Systems of Linear Equations: Solving by Substitution

The instructions tell me to solve "by substitution". This means that I need to solve one of the equations for one of the variables, and plug the result into the other equation in place of the variable I've solved for. It does not matter which equation or which variable I pick. There is no right or wrong choice of equation or variable; 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 of that equation? 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:

Now I'll plug this in (that is, I'll "substitute " this) for y in the first equation, and solve the resulting one-variable equation for the value of x:

Now I can plug this x-value back into either equation, and solve for the corresponding value of y. But since I already have an expression for "y =", it will be simplest to just plug into this:

Then my solution is the following point:

4x + (−4x + 24) = 24
4x − 4x + 24 = 24
24 = 24

We already know (from the previous page) that these equations are actually both the same line; that is, that this is a dependent system. We know what this looks like graphically: we get two identical line equations, and we get a graph that displays just the one line. But what does this look like algebraically?

The first equation is already solved for y, so I'll substitute that into the second equation:

Well, um... yes, twelve does equal twelve, but so what?

I did substitute the result from 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 other line.

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:

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.

Now I'll plug this into the other equation:

Um... I don't think that's right....

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. But 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 page, 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.