Settle in, folks, today's a long one.

In The Function Machine, we learned that functions can be depicted as curves graphed on a coordinate plane. In What Does the Function Look Like?, we learned how to tell the general shape of a function's graph based on characteristics of its equation, and vice versa. Today, we'll be focusing on linear equations (meaning any equation that graphs into a straight line).

The defining characteristic of a linear equation is that the highest power of x in the equation is x to the first. This denotes that for every y value, there is exactly one corresponding x value. Of course, there is always exactly one corresponding y value for every x, but this is one of those “square is a rectangle; rectangle is not necessarily a square” moments. We know there's exactly one y for every x because we choose our x's independently and the y's are dependent on them. There can't be more than one y for any given x; you've only got one output slot in your machine. We don't necessarily know that our y values won't repeat, though, unless the function only has x to the first power and no higher.

So let's look at a couple of linear functions. I'll pick two at random.

y = 2x + 3

y = 6x – 5

Now, let's think for a moment about how those two lines might relate to each other. There's only a couple of ways a pair of lines can interact. They could be parallel, meaning that they never intersect. Parallel lines would have the same slope (you can find the slope as the coefficient on the x value in the equations above). These two lines do not have the same slope, so they're not parallel.

If they're not parallel, then by definition, they must intersect. And since they're straight lines, they intersect at exactly one point. Let's think for a moment about how we are able to graph equations in the first place. For a point to be on the line of an equation, that means that when you choose that point's x value as your input, you get that point's y value as your output. The x and y values of that point, plugged in to the equation, would make it true.

Well, if the two equations above intersect, then that means there's exactly one point where the same set of x and y values would make both of these equations true at the same time. That sounds like something that'd be handy to know, huh? Wonder how we could do it...

What we've got here is known as a system of equations. That just means a set of more than one equation that we're looking at together. And finding that intersect point requires what we refer to as “solving the system.” There are a couple of ways to do it – we'll just look at the two most common.

Method 1: Substitution Gone Wild

We've already established that if you have a given value for one of the variables, you can plug that value in for that variable in an equation to find the remaining one. Extrapolate that – if you can substitute individual values for a variable, why couldn't you substitute whole expressions in for variables, if you happened to know an appropriate one? Well, turns out you can. Let's look at our equations again.

y = 2x + 3

y = 6x – 5

Now, the first step to substitution is to solve one of these equations for one variable. They're both already solved for y, but for the sake of demonstration, I'm going to solve the first one for x.

y = 2x + 3

y – 3 = 2x

½ y – 3/2 = x

So now we have x in terms of y. Let's take that newly-found expression and substitute it in for x in the other equation.

y = 6x – 5

y = 6(½ y – 3/2) – 5

Now we have an equation with only one variable in it. THAT we can solve to get a number. So let's do it.

y = 6(½ y – 3/2) – 5

y = 6/2 y – 18/2 – 5

y = 3y – 9 – 5

y = 3y – 14

14 + y = 3y

14 = 2y

7 = y

So y is 7. Now that we have that information, we can plug THAT back in to our first equation to find x. The trick with this method is very simple – ALWAYS SWITCH EQUATIONS! If you accidentally try to plug this number back into the equation you were just using, you'll end up in an infinite loop, going around in circles and getting more and more frustrated. That or you'll cancel all the variables out and end up with 7 = 7, which doesn't help at all. Remember to keep switching back and forth. Back to the first equation we go!

y = 2x + 3

7 = 2x + 3

4 = 2x

2 = x

So the intersect point of these two equations is at (2, 7). That's the value that makes both equations true at the same time.

I like this method for its simplicity and ease of use. But there is another method that's pretty popular, and it all hinges on opposing coefficients.

Method 2: Addition/Elimination/Subtraction method – Opposing Coefficients

This method goes by many names. I prefer the Elimination method, because it tells you what you're doing. The goal with this method is the same as the previous one – get one equation to only have one variable in it. But this time, we're going about it a little differently.

The basic idea of the Elimination method is to figure out a way to get the two equations to have opposite coefficients on one of the variables, so that if you were to add all the terms in those two equations, you'd cancel out one of the variables entirely. It requires us to establish at the outset that a couple of math moves are legal:

1. That we can multiply an entire equation by a single constant, multiplying each term by that constant, and we won't change the value of the equation itself. It's just like creating equivalent fractions back in middle school; as long as you do the same thing to each term, you're creating an equivalent equation that looks different so that we can work with it better.

2. That we can add two equations together, by adding corresponding terms, to create a new equation that will also be true. In practice we'll be adding straight down in columns.

Let's take a look at our equations again.

y = 2x + 3

y = 6x – 5

Okay, I see some potential here. If we were to multiply the first equation by -3, we'd end up with a -6 on the x that would cancel out with the positive 6 in the second equation. Let's try it.

-3(y = 2x + 3) → -3y = -6x – 9, so

In The Function Machine, we learned that functions can be depicted as curves graphed on a coordinate plane. In What Does the Function Look Like?, we learned how to tell the general shape of a function's graph based on characteristics of its equation, and vice versa. Today, we'll be focusing on linear equations (meaning any equation that graphs into a straight line).

The defining characteristic of a linear equation is that the highest power of x in the equation is x to the first. This denotes that for every y value, there is exactly one corresponding x value. Of course, there is always exactly one corresponding y value for every x, but this is one of those “square is a rectangle; rectangle is not necessarily a square” moments. We know there's exactly one y for every x because we choose our x's independently and the y's are dependent on them. There can't be more than one y for any given x; you've only got one output slot in your machine. We don't necessarily know that our y values won't repeat, though, unless the function only has x to the first power and no higher.

So let's look at a couple of linear functions. I'll pick two at random.

y = 2x + 3

y = 6x – 5

Now, let's think for a moment about how those two lines might relate to each other. There's only a couple of ways a pair of lines can interact. They could be parallel, meaning that they never intersect. Parallel lines would have the same slope (you can find the slope as the coefficient on the x value in the equations above). These two lines do not have the same slope, so they're not parallel.

If they're not parallel, then by definition, they must intersect. And since they're straight lines, they intersect at exactly one point. Let's think for a moment about how we are able to graph equations in the first place. For a point to be on the line of an equation, that means that when you choose that point's x value as your input, you get that point's y value as your output. The x and y values of that point, plugged in to the equation, would make it true.

Well, if the two equations above intersect, then that means there's exactly one point where the same set of x and y values would make both of these equations true at the same time. That sounds like something that'd be handy to know, huh? Wonder how we could do it...

What we've got here is known as a system of equations. That just means a set of more than one equation that we're looking at together. And finding that intersect point requires what we refer to as “solving the system.” There are a couple of ways to do it – we'll just look at the two most common.

Method 1: Substitution Gone Wild

We've already established that if you have a given value for one of the variables, you can plug that value in for that variable in an equation to find the remaining one. Extrapolate that – if you can substitute individual values for a variable, why couldn't you substitute whole expressions in for variables, if you happened to know an appropriate one? Well, turns out you can. Let's look at our equations again.

y = 2x + 3

y = 6x – 5

Now, the first step to substitution is to solve one of these equations for one variable. They're both already solved for y, but for the sake of demonstration, I'm going to solve the first one for x.

y = 2x + 3

y – 3 = 2x

½ y – 3/2 = x

So now we have x in terms of y. Let's take that newly-found expression and substitute it in for x in the other equation.

y = 6x – 5

y = 6(½ y – 3/2) – 5

Now we have an equation with only one variable in it. THAT we can solve to get a number. So let's do it.

y = 6(½ y – 3/2) – 5

y = 6/2 y – 18/2 – 5

y = 3y – 9 – 5

y = 3y – 14

14 + y = 3y

14 = 2y

7 = y

So y is 7. Now that we have that information, we can plug THAT back in to our first equation to find x. The trick with this method is very simple – ALWAYS SWITCH EQUATIONS! If you accidentally try to plug this number back into the equation you were just using, you'll end up in an infinite loop, going around in circles and getting more and more frustrated. That or you'll cancel all the variables out and end up with 7 = 7, which doesn't help at all. Remember to keep switching back and forth. Back to the first equation we go!

y = 2x + 3

7 = 2x + 3

4 = 2x

2 = x

So the intersect point of these two equations is at (2, 7). That's the value that makes both equations true at the same time.

I like this method for its simplicity and ease of use. But there is another method that's pretty popular, and it all hinges on opposing coefficients.

Method 2: Addition/Elimination/Subtraction method – Opposing Coefficients

This method goes by many names. I prefer the Elimination method, because it tells you what you're doing. The goal with this method is the same as the previous one – get one equation to only have one variable in it. But this time, we're going about it a little differently.

The basic idea of the Elimination method is to figure out a way to get the two equations to have opposite coefficients on one of the variables, so that if you were to add all the terms in those two equations, you'd cancel out one of the variables entirely. It requires us to establish at the outset that a couple of math moves are legal:

1. That we can multiply an entire equation by a single constant, multiplying each term by that constant, and we won't change the value of the equation itself. It's just like creating equivalent fractions back in middle school; as long as you do the same thing to each term, you're creating an equivalent equation that looks different so that we can work with it better.

2. That we can add two equations together, by adding corresponding terms, to create a new equation that will also be true. In practice we'll be adding straight down in columns.

Let's take a look at our equations again.

y = 2x + 3

y = 6x – 5

Okay, I see some potential here. If we were to multiply the first equation by -3, we'd end up with a -6 on the x that would cancel out with the positive 6 in the second equation. Let's try it.

-3(y = 2x + 3) → -3y = -6x – 9, so

-3y = -6x – 9

+ y = 6x – 5

-2y = 0x – 14

So we now have zero x's – they've cancelled themselves out. Let's solve the remaining equation for y to see what we get.

-2y = 0x – 14

-2y = -14

y = 7

Same as before. That's a good sign! Now, here's the only tricky part. We have to keep switching equations, so we need to know which one we were technically using up there. Which equation were we actually using? Well, the way I think of it is this: we added the second to the first, but we didn't DO anything to the second equation, so that one's still fair game. Think of it as using the second equation as a key to unlock the first one. We haven't actually worked with the key equation yet, just used it to unlock others. Let's plug our new value for y back in to the key equation.

y = 6x – 5

7 = 6x – 5

12 = 6x

2 = x

So our point is (2, 7) – just like with the other method!

To be honest, I'd rather use the substitution method. I think the process is simpler to remember, and with systems of equations, there are so many different ways to get from start to finish that I'd rather have a simple process I can remember than try to keep straight which equation is “active.” Plus, if Calculus is in your future it helps to learn the basic idea of substituting an expression in for a variable now, while it's still relatively simple, so that later when you're writing derivatives that stretch all the way across your notebook page you'll have a more instinctive sense of how to accomplish the substitution part of it.

Elimination also has a tendency to turn into big ugly fractions – notice I chose coefficients for the terms that were both small and factorable? This is a nice, simple example, but math textbooks have a tendency to make problems unnecessarily complicated simply by using ugly numbers.

Speaking of which, stay tuned for next month's Math Journey, when we tackle a big ugly system in three variables!

+ y = 6x – 5

-2y = 0x – 14

So we now have zero x's – they've cancelled themselves out. Let's solve the remaining equation for y to see what we get.

-2y = 0x – 14

-2y = -14

y = 7

Same as before. That's a good sign! Now, here's the only tricky part. We have to keep switching equations, so we need to know which one we were technically using up there. Which equation were we actually using? Well, the way I think of it is this: we added the second to the first, but we didn't DO anything to the second equation, so that one's still fair game. Think of it as using the second equation as a key to unlock the first one. We haven't actually worked with the key equation yet, just used it to unlock others. Let's plug our new value for y back in to the key equation.

y = 6x – 5

7 = 6x – 5

12 = 6x

2 = x

So our point is (2, 7) – just like with the other method!

To be honest, I'd rather use the substitution method. I think the process is simpler to remember, and with systems of equations, there are so many different ways to get from start to finish that I'd rather have a simple process I can remember than try to keep straight which equation is “active.” Plus, if Calculus is in your future it helps to learn the basic idea of substituting an expression in for a variable now, while it's still relatively simple, so that later when you're writing derivatives that stretch all the way across your notebook page you'll have a more instinctive sense of how to accomplish the substitution part of it.

Elimination also has a tendency to turn into big ugly fractions – notice I chose coefficients for the terms that were both small and factorable? This is a nice, simple example, but math textbooks have a tendency to make problems unnecessarily complicated simply by using ugly numbers.

Speaking of which, stay tuned for next month's Math Journey, when we tackle a big ugly system in three variables!

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