Algebra problems can relate to any area of maths. Problems often include a mix of algebra, numbers and geometry. A framework can be used to tackle these problems. Most algebraic problems require one or two steps in order to arrive at the correct answer. Most of the difficult problems involve forming equations and solving them before being able to use the answer in some way.

Algebra is one of the biggest topics in math and one which students are going to be coming across the most throughout the mathematical careers. Algebra really does take the essence of maths in that it essentially distils down all information to variables which you can form an equal sign out of. So whilst we're going to be going over the same problem solving framework I outlined in my last video, which you can see a link up in the description, we're not going to be really going through this problem solving framework as much as we're going to be applying it pretty much right away and seeing how it's most relevant to algebra. So to begin, as with all questions and maths, you got to read the question and we're still going to be focusing on this idea of highlighting important information and crossing off as it's used. But we'll of10 see in algebra in particular is that unless you've got a very complex problem, usually going to be using most pieces of information fairly immediately in a question.

One other thing about it is that you're going to usually need to use information which isn't necessarily on the paper. For example, something like the area of a circle is pi r squared or the sum of all angles in a triangle is 180 degrees. These are things which you're just going to have to learn to practise and actually understanding the maths you're doing. Unfortunately, there's no real way to cheat it. What do I need to do?

Is the second part. This is usually the most important part generally when you're doing algebra is if you've got a question that says show this result or find this out, your end result is going to have to be that and you're going to have to figure out what techniques to use along the way. In more complicated problems, there might be more steps. But in general this is the part where you need to really understand what is your final result. Because only when by knowing that, do you know what you need to do.

What don't I need this one's a little bit more insidious when it comes to algebra problems because although the question doesn't have as much information in it as, say, something like a literacy based question, it's still going to have quite a few things that you're going to need to interpret. So for example, like I said earlier, the sum of all angles in a triangle is 180 degrees. It might not be therefore immediately apparent what information is going to be useless until you're actually part of the way through a question. For example, you might not have actually needed some of all angles and triangles 180 degrees and says you might have needed to know about equilateral triangles instead. And they all have an angle of 60 degrees instead.

And it's only really by understanding the question, do you know which piece of information is useful which isn't? fourth, what math technique should I use again? We've been over this already, which is just a case of really selecting which kind of techniques. If you have a piece of information you've yet to use, more than likely you're probably going to need to use it at some stage. So if you ever feel stuck while doing an algebra question, try to find what you should use.

Is usually your first step forward. Is this answer correct? Which is usually fairly easy to do in algebra because algebra is very easy to reverse engineer. Just substitute answers back into the beginning of the question, see if you get the opening information. Once again, see if that's anything similar.

And lastly, have I done everything? Same as before. If you've been asked to find something which is a measurement of distance, have you made sure to put metres in front of it? If you're asked to find an angle, have you put degrees in front of it? Things like that. Let's jump in and fairly immediately try and actually use these techniques and solving why I'd say it's a fairly typical algebra question.

So here we have the perimeter of this rectangle is 46 centimetres. Find the area of the rectangle. This is far more common for an algebra style question will be like, we can highlight the information. We've been told the perimeter of this rectangle is 46 centimetres. So obviously, we can see fairly plainly that this is a rectangle.

So fortunately, we can cheque that visually and then to find the area of the rectangle. So one thing which isn't even mentioned in this question is the fact that you need to know how to calculate the area of a triangle and indeed the perimeter, and indeed the perimeter of a rectangle, as well as the area of a rectangle. My apologies. So to find the area of a rectangle, that means we have to do the multiple of both of these values here, these 3 x's. Okay?

So 3 x minus 2 and 2 x plus 5. And we also need to know what the perimeter is. And the perimeter is the entire distance if you were to go along the outside of the rectangle. So you need to know these pieces of information before you even get into the question. That's what I mean when it comes to the practise and slowly building up your mathematical skills.

So let's try and make an expression, because, as I say, the final thing we need to do is work out what the area is. And now the area in this case we know is going to be equal to 2 x plus 5 multiplied by 3 x minus 2. But we don't know what x is at the minute. Now, if we wanted to, we could immediately simplify this here and multiply it out, or we could leave it till later. Now, I'm going to leave this till later, just because it could turn out that one of these values is something really simple, which could then make the math easier, just turn this into two numbers.

But you could also expand this out. Now, that's up to pure personal preference. We're going to build our way up to that, though, because that's our final answer. But we need to figure out what else we're going to need to do. So, first of all, information we need, everything is fairly immediately already there.

So in this case, the perimeter of the rectangle is going to be two lots of 2 x plus 5 plus two lots of 3 x minus 2. And so once we have that, we know that's going to equal 46 centimetres, and we can immediately cancel out the information that we've used in the question already. We know that you use the perimeter, we know that this is a rectangle, we know we use 46 centimetres. We're not going to need to solve this algebraically. And this is a fairly common one line equation.

We've got one unknown, which is x, which means that we can solve it with the rest of the information that we have here. So we're going to do that now. So 2 times 2 x plus 5. And this is where some of the skills are. Multiplying at brackets comes in useful, and that's something which you can find in other videos.

But here, we're just going to multiply it out to be 4 x plus 10 plus multiplying two different bits, 6 x minus 4. And then that means we can combine these. So we've got 10 x plus 6 equals 46 centimetres. So then we know that 10 x equals 40 centimetres. So we know that x equals 4 centimetres.

Okay, now, we can do this at the end if we want to, but I'm just going to show right now how we could then use this just to double cheque with the right answer. Because we know that if x equals 4 centimetres, then this top bit is going to be 2 x plus 5. So it's going to be 2 times 4, which is 8 plus 5. So that's 13 centimetres up there. And then down here, we're going to have 3 x minus 2.

And then that means we're going to have 3 multiplied by x. In this case, that's going to be 12 minus 2, which is then going to be 10. Then if we figure out the perimeter of this is going to be 2 times 10, which is 20 plus 2 times 13, which is 26. 20 plus 26 equals 46, and that is the perimeter. So we know these are good, reliable answers.

Now, to find the area which, as we said before, these two numbers multiplied together, if we wanted to, we can multiply these outright and turn them into a series of squares, or we could substitute and do each bracket individually. I'm going to do each bracket individually, just because it minimises the chances of errors. But again, if your personal preference is to multiply out, go ahead and do so. If we substitute in for x, we've now got 2 times 4 plus 5, which, as we found out above, was 13 centimetres, so that shouldn't be any surprise. Multiplied by 3 x minus 2.

So it's going to be 3 times 4 minus 2, that's going to be 10. So we have 13 times 10, and that means that our final answer is 130. So we already checked our answer for perimeter and we can cheque our answer for the area. Does this sound reasonable? Well, I'd say so.

We've got two digit numbers multiplying together, so we're probably going to end up with either something like a 3 or 4 digit number. And in this case, we have two very small two digit numbers. So 130 is reasonable. Is there anything else we have to add to this? I would say that we do, because we're asked to find an area, and areas have units.

In this case, we're told the perimeter is 46 centimetres, so these are actually 13 centimetres and 10 centimetres. That part matters less because we're not being judged on those, but when we multiply those two guys together, we're going to make 130 centimetres squared. Okay, there's our final answer. And we could then double cheque this if we want to, and we would do that with perimeter, so it's less necessary for the area. That shows you how you'd find this out for a fairly standard one line algebraic issue, trying to turn it into an equal sign and then using techniques to break it down.

Let's try this again for a trigonometry question. In this case, find the value of Y. So, as I alluded to earlier, knowing things like the sum of all angles in a shape is just information you're expected to know. So we're going to need to turn this into an equal sign at some point. So we're going to need Y equals something degrees.

Okay? So we're going to need to set up a couple of equations here, because we've got one piece of information not sure about, and this is where you just expected to know certain piece of information. So, for example, we know for a fact that y plus 2 x minus 3 should equal 180 degrees. Okay? So we can fill that in now.

So we know that Y plus 2 x minus 3 equals 180 degrees. Okay. And this is also A degrees. So what that means is in order to find out what Y is, we're going to need to just find out what x is and then substitute from there. So to find out what x is, we need to use another piece of information.

And this is where it's just assumed pieces of information you're expected to know. So we're thinking about what from the question, have we not yet used? And that's these other angles here. So that hasn't been even said in the question what the value of these angles are, we have to read it off. So if we highlight the important information, is it these things here and we've used 2 x minus 3, but we're actually going to have to use it again, so although we would have ticked off 2 x minus 3, we're going to kick it off again when we use all 3 of these pieces of info.

When we add these together, we get 4 x plus 29 plus 5 x plus 11 plus 2 x minus 3 should all add up to equaling 180 deg. Okay. Now fun fact, because of the equation that we made previously, we know that this here is also going to equal why? So that's another potential simultaneous equation we could have set up. But less interest in that for now.

By adding all of these together, we end up getting 11 x and then we've got 29 plus 11. So it's going to be 40 minus 3. So that's going to be plus 37 degrees equals 180 degrees. This one looks like it's not going to be as nice of a number immediately. So we get 11 x is then going to be equal to 180 -37 that's going to be 143 deg okay.

Now we can then divide this by 11 right away if we want to. Or we could leave it in this form and then substitute it in later. But we'll do that now just because it leaves us a nice value of x here, even if we are going to have to multiply it a little bit. So it's equal to 143 divided by 11 you can either do this in a calculator or think about what it's going to look like for 11 times 11, usually follows the pattern of being 11, 22, 33 and things like that. But when you reach 11 squared, that's 121.

And this is one of those quick tricks that you can pick up when you get enough practice. And we know that is actually going to equal 13 degrees. Okay. And then from here it's a case of substituting into this lower value here and we'll cheque our value of x in just a moment and see if it cheques out. But actually we're 13 degrees.

We've got Y plus 2 times 13 minus 3 equals 180. So that means we've got Y plus 26 minus 3 equals 180. So y plus 23 equals 180 subtract from both sides. So we get the Y is equal to 157 degrees. So one thing which I've immediately added at the end there, not even thinking I probably should have stopped to mention this, but this is the habits that you should get into is I've added degrees right away.

So that is the form of our answer. Does 157 degrees look reasonable? Well, we're expecting a relatively large angle here because clearly from the diagram, we're going to have a situation where Y is going to be obtuse greater than 90 degrees. So that does seem like a reasonable value. One last thing we could do to cheque our values is to substitute in this value of X to see if that makes up 180 degrees when put in here.

And if we put in that do 13 times 4 plus 29 plus 5 times 13 plus 11 plus 2 times 13 minus 3, that all does come out to equal 180 degrees. And I encourage you guys to give that a try at home to recheck it. But that is a fairly longwinded solution. This gives you an idea of the kind of problem solving you'll be doing in algebra. Both between this and the previous problem.

It's far less about extracting all the information from questions such as having a really big toolbox at your disposal of information you learn through practise, and then understanding what to use, setting up an equal sign and solving it from there. Next time we're going to be speaking about statistical problems, but otherwise, hopefully you guys have learned something. Any questions, please put them in the comments below. And if you want to see any more of my videos, then please just have a click or ask me any questions, in which case, hopefully I will see you guys soon. Thank you.

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An enthusiastic young teacher who uses his technical wizardry to create fun & engaging lessons on our whiteboard. Lewis can help students of any age or SEN tackle the mysteries of maths.

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