Sherpa In this video, we take a look at linear equations and how to solve them using BIDMAS (Brackets, Indices, Division, Multiplication, Addition and Subtraction) and why it is so important to follow the golden rule of linear equations "Always do the SAME thing to both sides of the equation".

So we are looking at linear equations. We're going to look at what are they? So what do we mean by linear? We're going to look at bid Mass and how it affects the order in which we do things. And finally, we're going to see how the rule works with these things.

So we always have to do the same thing to both sides, and we're going to look at why we do that and what it means. So first of all, what are they? Well, the linear in linear equations, basically it refers to the highest power of x that is in our equation. So as an example, let's say we've got five x plus four equals 49. Well, in this case, we've got an x on its own, right?

There's no x squared or x cubed. We could think of this as x power of one. Anything from power of one is just itself. And so we could add a power of one there. It doesn't change anything, but it's just it allows us to think of this as in terms of powers, right?

So we would say this x has a power of one. And because it's got a power of one, because there are no x squares or x cubes or anything higher than that, we would say this is a linear equation. As an example, contrary to this, if we had say, I don't know, three x squared plus four x minus seven equals twelve. Well, in this case, because we've got an X squared here, we've got x to the power of two. The fact that we've got a power of two in our equation means that this equation is no longer linear, even though we've got x to the power of one in the equation.

That is trumped by the fact that we've got an X squared or an x to the power of two. And so we'd actually say this equation is quadratic.

You may have heard of the quadratic formula. That's why we call it a quadratic formula, is because it applies to quadratic equations. The equation part, the equation just means anything with an equal sign, right? So it's anything where we have to solve for a particular value for x and there are only certain values of x that will satisfy that. Right?

Bid Mass. The easiest way to understand how Bid Mass relates to solving linear equations is probably by thinking of a fairly simple example of a number machine, right? Let's go back to this simple example. We've got five x plus four equals 49.

And we know from Bid Mass that we do multiplication first and then we do addition afterwards, right? So we would do five times x and then we take that whole result and then we'd add it on to four. Well, if we think of this in terms of a number machine, we start off with x, we times x by five, and then we take that result and then we add fall onto it, and that then gives us the final result of 49. Well, that's how we can represent this equation in a kind of number machine. The tricky thing is, though, we want to find out what X is, right?

We want to go back down the opposite way. We want to kind of fight against the current, almost. We want to undo everything we've just done. And so we're kind of starting off with 49, we want to go back down the other way and we want to reverse this adding four. What's the opposite of adding four?

We know we have to minus four and then we take that result and we have to do the opposite of times by five. What's the opposite of timing by five? It's going to be dividing by five and that then undoes that times by five. And then once we do those two things, we should get back to where we started, right? We should get back to that value for X.

And whatever value we have here is going to be our value for X. We've kind of inadvertently solved the equation. We'll look at this in a couple of examples in the next slide. So that's been massive. That's how it relates to equations.

Finally, the rule. So the rule is we will do the same thing to both sides. And this becomes very obvious when we actually look at a very simple example, right? So we could do, say, I don't know, five x equals ten, right? Well, we know from our five times tables, right?

Five times what gives us ten. We know five times two gives us ten, right? We know X has to be two. What if we were to do the proper math way of solving this? Well, I want to get X on its own, right?

I want to end up with X equals something. And to do that, I need to therefore get rid of this five. And just like with our number machine, I always do the opposite. To do the reverse, I want to get rid of this five. And we can imagine that we're actually doing five times next year.

Well, the opposite of times by five is dividing by five, right? So I divide that side by five. And the whole reason I do that is to then cancel out that times by five. So if I cancel that out, I might just put it in red. Actually, if I cancelled out that times by five, the only thing I'm left with is going to be x, right?

I'm doing X times five divided by five. And obviously, if I make it five times bigger, then five times smaller, I get back to my original X. Well, if I don't do the same thing to the right hand side, I end up thinking that x equals ten, right? Which we know isn't true. We know the solution to this equation is x has to equal two so I have to do the same thing to both sides.

I have to divide this side by five as well. And if I do that, I do ten divided by five or ten divided by five is going to give me two. And again, we get our answer. So that's why we have that rule. We always do the same thing to both sides, right?

Let's look at a couple of examples. So, first of all, let's do question eight. So we want to solve five brackets. X minus six equals 65. Well, another way to think about this is we want to work our way in towards X, right?

Just like on the number machine, we work our way back towards X. We do the same thing here, right? So we kind of start off outside, further away from X, and then we move in and then we take care of the 6th and then we end up at X. Again, we're doing the opposite to bid mass, because with bid mass, we would take care of the minus six first and then times that whole thing by five. When we're doing the reverse, when we're working our way back towards X, we do the opposite of Bitmaster, we start outside the brackets.

We do that first and then we work inside the brackets at the very end. So the first thing we do is we take care of this five. Well, if I write this out slightly bigger, we're doing five times X minus six equals 65. And if I want to get rid of this five times, here what's the upset of X five. The opposite of times in by five is dividing by five.

And I do the same thing to both sides because it's a rule that we just covered. And the whole reason I do that is to cancel out that timing by five. So all I'm left with on the left hand side is going to be X minus six. On the right hand side, I've got to be 65 divided by five. Well, that's going to give me 13.

And the final thing we need to do is we need to get rid of this minus six, right? I want to end up at the end of the day, I want to do a load of maths and then I want to end up with X equals something, and that thing in that box is going to be my answer. So how do I get rid of that minus six? Well, what's the opposite of minus six? The opposite of minus six is plus six.

So I plastic both sides and again, the reason I do that is to cancel out that minus six. And so all I'm left with on the right hand side, on the left hand side is going to be X. Well, we also have to do plus six to the right hand side, we do 13 plus six and we get 19, and that is our answer. We kind of inadvertently arrived at our answer just by setting up these very simple rules. We get X equals 19 as our answer, right?

Final question, question 20. In this case, it's not quite as obvious, right? Which fully above question, we're knew that this left hand side was going to be the side that we wanted all the X's to be on, right? We said this is going to be on X side. We said we want to end up with X equals something.

Well, in the case of question 20, it's not as obvious, right? We've got X's on both sides and so do we want to get X equals something or do we want to get something equals X? Both are going to be valid, but we want to decide now what we're going to be aiming for. And I think because we've got more X's on the right hand side already, we've got five here and three here, we're going to say this is our X side. So if we want to get all the X's onto this side well, at the moment I've got a few exes over on the left hand side, so I want to get rid of this three X floating around here.

Well, to get rid of that three X, at the moment this three X is adding on to twelve, right? So what's the opposite of adding on three X? I just subtract three x So I subtract three X from both sides and I know therefore that's going to cancel out with that three X and I'm going to end up with just twelve on the left hand side and two X on the right hand side. And I saw that plus four lingering there. I can also try and get rid of this plus four, right?

Because I want to get at the end of the day, if this is my ex side, I want to end up with doing a load of maths and eventually ending up with something equals X. And so I need to get rid of this plus four. How do I get rid of that? Well, what's the opposite of plus four? It's going to be minus four, so I minus four from both sides.

Twelve minus four is going to give me eight. And again, the reason we do that is to cancel out that plus four. So we're left with eight equals two X. The final thing I need to do is get rid of two. Well, what's the two doing to X?

The two is timing the X. You can imagine there's a very small time sign in between the two and the X the opposite of times by T is going to be dividing by T, and so I can divide both sides by T and that's going to give me four on the left hand side. Again, the reason we do that is to cancel out that times in II on the right hand side and so we're left with four equals x. And that is my answer. That is the end of linear equations.

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