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Three Equations with Three Unknowns

Now we’ll talk about the three equations with three unknown case. The very first thing I’ll say is that this is an awfully rare scenario. It’s unlikely, unless you’re doing very well, that you’re gonna be handed something like this but. It could show up. So it’s worth while talking about this.

So, suppose we have this particular system. We have three equations and three unknowns. Now of course, I haven’t given a question here. But let’s just assume that for whatever reason, we need to solve for the individual values of all three variables. Well, a few thoughts first of all.

Big idea number one, if the number of equations is fewer than the number of variables, you will not be able to solve for the individual values of the variables. If you have two variables and only one equation, you can’t solve for those individual values. If you have three variables and two equations, you can’t solve.

In order to solve for individual values, you need to have three equations with those three variables. Big idea number two is quite subtle. Depending on what the question asks, you may not have to solve all the way. Now this is tricky. Sometimes the question will say what’s the value of X?

What’s the value of Y? They’ll want to know a specific variable. Sometimes though they’ll ask about an expression. What’s the value of X plus Y? What’s the value of X minus Z? What’s the value of X plus Y plus Z?

Something like that. Often, when the test does that, there’s a much shorter way to solve directly for the value of the expression without solving for the individual values of the variables. So always keep your antenna up for these kinds of shortcuts. If you can go directly to the answer without doing all the work of solving all the equations all the way, by all means, do so.

Finally, big idea number three. Inside every big problem is a little problem struggling to get out. Now, of course, this is kind of a playful way to say it. But it turns out in mathematics, much of mathematic problem solving consists of reducing problems you don’t know how to do to problems you do know how to do.

Now this is true not only at the level of the test, but even at the level say of calculus, and beyond calculus analysis, topology, even all the way up to PhD levels in mathematics. The way that new math gets done at the PhD level, is that mathematicians are reducing problems that we, the human race, doesn’t know how to do, reducing them to problems we do know how to do.

So it turns out this is a very sophisticated strategy. And it’s, it’s a very important perspective in your problem solving abilities on the test. Now in this particular case, I’ll show you what I mean. So here we have three equations with three unknowns and I’ve labeled the equations. Well, theoretically, you’ve watched these videos and so you know how to solve two equations with two unknowns, that we’ve covered already.

So now we wanna know how to do this. So what we’re gonna do is take a series of steps that will allow us to change three variables with three unknowns into two variables with two unknowns because that’s something we know how to do. Well, first thing that catches my eye, notice we have a plus X and a minus X. That really just kind of jumps out of me.

It’s kind of a screaming sign as far as I’m concerned, in these equations. What that means is I could add these two equations and eliminate an X. So I’m going to add B and C. B is 2W plus X minus 4Y equals negative 14. C is 3W minus X plus 2Y equals eight. Add them up, add everything straight down, we get 5W minus 2Y equals negative six.

I’m going to call this equation D. This is now one equation that has two unknowns. So by itself it’s not very useful, but if we could get another equation with the same two unknowns, then we’d be in business. So we wanna pick another pair of the three original equations. A different pair.

So not B and C this time. A different pair, and see if we can eliminate X again in this new pair. Because if we eliminate X again, we’ll get another equation that has just W and Y in it. So again, this plus X look, looks like it would be pretty easy to cancel it with this minus 2X.

If I just multiply B by two, then that should be able to cancel them. So I’m just gonna have A by itself. A is W minus 2X plus 3Y equals 13. I’m gonna multiple two times B. So everything on both sides of the equation gets multiplied by two. Of course, you’re always allowed to multiply both sides of an equation by the same thing.

That’s perfectly legitimate. That will give me 4W plus 2X minus 8Y equals negative 28. All right. So very good. So now, again, add straight down, I get 5W, the X’s cancel as predicted, I get negative 5Y equals negative 15.

Now ordinarily, I would say G, all of these are multiples of five, wouldn’t it be easy to divide by five and get smaller numbers then that might make it very easy to cancel, to substitute to do all kinds of things. But I’m not gonna that only because I’m gonna call this equation E as is, only because they noticed I have a 5W here, and a 5W here.

Wow. Those are kind of screaming out to me now. Can I just solve for 5W for both of these equations? 5W equals 2Y minus six. 5W equals 5Y minus 15. So, if these two things both equal 5W, they must be equal to each other.

So this is an example of substitution. We’re gonna set this equal. 5Y minus 15 equals 2Y minus six. Well now I’m down to one equation with one unknown. This is great. I know how to solve this.

So subtract 2Y from both sides, we get 3Y minus 15 equals negative six. Add 15 to both sides, 3Y equals nine divide by three, we get Y equals three. Okay. And of course we know, with the two equation two unknown case, once we get the value of one variable, we just plug it in. We could plug it into this one or this one, it doesn’t matter.

This one looks like it has smaller values. So I’ll just plug it in there. 5W equals 2Y which is six minus six equals zero. That’s funny. If 5Y equals zero, 5W equals zero, that means that W must equal zero. Okay.

Very interesting. That can happen sometimes. W equals zero. So now, I have the value of W, the value of Y. I’m almost done but I need the value of X. So now I actually have to go back to one of the original equations.

Let’s take the top one, A. W that’s zero minus 2X plus 3Y. So three times three is nine. Equals 13. Subtract nine from both sides. I get negative 2X equals four or X equals negative two.

And those are the three values, of the three variables from these equations. Again, notice that the strategy that I used, was to reduce the three equation, three unknown scenario, to a two equation, two unknown scenario. So just to summarize. First of all, I’m gonna pick two of the three equations using either substitution or elimination, eliminate one variable.

Incidentally, 99% of the time it’s gonna be much easier to use elimination rather than substitution. Then I’m gonna pick another pair of those original equation, and similarly eliminate the same variable. So at this point, I will be down to the two variable with two equation case. Now just use the two equations, the two unknown techniques, we’ve learned those already.

Solve for those two variables. When you have the numerical values of those two variables, finally plug into any of the original equations to find the value of the third variable. This is exactly how you solve this scenario