Study Guide

Systems of Equations and Inequalities - Solving Systems of Linear Equalities

Solving Systems of Linear Equalities

When in the course of algebraic events, it becomes necessary for a student to finish their homework, and to solve a system of linear equations, a decent respect to the opinions of their math teacher requires that they should know multiple ways to find the solutions.

…Maybe that's a bit much? We don't want to declare a revolution or anything like that.

Anyway, a system of linear equations is just a set of two or more different linear equations. You'll occasionally be given two linear equations and asked to solve for x and y. We want to know all of the values for x and y that make both equations true. There are numerous ways to do this, including graphing the lines, using substitution, some clever elimination, and guessing.

Just kidding on that last one.

  • The Graphing Method


    When we're asked to solve a system of linear equations for x and y, we're really being asked to find the (x, y) point where these lines meet. Do they meet at (1, 1)? Pittsburgh? The fourth moon of Neptune? How on Earth (or Neptune) do we find the answer?

    We could guess (x, y) values—easy if the answer is (1, 1), but tough for . Besides, we already said we were kidding about guesswork. Let's try graphing instead.

    A Plot for Successful Graphing

    First, find some graph paper—we don't want anyone scratching a grid into the kitchen table. Then graph both lines on the same coordinate grid. When they're both in slope-intercept form, we can do this by plotting the intercepts and then using the slope to find another point. Don't try to intercept the plots. Let the spies do that. It's their job.

    If we graph two equations and both of them happen to be the exact same line, then that system has an infinite number of solutions. That's because one line is exactly on top of the other and both lines will extend, in the words of Buzz Lightyear, "to infinity and beyond." Except there's really no "beyond infinity." You get the idea, though.

    On the other hand, if we graph two equations and they happen to be parallel to each other, there is no intersection. They never meet up. That is the definition of parallel lines, after all. We can plug in numbers for three months and we won't find an intersection, and we'll waste our entire summer.

    Let's not completely diss plugging in numbers, though, because it's a useful method for checking our answers. To see if a point is a solution for a system of equations, simply plug the point into each equation individually and check. The point must satisfy both equations. No exceptions allowed. We're very picky.

    Sample Problem

    Solve the following system of equations.

    y = 2x – 2

    First, plot both lines on the same coordinate grid.

    This seems clear enough: these lines meet at the point (2, 2). However, we have to double-check that it's actually a solution for the system. There's no shortcut to success, after all.

    Did you hear about Billy? He tried to take a shortcut to success, but he took a shortcut into the Forbidden Zone instead. True story. Let's plug (2, 2) into our equations to see if it satisfies them both.

    2 = 2(2) – 2

    2 = 4 – 2

    2 = 2

    Why yes, 2 always does equal itself. Well done.

    2 = -3 + 5

    2 = 2

    We can now confirm that 2 continues to equal itself even now, ten seconds into the future. That means that (2, 2) is a solution to this system of equations.

    Sample Problem

    Solve the system of linear equations:

    3x + 3y = 9

    x + y = 3

    Maybe by graphing? Let's do it by graphing.

    Uh, didn't we start with two equations? Maybe one of them forgot to take the left turn at Albuquerque? Or maybe both of them describe the same line? We'll check that plugging some points into both equations, to see if we get the same results. If that doesn't pan out, we'll ask a wascally wabbit for directions.

    Peeking at the graph, (0, 3), (1, 2), and (3, 0) all look to be on the line:

    3(0) + 3(3) = 9

    0 + 3 = 3

    Check.

    3(1) + 3(2) = 3 + 6 = 9

    1 + 2 = 3

    Double-check.

    3(3) + 3(0) = 9

    3 + 0 = 3

    T-t-t-triple-check. We can confidently say that there are infinitely many solutions to this system of equations. We could keep plugging numbers in, but what's the point? They're the same line. Guess we won't need these carrots after all.

  • The Substitution Method


    Doesn't the word 'substitute' conjure up some sort of image of a teacher with pointy eyeglasses and a ruler?

    Hmm…disregard that, we're talking the substitution method, not substitute teacher. But the substitution part does mean the same thing as the substitute in substitute teacher—replacing something with another thing.

    Sample Problem

    Use the substitution method to solve this system of equations.

    y = -x + 3

    2x + y = 4

    Our first question is, why can't we be friends? Next we ask ourselves, what do we want to substitute? Well, the first equation has been conveniently solved for y already. At any spot where there's a y, we can substitute in (-x + 3). And we know where to stick that.

    "(-x + 3) can't we be friends, (-x + 3) can't we be friends?" Sing along now.

    We suppose it could also go in the second equation too.

    2x + (-x + 3) = 4

    Well, well. That substitution made y disappear, which will let us solve for x. Usually when things disappear, either a magician or large sums of money are involved.

    x + 3 = 4

    x = 1

    Now we can plug 1 in for x in either of the original equations and solve for y. We'll put it in y = -x + 3.

    y = -1 + 3 = 2

    We now have a solution to the system of linear equations: (1, 2). That's where the two lines meet. Or is it? Maybe we should double-check.

    y = -x + 3

    2 = -1 + 3

    Our first equation checks out. The left- and right-hand sides agree.

    2x + y = 4

    2(1) + 2 = 4

    Same here. Both sides of both equations agree, so we're in the clear. We're still wondering why (-x + 3) doesn't want to be our friend, though. And (-x + 3), why won't you return our calls? Was it the surprise water balloon? We already said we were sorry.

    The Substitution Method

    Now that we've done a problem using it, here is the substitution method laid out. Believe us, it's much less scary than that image of a substitute teacher.

    Step 1: Isolate one of the variables in one equation. Make sure the variable knows it's not personal, it's just algebra. We don't like to hurt any variables' feelings.

    Step 2: Substitute the expression from Step 1 into the other equation. Just cram the whole thing right in there. We should now have an equation with only one variable. Solve for it.

    Step 3: Take the answer from Step 2 and substitute it back into either one of the original equations. Solve for the other variable.

    Step 4: Check the solution in both of the original equations. Mistakes will be caught, quarantined, and given a good scrub down at this stage.

    Optional Step 5: Once everything checks out, bask in the warm glow of algebraic achievement.

    That's a lot of words to explain what we're doing, but we don't really need all of them. Let's toss out, say, all but one word for each step: Isolate, Substitute, Solve, Check, Bask. See, it still makes sense.

    Sample Problem

    Solve using the substitution method.

    -2x + y = -1

    -x + y = -3

    Which variable will we stick in the isolation chamber? That y in the second equation looks like it could use some alone time.

    y = x – 3

    Now we can substitute (x – 3) in for y in the first equation. We'd substitute it somewhere else, but that joke is already old.

    -2x + (x – 3) = -1

    -x – 3 = -1

    -x = 2

    x = -2

    We're halfway to a solution to this little problem. All we need is the cooperation of one of our initial equations. Sorry, -2x + y = -1, you've been volunteered. "Conscripted" would be the better word, we suppose.

    -2(-2) + y = -1

    4 + y = -1

    y = -5

    Looks like (-2, -5) is our solution. We're so close to basking in our achievement, but we can't until we check our answer. It would be awful if we basked with the wrong answer. Just the absolute worst.

    -2x + y = -1

    -2(-2) – 5 = -1

    4 – 5 = -1

    This is good. We're already at half bask; we can't help ourselves.

    -x + y = -3

    -(-2) – 5 = -3

    2 – 5 = -3

    Ah, there we go. The achievement washes over us. Excuse us for a moment; we're going to do our best "lizard sitting on a hot rock" impression.

    Sample Problem

    Solve using substitution method.

    -x = 3y + 6

    6y + 2x = -12

    We haven't given x any time in the isolation chamber yet, so let's get it alone in the first equation.

    -x = 3y + 6

    x = -3y – 6

    We now know x's dirty little secret. Don't worry, x, that isn't anything to be ashamed of. In fact, we'd like to introduce you to the x in the other equation; you two have a lot in common.

    6y + 2(-3y – 6) = -12

    6y – 6y – 12 = -12

    0 = 0

    Well, that's different. When all of our variables cancel out and we're left with a true statement like 0 = 0, that means that the lines are actually the same line. They lie on top of each other, so every point on one line is also a solution to the other line. That's an infinite number of solutions. We couldn't count that high, even if you gave us a week.

  • The Elimination Method


    This method for solving systems of equations goes by many names: the addition method, elimination method, or solving by "linear combination." Doesn't that sound like fun?

    It's not that bad, though. Once you get the hang of it, it's amazingly quick. It's quicker than your little brother chasing the ice-cream truck.

    Sample Problem

    Solve this linear system of equations using the elimination method.

    3x – 3y = -1

    2x + 3y = 6

    The secret to the elimination method is that we can add our two equations together top-to-bottom, and if we can solve the new equation for x or y, it will be part of the solution to the original equations. It's not Houdini-level magic, but it's a neat trick nonetheless.

    Looking at our system of equations, we see that we have 3y and -3y. If we add the equations together, the y's will eliminate each other out, leaving us with only x's. The y's will have to sit out until the next round.

    (3x + 2x) + (-3y + 3y) = (-1 + 6)

    5x = 5

    x = 1

    All right, y, you can come back into the game. Now, 1, we want you to take x's spot on the field. Together, try to take on 2x + 3y = 6 and score a few points. Maybe find the true meaning of y while you're at it.

    2(1) + 3y = 6

    3y = 4

    Oh, y was a fraction all along. Our solution is . We need to do a background check on this.

    3x – 3y = -1

    3 – 4 = -1

    So far, so good.

    2x + 3y = 6

    2 + 4 = 6

    So farther, even better.

    The elimination method works great when all of the variables have coefficients attached to them. If those numbers are harder to get off than cling wrap, then just eliminate them. Trying to use substitution would be too much of a hassle. Not that we can stop you from trying.

    Sample Problem

    Solve this linear system using the elimination method.

    x + y = 2

    2x + 3y = 4

    Even in a case like this, where the substitution is practically done for us, we can still use elimination.

    In this case, though, we need to take an extra step first. If we add our equations together now, nothing will be eliminated. That helps no one, except the Advocates for Unsolved Equations.

    We need to multiply one or both of our equations to create some additive inverses. Multiplying x + y = 2 by -2 will do the trick. Now our system of equations looks like this:

    -2x – 2y = -4

    2x + 3y = 4

    Let's get eliminating.

    (-2x + 2x) + (-2y + 3y) = (-4 + 4)

    0 – y = 0

    Solving for y is so easy at this point, that y = 0. Oh, we couldn't even wait until we finished that sentence in order to do it.

    Now we plug y into one of the initial equations and get our answer. Use the originals, not the ones we multiplied by, just in case we made some mistake before.

    x + 0 = 2

    x = 2

    Our solution is (2, 0). But now we check it, because we're super careful with this kind of stuff.

    x + y = 2

    2 + 0 = 2

    That checks out.

    2x + 3y = 4

    2(2) + 3(0) = 4

    And we're golden. Not literally, of course. Otherwise, we would be constantly fighting off people wanting a piece of us. More than usual, anyway.

    Sample Problem

    Solve this linear system of equations using the elimination method.

    y = x + 7

    xy = 8

    We're feeling a little dangerous right now. We're just going to add up each side of these equations right now, as they are. "Oh no, the terms aren't all lined up." Well boo-hoo, we'll sort that out later.

    (y) + (xy) = (x + 7) + (8)

    x = x + 15

    See? It worked out all right in the end. We eliminated y, didn't we? Now let's get all our x's on the same side and solve.

    (xx) = 15

    0 = 15

    Oh no, we take it back, we take it back. Did we make a mistake in our hubris? No, the math checks out, again and again and again and again. Yes, we were so freaked out we checked it four times.

    Wait, this kind of nonsense is what we get when a system of equations has no solutions. The graphs never cross, so they should be parallel. We can check that.

    To graph y = x + 7, we can use the y-intercept and then plug in a value for x to get another point. For the y-intercept, that's b = 7, or (0, 7) since we want to graph it. When x = 1, we have y = 8. This makes the line look like:

    To graph xy = 8, we can find the x- and y-intercepts. For the x-intercept, let y = 0. That means x – 0 = 8, so x = 8. This gives us the point (8, 0) where this line crosses the x-axis. For the y intercept, x = 0. That gives us 0 – y = 8, so y = -8.

    We almost have enough intercepts to film a movie. Interception: You have to graph deeper.

    Just as we said, our system has no solutions because the graphs are parallel. It's nice to know we didn't goof anything up.

    Sample Problem

    Solve using the elimination method.

    3x – 5y = 6

    6x – 10y = 12

    We see immediately that multiplying the first equation by -2 will eliminate x. Go go go for it.

    -2(3x – 5y) = -2(6)

    -6x + 10y = -12

    Let's just add our equations together and—huwawa?

    6x – 10y = 12

    -6x + 10y = -12

    0 = 0

    Surprise! Our lines are secret twins. See, 0 will always be equal to 0. Therefore, there are infinitely many solutions. The lines sit on top of each other, so any point found on one line has to be on the other one too.

    Elimination Summary

    Let's review what to do when we want to use the elimination method. None of the steps require anyone to sleep with the fishes.

    Step 1: Multiply one or both of the equations by a constant so that the coefficients for the same variable in both equations only differ by sign. Like STOP and YIELD? No, silly. Like plus and minus.

    Step 2: Add together the revised equations for Step 1. Combining like terms will eliminate one of the variables. How about that—one variable just drops out. (Don't feel bad, we'll bring it back in the next step.) Now solve for the remaining variable.

    Step 3: Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable. Follow with more achievement-based glow-basking. You've earned it at that point.

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