Let's explore the basics of inequalities - what they are and how to solve them.
Inequalities show the relationship between two expressions that are not equal. They are useful when looking to project profits and breakeven figures.
We look at examples on a number line and see how the solution is found by moving along the line until we reach a point where the inequality symbol is valid.
We are looking at inequalities and we're going to focus on three subtopics. First of all, we're going to look at number line representation. Secondly, we're going to look at solving inequalities, so finding certain values for x which would satisfy this kind of inequality equation. And thirdly, we're going to look at graphs and how we can represent inequalities as regions on the graphs. So let's start off with the number line representation.
So I've got a simple example here. First of all, I'm going to draw my number line out. So I put zero in the middle. We're going to have four up top and minus three down the bottom. And one way to remember how we actually draw these inequalities is this inequality here.
They're greater than or equal to inequality. Takes much longer to say than just that inequality, just the greater than inequality. And so when I've got this extra line beneath my inequality because it takes me longer to say, I know it's going to take me longer to draw that circle. And so I have to draw the circle and fit it in. And that's a quick cheat way of remembering which inequality is which.
So just this inequality here, because it takes me a very short amount of time to say, it just takes me half a second to say less than I know that it's only going to take me half a second to draw that circle.
That is our inequality. Simple as that. Hardest thing is just knowing if we're doing a filled in circle or an empty circle and it's just if it takes longer to say or shorter to say, let's go and do solving inequalities. So this is almost exactly the same as just solving regular equations. So we want to get x on its own.
First of all, I'm going to add two x to both sides. We're going to end up with 14 is greater than or equal to eight plus two x. Let's get rid of this eight now. So I'm going to subtract eight from both sides. We're going to end up with six is greater than or equal to two x.
And finally we're going to divide both sides by two. So three is greater than or equal to x. Or we could think of this as x is less than or equal to three. Either one is either one is correct. There's one caveat to this.
So there's one thing, there's one kind of trap that people often fall into. So I'm going to copy this out down here and I'm going to show you first of all what the trap is and then how we can avoid it. So if we start off at the beginning again, this time I'm not going to do the same method, I'm going to do a slightly different method. So I'm going to decide to have all my x's on the left hand side rather than on the right hand side. In that case, I'm going to subtract 14 from both sides, so I'm going to end up with two.
X is greater than or equal to minus six, and then I'm going to divide both sides by t, so minus x is greater than or equal to minus three. And finally I'm going to times both sides by minus one or divide both sides by minus one. Either way is the exact same. The effect that it has is obviously just to cancel those minuses out because minus minus gives us a plus and we get x is greater than or equal to three. Well, we can see instantly that these two solutions are not the same, right?
We can't have x is less than or equal to three and x is greater than or equal to three. And so we know we've slipped up somewhere. And the mistake that we've made, the trap that we've fallen into, is when we times by minus one, times by minus one, we have to remember to flip this sign around. And so whenever we times by negative number, we always flip the sign around and that allows the inequality to stay true.
Once we do that flip, we can see again we end up with the exact same.
I just realised I've not flipped the sign, I just kept the sign where it is. We flip the sign and we end up with the same inequality. So basically, those two are the same. Finally, we'll look at subtopic number three graphs. So this is a way of kind of graphically representing an inequality.
So I'm using the same example as in the first sub topic in number lines. So this time we're going to actually put it onto a graph. So an x y graph. So we've got x here and y here, and we're told that x is less than four, so that's what four here and x is greater, so they're not equal to minus three. So I'm going to have minus three down here.
So it's very similar to the number line version. Well, I know that when I've got my number inversion, greater than or equals to means I have a filled in circle. And just like that, very similar to that, I have a filled in line that goes through my minus three. Well, because x is less than four, just like my empty or my kind of half circle, I've got a dotted line or a line with half of itself missing. And that inner region between these two lines is going to represent the valid region where x will be true for this in a quality.
Right, let's go ahead and do some example questions. So we've got three questions, one for each topic. So first of all, we're going to look at showing in the quality on a number line. So we've got an inequality, x is greater than minus three. Well, because greater than doesn't take us very long to say.
We know it's going to be an empty circle, right? It doesn't take us long to draw the circle if this was greater than or equal to, we know it will take us longer to draw the circle and therefore we would fill in that circle. This is not different what we did in the first example, right, because we don't have this second bit, we don't have anything here. And so what we do in that situation is we just draw an arrow. So I know that x is greater than minus three.
So anything above minus three is going to be valid for this inequality right? Question b solve the inequality of seven y -34 is less than or equal to eight and so just like with a regular equation I can rearrange this. I can add 34 to both sides. And so I get seven y is less than eight 4242. And we're going to be testing our seven times tables.
We divide both sides by seven and we end up with y is less than or equal to six and so that's my final answer. Finally, we want to look at Shading regions to represent inequalities, and in this instance, we bought three different inequalities. And so first of all we got Y is greater than minus four. Well, minus four is down here. I'm looking at the why variable.
So I know I'm going to be concerned with blocking off a full region either above or below minus four. And so anything above minus four is going to satisfy that inequality. Next up, your x is less than two, and so I'm going to be blocking off a horizontal region.
Anything above this line and anything to the left of this line is at the moment going to satisfy these two inequalities. Finally, why is less than two x plus one? Well, if we imagine drawing a line of Y equals two x plus one, it's going to make this much easier to understand, right? So if I just plot Y equals two x plus one, well, I know it intersects at one, right? That's my I've got kind of Y equals MX plus C, right?
And therefore that one is going to be my Y intercept. So it goes through one, and it's got a gradient of T. And so if I go along one, I have to go up T, I go along one, I go up T and I go down back one, and I go down to and we start to build up our straight line to that equation. We end up with a line that looks a little bit like this. Well, we're told that Y doesn't equal this line, y has to be less than this line.
And so anything below this line is again going to be valid. And so what we're left with is this small region between these lines, where it's to the left of this vertical line, it's above the horizontal line, and it's below this diagonal line. And that's where we can represent inequalities as regents. And that brings to a close inequalities.
Book a lesson with this tutor
A young, gifted tutor who uses his engineering experience to ignite your curiosity in Maths. A straight 'A' student, he knows exactly how you need to learn to help you ace your exams.