Study Guide

Systems of Linear Equations

For most of this unit we'll be concerned with linear systems of equations, or systems of equations in which all the equations are linear.

These systems are nice because we can graph each equation as a line. A solution to a system of linear equations is a point which lies on all lines in the system. Funny...we just got off the phone with a telemarketer who was also lying on the line.

We'll mostly be looking at systems that have only two equations, meaning we're looking at only two lines at a time. We're starting off simple because we don't want you to get dizzy. We love watching people fall down as much as the next website, but we don't want it to happen to you.

There are three things that can happen when we graph two lines.

1. The lines may intersect exactly once.

2.The lines may be parallel (have the same slope) and not intersect at all.

3. The lines may be the same and intersect at every point.

A solution to a system of two linear equations is a point that lies on both lines at once. If the lines intersect exactly once, the system has one solution. If the lines don't intersect at all, the system has no solutions. If the lines are the exact same line, the system has infinitely many solutions. If the lines decline comment and refuse to tell us how many times they intersect, they're probably hiding something. We should launch a full-scale investigation.

These are the only possibilities for a system of two linear equations: we either have 1, 0, or infinitely many solutions, with the solutions being points. Note that it's a different story if we're able to use curved lines, but here we're only dealing with linear equations. Curved lines have not been invited to this party. They put a hole in the living room wall last time they were over, and they won't be invited back.

To solve a system of equations means to determine if the system has 1, 0, or infinitely many solutions, and to find the solution(s) if they exist. If they don't exist, you can write a short sci-fi story about them.

There are three main ways to solve a system of linear equations:

  1. Graphing
  2. Substitution
  3. Addition

Be careful: Whichever method you use to solve a system of equations, check your answer to make sure it really is a solution to the system of equations. Some decoys will masquerade as correct solutions only to gain and exploit your trust. Until you've double-checked, don't trust any solution further than you can throw it. Which, considering that we're dealing with abstract mathematical concepts and not tangible objects, isn't far.

  • Solving Systems of Linear Equations by Graphing

    The solution(s) to a system of linear equations are all the point(s) where the lines intersect. To solve a system by graphing, we graph the lines and see where they meet up. Are they grabbing a couple of lattes at Starbucks, or is this more of a public park rendezvous?

    Of course, if we want to get the right answer, we need to draw the graphs carefully. Now is not the time to challenge yourself by attempting to draw the graph with your non-dominant hand.

    Sample Problem

    Solve the following system of equations by graphing:

    If we graph these equations, we get this picture:

    We see that the lines intersect at (1, 1). Let's make sure this is actually a solution to the system of equations. First, we check the values in the first equation:

    y = x
    1 = 1

    Yup, the values x = 1 and y = 1 are definitely solutions to this equation.

    Now, let's check the values in the second equation:

    y = 2 – x
    1 = 2 – 1

    When x = 1 and y = 1, the left-hand side of this equation is 1 and the right-hand side is 2 – 1, which is also 1. It's true; you can use your fingers to check our math.

    The point (1, 1) is a solution to each equation in the system. In other words, the point (1, 1) is on both lines. This means (1, 1) is a solution to the system of equations. It is also a common final soccer game score.

    Sample Problem

    Solve the following system of linear equations:

    4x + 3y = 24
    y = (-4/3)x + 8

    First we graph the equation 4x + 3y = 24 by looking at the intercepts:

    Now we graph the equation y = (-4/3)x + 8, which also has a y-intercept of 8 and a slope of -4/3:

    Somebody's a copycat. The second line we graphed landed right on top of the first one. The first line wasn't ready for it, either. It jumped a mile. It was great.

    What the presence of overlapping lines means is that the two lines are actually the same and intersect at every point along the line. These two are always shaking hands and slow dancing. If they're being honest, it's getting a little old. What, they can't groove to something up-tempo every once in a while?

    All points on the line y = (-4/3)x + 8 are solutions to this system of equations.

    We could also say all points on the line 4x + 3y = 24 are solutions, since that's a different way to write the same line. Two ways to write the same line...a screenwriter with writer's block would be envious.

  • Solving Systems of Linear Equations by Substitution

    To solve linear systems by substitution, we solve one equation for one variable and then use that information to solve the other equation for the other variable. It's exactly the same as when a basketball team makes a substitution, except with less basketball and more math.

    Let's do a couple of examples and see what happens.

    Sample Problem

    Solve this system:

    The first thing we need to do has already been done: the first equation has been solved for y. Don't you love it when someone's already come by and done the work for you? Shmoop Algebra: we're a river to our people.

    We know that

    y = 6x – 4,

    so we can substitute (6x – 4) for y in the second equation:

    Now we can solve the new equation for x. Start by subtracting 3x from both sides:

    6x – 4 = 3x + 5
    3x – 4 = 5

    Then add 4 to both sides and divide by 3:

    3x = 9
    x
    = 3

    Since a solution to a system of linear equations is a point, we need to know what y is. Until we know y, all we have is half a point, and it's difficult to win an argument with one of those.

    To find y, we take our value for x, stick it into either equation we like, and solve for y. If x = 3 and y = 6x – 4, then

    We think the point (3, 14) is the answer. To confirm this, we need to make sure this point satisfies both of the original equations. If it fails either test, we can toss it out with yesterday's garbage. Hope it likes day-old sushi.

    Is the point (3, 14) on the line y = 6x – 4?

    When x = 3 and y = 14, the right-hand side of this equation is

    6(3) – 4 = 18 – 4 = 14,

    which agrees with the left-hand side of the equation. The point (3, 14) is on the first line.

    Is the point (3, 14) on the line y = 3x + 5?

    When x = 3 and y = 14, the right-hand side of this equation is

    3(3) + 5 = 14

    which agrees with the left-hand side of the equation. The point (3, 14) is on the second line.

    Since the point (3, 14) is indeed on both lines, it's the solution to the system of equations and the answer to all our dreams. Well, except for that one dream where our hands are giant meatballs. We still don't have an answer for that one.

    A Summary, So Far

    We've now used substitution to successfully find the point of intersection for two lines that intersect exactly once. Let's tidy things up a bit and figure out the general steps we need to take for this sort of problem. Once we're done, we should also tidy up the living room. It's great that you wanted to build a fort out of the couch cushions, but people have to live here.

    1. Solve one equation for one variable.
       
    2. In the other equation, perform substitution to get rid of the variable we solved for in step 1.
       
    3. After substituting, solve the other equation.
       
    4. Find the value of the variable we solved for in step 1.
       
    5. Check that the answer works in both original equations.

    Sample Problem

    Solve the system of equations:

    1. Solve one equation for one variable.

    The first equation has x all by itself (with a coefficient of 1), so it's easiest to solve that equation for x. Here we go:

    x = 7 – 4y

    2. In the other equation, perform substitution to get rid of the variable we solved for in step 1.

    The other equation is 2y – 3x = 8.

    Performing substitution gives us 2y – 3(7 – 4y) = 8.

    3. After substituting, solve the other equation.

    We need to solve 2y – 3(7 – 4y ) = 8.

    Simplify that thing to find 2y – 21 + 12y = 8.

    Simplify a bit more to get 14y = 29, and divide by 14 to track down y:

    Ugh, we're left with a fraction. However, it's the best we can do in this instance. Let's try to overlook our dislike of fractions, though, and make the most of a bad situation. Where are you from, fraction? Oh, really? Well did you...okay, we can't do this. We tried.

    4. Find the value of the variable we solved for in step 1.

    We know that x = 7 – 4y, so plug in .

    Another fraction. A negative one this time. Oh joy.

    5. Check that the answer works in both original equations.

    We think the answer is .

    Oy, we almost hope we're wrong. Do these values work in the equation x + 4y = 7?

    When  and , the left-hand side of the equation is

    which is indeed 7.

    Do these values work in the equation 2y – 3x = 8?

    Let's see if 2y – 3x really does equal 8 for these bizarro values of x and y.

    How about that; it actually worked! There may be a place for fractions in the universe after all.

    We were right. The answer is .

    So far, each of the systems we've solved using substitution has had exactly one answer, but a system of equations could have no solutions or infinitely many solutions. How's that for a wide range of options? Just somewhere between "none" and "infinity," that's all.

    Sample Problem

    Solve this system:

    Let's do substitution. All the cool kids are doing it.

    1. Solve one equation for one variable.

    The first equation is already solved for y, which makes our lives better.

    2. In the other equation, perform substitution to get rid of the variable we solved for in step 1.

    We substitute for y in the equation 2yx = 8 to get:

    3. After substituting, solve the other equation.

    To solve , first we simplify to find x + 2 – x = 8.

    Then we run into trouble. Even if "Trouble" is your middle name, you're not going to like what comes next. When we combine the x terms, we're left with the statement 2 = 8.

    Uh-oh. We know 2 doesn't equal 8, or else the Raptors got royally ripped off by the official scorers at last night's game.

    What does this incorrect equation tell us? There's no solution to the system. For these lines to intersect, 2 must equal 8, which is ridiculous. Our calculator agrees.

    Whenever substitution leads us to such a ridiculous and impossible statement, it means the system of equations has no solution, which leads us to an important life lesson: being ridiculous and impossible never solves anything.

    On the other hand, if substitution leads us to a statement that's always true, such as 1 = 1, it means that the lines are actually the same, and every point on either line is a solution. Everyone is happy, and order is restored. Now the only thing that's ridiculous is how much we're enjoying solving substitution problems.

    If one or both of the equations in a system contains fractions, we get rid of the fractions and then proceed as usual. We know you like this news. It's actually a little scary how much you enjoy getting rid of fractions; we only hope it doesn't give way to more destructive behavior, such as torturing polynomials.

    Sample Problem

    Solve the system of equations:

    Since each equation has fractions, we'll get rid of the fractions first. Multiply both sides of the first equation by 2:

    6y = x + 2

    And multiply both sides of the second equation by 4:

    8xy = 4

    Now, instead of solving the original system of equations, we can solve the system

    We know how to do this: we just follow the same steps we've been following. They have to lead somewhere.

    1. Solve one equation for one variable.

    Since x has a coefficient of 1 in the first equation, we'll solve the first equation for x to get:

    x = 6y – 2

    2. In the other equation, perform substitution to get rid of the variable we solved for in step 1.

    We replace the x in the second equation:

    3. After substituting, solve the other equation.

    We need to solve the equation 8(6y – 2) – y = 4.

    Simplify to find 48y – 16 – y = 4, and rearrange to get .

    4. Find the value of the variable we solved for in step 1.

    Since we found that x = 6y – 2 when , we get

    We're not thrilled about these fractions, but we also weren't thrilled after seeing the previews for Adam Sandler's Jack and Jill, and look how great that turned out! (Psych. It was awful.)

    5. Check that the answer works in both original equations.

    We think the answer is .

    Let's make sure these values work in the original equations—you know, the ones with the fractions in them. First, let's check that we have a solution to the equation .

    When we substitute in the values we found for x and y, the left-hand side of this equation is

    .

    The right-hand side of the equation is

    Since the left-hand and right-hand sides of the equation agree, we have a solution. It gives us even more confidence in our solution to know that four out of five dentists also agree.

    Second, let's check the equation .

    When we substitute in the values we found for x and y, the left-hand side of this equation is

    which agrees with the right-hand side of the equation. Houston, we also have a solution to the second equation.

    We can safely say that the solution to the system of equations is .

    When solving systems of equations that have fractions in them, it's best to check the answers in the original equations. Although it's tempting to check the answers in the nicer equations, what if we made a mistake when getting rid of the fractions? Almost inconceivable, we know, yet possible. Then we'd be finding the right solutions for the wrong equations, which wouldn't help us any more than if we were to be in the right place at the wrong time. For example, at the Kodak Theatre four months before the Oscars, or in Central Park at 4 a.m.

  • Solving Systems of Linear Equations by Addition

    When we have two equations, we can add them together to get a new equation. We don't even need to ask for their permission. Let's say we're trying to solve the following system:

    a = b
    c
    = d

    We can add the left-hand sides of the equations and the right-hand sides of the equations to get:

    a + c = b + d

    When we do this, we say we're adding the equations. Come to think of it, that term may be too self-explanatory for bolding. Oh well, too late now.

    We can also add more complicated equations. If we've got this system:

    3x + 2y = 7
    y
    = 3x – 2

    ...then we can add 'em up vertically to get:

    3x + 2y + y = 7 + 3x – 2

    We'll use this concept to solve systems of linear equations.

    Sample Problem

    Solve the system of linear equations:

    We add the equations to find that

    y + xy = 3x + 2 + 4.

    This simplifies to x = 3x + 6.

    All of a sudden the variable y is gone. We don't know where it went, but hopefully it's in a better place. We now have an equation we can solve for x. Doing this yields x = -3. Now we can put -3 in for x in the first equation to find y:

    y = 3(-3) + 2 = -7

    The solution to the system appears to be (-3, -7). Let's check this answer in the original equations. We want to check it before we wreck it.

    For the first equation, when x = -3 and y = -7, the left-hand side is -7 and the right-hand side is 3(-3) + 2, which also happens to be -7. These values work in the first equation.

    For the second equation, when x = -3 and y = -7, the left-hand side of the equation is xy = (-3) – (-7), which is 4, the same as the right-hand side of the equation.

    These values work in both the equations, so the solution really is (-3, -7).

    The previous example worked because the the coefficients of y in the two different equations were additive inverses of each other. The first equation had a positive y, and the second equation had a negative y.

    When we added the equations, we eliminated the variable y. Now y sleeps with the fishes.

    This example was one of the kindest ones we could have given you, because the coefficients of y in the two equations were already additive inverses. In general, we need to do much more legwork to get coefficients that are additive inverses. Once we're done with our legwork, we can focus on our glutes.

    Sample Problem

    Solve the system of equations:

    The coefficients of x aren't additive inverses, but we can be sneaky and mold them into what we want them to be. Take the first equation and multiply each side by -2. This gives us

    -2(x + 5y) = -2(11),

    which simplifies to

    -2x – 10y = -22.

    Write this on top of the second equation:

    Now the coefficients of x are additive inverses of each other, so we can work our system-solving magic. Nothing up our sleeves here...

    Add the equations to eliminate the x:

    -10y – 3y = -18

    Then solve to get:

    Now we can stick that value in for y in the first original equation to find x:

    We think the solution to the system of equations is .

    As always, we're going to cover our hineys and double-check this in the original equations. For the first equation, when and , the left-hand side is

    which is 11, exactly as it should be.

    For the second equation, when  and  the left-hand side is

    This agrees with the right-hand side of the second equation, so we've got it. Man, being right never gets old.

    When we solve a system by addition, we could also say we're solving by elimination. We eliminate one variable, find the value of the other, and then find the value of the variable we eliminated. It's like the NFL playoffs. These variables are one and done.

    The general steps we're using here are similar to the ones we used with the substitution method.

    1. Eliminate a variable to get an equation in one variable.
       
    2. Solve the equation from step 1.
       
    3. Use one of the original equations to find the value of the other variable.
       
    4. Check your answer in both original equations.

    So far, the first step ("Eliminate a variable'') hasn't been too difficult. We've only needed to multiply one equation in the system by a number in order to eliminate a variable. Now it's time to up the ante. Of course, we don't mean "ante" in the poker sense. We can't condone gambling, unless it's on the stock market.

    We frequently need to multiply both equations by something in order to eliminate a variable.

    Sample Problem

    Eliminate the variable x from the following system of equations to get an equation in terms of y.

    The coefficients on the x are 3 and 2, which have a common multiple of 6. If we multiply the first equation by 2 and the second equation by 3, we're almost there:

    Now the coefficients on x in the two equations are the same, which is close, but no cigar. Again, we're not referring to a literal cigar. We can't condone smoking.

    We want our coefficients to be additive inverses, so we can multiply either equation by -1. In other words, we need to flip all the signs in one equation. We could flip a coin to determine which one we should choose, but how do we decide which coin to flip?

    We'll make things easy on ourselves and go chronologically. Emphasis on "logically." If we pick the first equation, we have

    Now we add the equations to find -7y = 7, and the x has been eliminated.

    There's one other thing that can happen: our linear system can contain fractions. When this happens, we just get rid of the fractions first and then solve the system as we've been doing. If they don't go quietly, don't be afraid to get a bit rough. Show those fractions who's boss.

    Sample Problem

    Solve the system of equations:

    First, we get rid of the fractions. Multiply the first equation through by 6 and the second equation through by 5 to find

    We know what to do from here. In this case, we can eliminate either variable without too much trouble, assuming they don't get frisky. Let's eliminate y, since its coefficients already have opposite signs. Thank you, y, for cooperating.

    Multiply the second equation by 2, so now we're looking at the system of equations

    We add these equations to find 33x = 26, which means, unfortunately, that .

    Use the first original equation to find y. We plug in our value of x into , which gives us

    .

    This simplifies to

    ,

    and then we multiply both sides by the reciprocal of y's coefficient to get

    Now we think the solution to the system is , but we need to check our answer in the other equation, which was .

    Let's plug in the values we found for x and y and see if they work. If they don't, we'll try to keep from throwing a hissy fit. Look at things through a clean pair of perspectacles.

    Yep, that's the same thing as 2. These values work, so we found the solution to the system of equations. Yeesh, finally.