel:

1. When using one of the three methods for solving a system of two linear equations, when you simplify and get to the end of the problem, you will get one # that does not equal the other #. It will look something like:

7 = -5   or  0 = 4,  which clearly isn’t true.

2. The slopes (m) of the two lines are identical. In order for you to see this, you must convert the equations into slope-intercept form (y = mx + b), and then simply look at the slopes. The equations of lines may or may not originally be in slope-intercept form (y = mx + b). If they are not, convert them to slope-intercept form by solving for (isolating) y.

Also, be sure each equation is in simplified form. If there is a Greatest Common Factor in an equation, you must factor it out. If you don’t, the slopes may appear different, even though, by proportion, they are actually the same.

3. Graph and look. You can find out if lines are parallel without graphing, as described in the last paragraph. But this method (graphing & looking) should act as a backup to the two methods above, to confirm your answer. It may also be a great help if you are a more visual learner. For a reminder on graphing from an equation, see: How to Graph a Linear Equation. Once drawn, look at the lines to see if they appear to cross.

Infinite Solutions - Dependent

The entire lines overlap… because they are essentially the same line. The graph actually looks like one line. This is called a dependent system. There are 3 ways to tell this:

1. When you use the Substitution Method for solving a system of two linear equations, the equation you substituted into will simplify to something like this:

4 = 4, or

-8 = -8, or

0 = 0.

It won’t even let you get to the point where a variable equals a number, revealing that the system is dependent.

Note: When this happens, students tend to think they made a mistake because this outcome seems so awkward, but usually they haven’t made a mistake… it’s supposed to turn out this way to indicate that it’s a dependent system.

2. When reduced to simplest terms and converted to same form, the equations are identical. If you are going to compare equations, they must both be in the same form as each other (either slope-intercept form or standard form).

Note: Often times, these equations may look similar before they are completely simplified. If they are in the same form, you may notice the coefficients are different, yet proportional. This can be a sign that dividing one or both equations through by a certain factor will then reveal the equations to be identical.

This is why it is so important to try to simplify any equation by looking to factor out a GCF and arranging into standard form in the beginning of every problem. Doing this here would instantly reveal that the system is dependent.

3. Graph & Check – Graph both lines and look at the graph. It should be pretty obvious that the lines overlap. Actually, it will just look like one line.