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 one of the systems from the "solving by graphing" page.)

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- Solve the following system by substitution.

2*x* − 3*y* = −2

4*x* + *y* = 24

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.

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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*:

4*x* + *y* = 24

*y* = −4*x* + 24

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*:

2*x* − 3(−4*x* + 24) = −2

2*x* + 12*x* − 72 = −2

14*x* = 70

*x* = 5

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:

*y* = −4(5) + 24 = −20 + 24 = 4

Then my solution is the following point:

solution: (*x*, *y*) = (5, 4)

There is no particular rule about which equation to plug into. You picked one of the equations and solved it for one of the variables. You can now pick any of the *other* equations that contains that solved-for variable, and plug in for that variable. The idea is to extract some information from the first equation, and then plug its info into one of the other equations, and see where that takes you.

Both the first equation you solve, and the different equation you plug into, are entirely your choice. There can be many correct ways to solve a system by substitution, but the most important thing to remember is to plug into a *different* equation rather than the one you'd started with.

For instance, in the above exercise, if I had substituted my "−4*x* + 24" expression into the same equation as I'd used to solve for "*y*=", I would have gotten a true, but useless, statement:

4*x* + (−4*x* + 24) = 24

4*x* − 4*x* + 24 = 24

24 = 24

Yes, twenty-four does equal twenty-four, but who cares? How does this help? It doesn't, other than to suggest that something might be wrong with your process or assumptions.

So, when using substitution, make sure you substitute from one equation into the *other* equation, or you'll just be wasting your time.

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- Solve the following system by substitution.

*y* = 36 − 9*x*

3*x* + *y*/3 = 12

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:

3*x* + (36 − 9*x*)/3 = 12

3*x* + 12 − 3*x* = 12

12 = 12

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:

solution: *y* = 36 − 9*x*

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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 page) 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 − 9*t*)", 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 − 9*x*".)

- Solve the following system by substitution.

7*x* + 2*y* = 16

−21*x* − 6*y* = 24

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 "2*y*") will divide evenly into the 16.

7*x* + 2*y* = 16

2*y* = −7*x* + 16

*y* = −(7/2)*x* + 8

Now I'll plug this into the other equation:

−21*x* − 6(−(7/2)*x* + 8) = 24

−21*x* + 21*x* − 48 = 24

−48 = 24

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.

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:

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 distinct parallel lines.

A useless result means a dependent system which has a solution (every point on the whole line, rather than just one point); a nonsense result means an inconsistent system which has no solution of any kind. Don't confuse the two!

https://www.purplemath.com/modules/systlin4.htm

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