System of Equations

Question

Solve the system of Equations. 

 

X+y+z=3

x-y+2z=9

5x+y+z=15

Answer 1

I love solving systems of equations because it's actually quite a freeform task, as long as you understand the goal and the general steps to get there. There are a few different methods taught, but I'm partial to the 'substitution' method – because it's what I learned in high school, it feels simpler to me, and it applies easily in more situations than the 'elimination' or 'addition' method. So let's approach this one with substitution.

The basic idea behind the substitution method is this: we can leverage the transitive property of equality learned way back in algebra 1. Remember 'if a = b and b = c, then a = c'? Well, in this method we're using that concept in the context of 'if we can identify one variable in terms of the others, we can substitute that resulting expression in for the variable in other equations in the system.' I always remind students that if we can get an equation down to just one variable type, we can solve it to get a number. If we have more than one type of variable (x's and y's, for example), we can only get it as far as one variable in terms of the other. For example, we might be able to get to x = y + 4, but no further. If we want to figure out what numbers correspond to x, y, and z in our problem, we'll first have to narrow it down to one variable so we can solve and get a number.

In broad terms, our strategy for solving systems using this method is to eliminate variables by figuring out what they are in terms of the others, then substituting those expressions in to remove that variable from the problem entirely. Once we're down to one variable, we solve that equation to find its numerical value, then begin working backward substituting that number in to find the other variables. Since we have three variables, we'll have to do the first step twice to get it all the way down to one variable before we can start finding solutions. The freeform part comes in here, because it doesn't really matter which variable you start with or what order you complete them in, as long as you're following the general strategy you'll get the correct answer.

The important thing to remember when working the substitution method is to ALWAYS SWITCH EQUATIONS. I'll explain what I mean by that as we go. So, let's finally – finally – take a look at our problem:

Solve the System:

x + y + z = 3

x – y + 2z = 9

5x + y + z = 15

So, our first step is to solve one of these equations for one of the variables. We can choose any equation, and any variable to solve it for. It'll work regardless, but I always suggest looking at your equations and finding the one that'll be the simplest to solve for a variable. In this case, I'd say it's probably simplest to start with the first equation, since the coefficients are simplest. Mark that equation so you remember you've used it, and let's go:

x + y + z = 3

Now, we could solve for x, y, or z, but I've chosen x because it's the first one up.

X + y + z = 3

x = 3 – y – z

Now we have an expression to represent x. Let's switch to one of the other equations (ALWAYS SWITCH EQUATIONS!) and get to work. I'm going to choose the second equation, since at this point they're all about equal difficulty and going straight down the list will be easier to keep track of. Mark that equation so you know you've used it.

X – y + 2z = 9

Well, we know what x is, right? It's 3 – y – z. So plug that in and let's go!

X – y + 2z = 9

(3 – y – z) – y + 2z = 9

3 – 2y + z = 9

–2y + z = 6

z = 6 + 2y

Now we have an expression to represent z as well. So let's turn to the third equation – the one we haven't worked with yet – and plug both of those in to solve for y! ALWAYS SWITCH EQUATIONS!

5x + y + z = 15

5(3 – y – (6 + 2y)) + y + 6 + 2y = 15

Notice we had to substitute our z expression in twice, since there was a z in our expression for x. But it's okay, because we've switched to an unused equation (more on that below). Oh, and don't forget to distribute the negative 1 through that parentheses!

5(3 – y – 6 – 2y) + y + 6 + 2y = 15

15 – 5y – 30 – 10y + y + 6 + 2y = 15

–9 – 12y = 15

- 12y = 24

y = -2

Progress! We now have the first phase of our solution: y is -2. Put a box around that last statement so you don't forget it. Now we begin phase 2: working backward to figure out the rest of the variables.

Again, make sure you switch equations here. Let's just go back to the top of the list again:

x + y + z = 3

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

x – 2 + 6 – 4 = 3

x = 3

Two down, one to go! X is 3. Put a box around that statement too, switch equations again, and follow me down the home stretch!

X – y + 2z = 9

We know both x and y now, so all we need to do is bring it home:

3 - (-2) + 2z = 9

3 + 2 + 2z = 9

5 + 2z = 9

2z = 4

z = 2

Put a box around that last statement, just for good measure. Hooray! We're done! The coordinate point of the solution to this system is (3, -2, 2).

Now what about that ALWAYS SWITCH EQUATIONS business?

The reason you want to be very careful to switch equations each time you start a new step is that if you use an equation to solve for a variable, and then plug the expression back in for that variable in the same equation right away, you wind up in an infinite loop. The variables cancel each other out and you wind up with something like 9 = 9, or even worse, 0 = 0, and those things are true, but profoundly unhelpful. It's like when you try to define a vocabulary word, but the only definition you can come up with uses the word itself – not helpful and not getting you any closer to solving the system. This can be quite frustrating for students who may find themselves going around in circles and not understanding why. This is why I kept saying you should be marking which equation you were working with as you started each step – it helps remind you to switch to a new equation each time.