The Purplemath Forums
of Linear Equations: Solving by
The method of solving "by substitution" works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable. Here is how it works. (I'll use the same systems as were in a previous page.)
2x – 3y
The idea here is to solve one of the equations for one of the variables, and plug this into the other equation. It does not matter which equation or which variable you pick. There is no right or wrong choice; the answer will be the same, regardless. But — some choices may be better than others.
For instance, in this case, can you see that it would probably be simplest to solve the second equation for "y =", since there is already a y floating around loose in the middle there? I could solve the first equation for either variable, but I'd get fractions, and solving the second equation for x would also give me fractions. It wouldn't be "wrong" to make a different choice, but it would probably be more difficult. Being lazy, I'll solve the second equation for y:
4x + y = 24
Now I'll plug this in ("substitute it") for "y" in the first equation, and solve for x:
2x – 3(–4x + 24) = –2
Now I can plug this x-value back into either equation, and solve for y. But since I already have an expression for "y =", it will be simplest to just plug into this:
y = –4(5) + 24 = –20 + 24 = 4
Then the solution is (x, y) = (5, 4).
Warning: If I had substituted my "–4x + 24" expression into the same equation as I'd used to solve for "y =", I would have gotten a true, but useless, statement:
4x + (–4x + 24) = 24
Twenty-four does equal twenty-four, but who cares? So when using substitution, make sure you substitute into the other equation, or you'll just be wasting your time.
y = 36 – 9x
We already know (from the previous lesson) that these equations are actually both the same line; that is, this is a dependent system. We know what this looks like graphically: we get two identical line equations, and a graph with just one line displayed. But what does this look like algebraically?
The first equation is already solved for y, so I'll substitute that into the second equation:
3x + (36 – 9x)/3 = 12
Well, um... yes, twelve does equal twelve, but so what?
I did substitute the first equation into the second
equation, so this unhelpful result is not because of some screw-up on my part. It's just that
this is what a dependent system looks like when you try to find a solution. Remember that, when
you're trying to solve a system, you're trying to use the second equation to narrow down the
choices of points on the first equation. You're trying to find the one single point that works
in both equations. But in a dependent system, the "second" equation is really just
another copy of the first equation, and all the points on the one line will work in the
In other words, I got an unhelpful result because the second line equation didn't tell me anything new. This tells me that the system is actually dependent, and that the solution is the whole line:
solution: y = 36 – 9x
This is always true, by the way. When you try to solve a system and you get a statement like "12 = 12" or "0 = 0" — something that's true, but unhelpful (I mean, duh!, of course twelve equals twelve!) — then you have a dependent system. We already knew, from the previous lesson, that this system was dependent, but now you know what the algebra looks like.
(Keep in mind that your text may format the answer to look something like "(t, 36 – 9t)", or something similar, using some variable, some "parameter", other than "x". But this "parametrized" form of the solution means the exact same thing as "the solution is the line y = 36 – 9x".)
+ 2y = 16
Neither of these equations is particularly easier than the other for solving. I'll get fractions, no matter which equation and which variable I choose. So, um... I guess I'll take the first equation, and I'll solve it for, um, y, because at least the 2 (from the "2y") will divide evenly into the 16.
7x + 2y = 16
Now I'll plug this into the other equation:
–21x – 6(–( 7/2 )x
+ 8) = 24
Um... I don't think so....
In this case, I got a nonsense result. All my math was right, but I got an obviously wrong answer. So what happened?
Keep in mind that, when solving, you're trying to find where the lines intersect. What if they don't intersect? Then you're going to get some kind of wrong answer when you assume that there is a solution (as I did when I tried to find that solution). We knew, from the previous lesson, that this system represents two parallel lines. But I tried, by substitution, to find the intersection point anyway. And I got a "garbage" result. Since there wasn't any intersection point, my attempt led to utter nonsense.
solution: no solution (inconsistent system)
This is always true, by the way. When you get a nonsense result, this is the algebraic indication that the system of equations is inconsistent.
Note that this is quite different from the previous example. Warning: A true-but-useless result (like "12 = 12") is quite different from a nonsense "garbage" result (like "–48 = 24"), just as two identical lines are quite different from two parallel lines. Don't confuse the two. A useless result means a dependent system which has a solution (the whole line); a nonsense result means an inconsistent system which has no solution of any kind.