Lesson video

Hello, Mr. Robson here, always a wise choice to join me for maths and you've made that wise choice today, well done.

Talking of wisdom, we're going to increase our mathematical wisdom today.

We're going to look at what it means to relate an algebraic solution to a graphical solution when solving inequalities.

I'm sure we can solve inequalities algebraically, but if we can make the link between our algebraic method and what we see on a graph for those inequalities, we'll become more powerful mathematicians.

The more links we can make between the mathematics we learn, the wiser we become.

So let's get going, let's get wise.

Our learning outcome is that we'll be able to solve inequalities graphically and relate this to solving algebraically.

Lots of keywords we're gonna hear throughout today's learning and inequality is used to show that one expression may not be equal to another.

A region is an area that graphically represents a solution to one or more inequalities.

Every coordinate pair within the region satisfies the inequalities that define the region.

Look out for those words throughout our learning today.

Two parts of this learning, let's begin with solving graphically.

We can solve this inequality, 3x plus four is less than 10.

Lots of ways we could do this, I'm gonna start by adding negative four to both sides of the inequality and then once we're at the position where we have 3x is less than six, I'm gonna divide both sides of the inequality by three and x is less than two.

We've solved the inequality algebraically and our solution is x is less than two.

We can graph this solution, it's that region there.

That's the region x is less than two.

Same inequality, same solution, same graph of that region.

When we compare this solution set to our original inequality, we notice something interesting.

That's the line y equals 3x plus four.

Are you spotting the connection yet? The inequality 3x plus four is less than 10, the solution x is less than two and what's represented on that graph there? Can you spot the link? Every point in the region x is less than two satisfies the inequality 3x plus four is less than 10 everywhere in this region.

Quick check you've got that.

Solve the inequality algebraically and compare your solution set to the graph.

Pause and do that now.

Welcome back, lots of ways we could solve the inequality but they will all lead us to the same solution.

I'm gonna add positive three to both sides.

From 2x is greater than four, I'll divide both sides by two.

I get the solution x is greater than two.

Comparing the algebraic solution to a graph, you might have said the graph shows that every point in the region x is greater than two satisfies the inequality, that line y equals 2x minus three.

We can see that it's greater than one everywhere in that region.

Graphing your solution can also help you validate your solution or spot an error.

Let's have a look at an example of what I mean there.

Seven minus 4x is less than negative five.

Lots of ways we could solve this.

I can add negative seven to both sides.

Negative 4x is now less than negative 12, well, we'll just divide both sides by negative four and the solution is x is less than three or is it? We can graph the solution that's the region x is less than three and the line y equals seven minus 4x.

Does the graph validate our solution or does it give us something extra to think about? Pause and have a think about that.

Welcome back, did you spot it? The graph confirms we've made an error.

It's not true that seven minus 4x is less than negative five in this region.

In fact, quite the opposite.

We divided by a negative without changing the sign when we solved algebraically, that's where our error came in.

Let's try that step again, from negative 4x is less than negative 12.

If we divide both sides of the inequality by negative four, we have to change the sign, x is greater than three.

This region, x is greater than three satisfies our inequality.

Graphing our solution enabled us to spot an error, reflect on our maths and try again to reach the right solution.

Spotting errors, understanding them and trying again is an important part of the learning process in maths and graphing our solution was part of that in this topic.

Quick check you've got that.

Does the graph validate the algebraic solution to this inequality? Pause, check on that algebraic solution, compare it to the graph, is everything okay here? Are we happy with that solution? See you in a moment.

Welcome back.

You might have said the graph confirms we've made an error.

At no point in the region x is greater than one is the inequality satisfied.

We need to spot the error and try again.

I'm sure you can spot where the error occurred, well done.

We divided by a negative without changing the sign.

By graphing this, we can see that the actual region should have been x is greater than one.

We can solve an inequality with unknowns on both sides.

2x plus one is less than 3x minus seven.

Lots of ways we could do this one.

I'm gonna start by adding negative 2x to both sides.

From here, I'll add positive seven to both sides and then x must be greater than eight.

We can graph this solution, that's the region x is greater than eight.

Same inequality, same solution, same region graphed.

When we graph the lines, y equals 2x plus one and y equals 3x minus seven, we notice something very interesting about our solution.

There are those two lines.

Can you see what I mean when I say something very interesting about a solution? Well spotted, the point of intersection.

The point of intersection are the two lines i.

e when x is equal to eight is the moment our inequality starts to become satisfied.

The inequality essentially asks us when is the line y equals 2x plus one below the line y equals 3x minus seven? The point of intersection is a vital bit of information if we want to answer that question.

We can see on the graph that y equals 2x plus one is above here in this moment, they are equal, those two lines at the point of intersection and then it's in this region that y equals 2x plus one is below y equals 3x minus seven, hence the region x is greater than eight is our solution to this inequality.

Quick check you've got that.

I'd like you to fill in the blanks.

There's a keyword missing and then there's a key symbol covered up.

Pause, what's going in those two spaces? Welcome back.

The sentence should read; the lines y equals x plus one and y equals 2x minus three intersect when x equals four.

Hence the solution to the inequality, x plus one is greater than 2x minus three is the region x is less than four.

We can use a graph to come up with algebraic solutions.

On a graph, we've got three linear equations, y equals 2x plus two, y equals five minus x and y equals x minus one.

We can use those linear equations to solve the inequality 2x plus two is greater than x minus one.

Essentially we're being asked when is the line y equals 2x plus two greater than y equals x minus one? Let's look for a point of intersection, there it is, they're equal at the intersection when x equals negative three and then y equals 2x plus two is greater here, i.

e those y values are higher on that line than they are on the line y equals x minus one, so our solution is x is greater than negative three.

That's that region there.

Everywhere in that region, the line 2x plus two is above the line x minus one.

One more example from me and I'll ask you to have a go at the problem.

I'm gonna use the graph to solve the inequality 2x plus two is less than five minus x.

Essentially I'm being asked when is the line y equals 2x plus two less than the line y equals five minus x.

I need the point of intersection.

Those two lines are equal when x equals one, so is my solution x is less than one, is it x is greater than one? If I look at the graph, I can see that the line y equals 2x plus two is below the line y equals five minus x in this region hence the solution is x is less than one.

Everywhere in that region, the line y equals 2x plus two is below the line y equals five minus x.

Your turn now, which of these is the solution to x minus one is greater than five minus x? You'll use the graph to pick the right one here.

Pause, see you in a moment, let's go through the answer.

Welcome back, hopefully you said it's option C.

The solution to x minus one is greater than five minus x is the region x is greater than three, that region there.

We can see in that region that line x minus one is above the line five minus x.

Practise time now, question one part a, I'd like you to solve algebraically the inequality 2x minus five is greater than negative one.

For part b, I'd like you to draw a graph to show why your solution to part a is correct.

You might choose to write a sentence to justify why your graph shows such a thing.

Pause and do this now.

Question two, I'd like you to draw a graph to show why this solution is incorrect.

I've solved the inequality five minus 2x is less than one, can't have a solution x is less than two.

You'll use a graph to show why that is not correct.

I'd like you to write at least one sentence to support your answer to this one.

Pause, try this problem now.

Question three, I'd like you to use the graphs to solve these inequalities.

On the graph there, we have linear equations, y equals one minus x, y equals 2x plus one and y equals six minus 3x.

Those lines are all you need to solve all four of those inequalities, pause and do this now.

Welcome back, feedback time, let's see how we did.

Solving algebraically the inequality 2x minus five is greater than negative one, we should have got to 2x is greater than four, hence x is greater than two.

Drawing a graph to show why the solution to part a is correct would look something like that.

You want the linear equation, y equals 2x minus five graphed and you want the region x is greater than two graphed.

And then you might have written a supporting sentence saying we can see that in the region x is greater than two, 2x minus five is always greater than negative one.

Question two, you were drawing a graph to show why this solution is incorrect.

Your graph would've looked like so, that's the line y equals five minus 2x and the region x is less than two.

I asked you to write a sentence to support this.

You might have written the graph shows that at no point in the region x is less than two is the inequality satisfied.

You might have gone on to say the inequality is actually satisfied in the region x is greater than two.

Question three, we were using the graphs to solve the inequalities.

For a, one minus x is less than two, you should have said x is greater than negative one, that's in that region there.

For part B, the solution was x is greater than zero as is shown in that region there.

For c, x is less than one, that's that region.

For d, x is less than or equal to 2.

5, that region on the graph and it's a solid line because x equals 2.

5 is included in this region.

Part two of the lesson now, we're going to look at double inequalities.

We can solve this double inequality.

First, we need to solve 2x is less than 3x plus one.

I can add negative 2x to both sides, add negative one and I've got x is greater than negative one.

Next, we need to solve 3x plus one is less than 13, so 3x must be less than 12, x must be less than four.

We can write that as one solution set.

Negative one is less than x is less than four.

We can graph this solution and look like so, that is the region where negative one is less than x is less than four.

That's the same inequality, the same solution and the same region.

When we compare this solution set to our original inequality, we notice something interesting.

Have you spotted the link? Every point in the region satisfies the inequality.

From there, we can see that this is the region where 2x is less than 3x plus one.

The line y equals 2x is below the line 3x plus one there, but also this is where 3x plus one is less than 13.

Quick check, you've got this.

I'd like you to solve this double inequality algebraically.

Pause and do this now.

Welcome back.

You should have started by solving negative x is less than 2x plus nine, so negative nine must be less than 3x, x must be greater than negative three.

Next, solving 2x plus nine is less than 11, 2x must be less than two, x must be less than one.

Writing that as one solution set, negative three is less than x is less than one.

Next, that's the same double inequality, the same solution you came up with.

I'd like you to explain how this graph relates to your solution.

You might choose to explain this verbally to somebody around you, you might choose to write a few sentences down.

Pause and do either of those now.

Welcome back.

I wonder what you said, I wonder what you wrote.

Hopefully, you said something like the graph shows that in the region negative three is less than x is less than one, the line negative x is less than 2x plus nine, we can see that there and 2x plus nine is less than 11, we can see that there.

Understanding this link between our algebraic solution to double inequalities and our graphical solution means that we can read the algebraic solution to a double equality from a graph.

We're gonna start by looking for intersections.

This moment is where the line y equals negative one intersect line y equals 2x plus nine and this moment is where y equals 2x plus nine intersects y equals x plus six.

Our solution is the region negative five is less than x is less than negative three as graphed there.

We can see that 2x plus nine is greater than negative one in this region and 2x plus nine is less than x plus six in this region.

Quick check, you can do that.

I'd like to use the graph to solve the inequality, negative one is less than 3x plus eight is less than x plus six.

Pause and use the graph to solve the inequality now.

Welcome back.

Hopefully you started with the intersections, that's where y equals negative one intersects y equals 3x plus eight.

That is the intersection of 3x plus eight with x plus six, so our solution must be this region, negative three is less than x, is less than negative one.

Practise time now.

Question one, I'd like you to solve algebraically this double inequality.

For part b, I'd like you to draw a graph to show why your solution to part A is correct.

Pause and try this one now.

For question two, we're gonna use the graph to solve these inequalities.

Part a and part b are double inequalities, for c, we're gonna do something slightly different.

You're going to explain why there is no solution to the inequality 10 minus 2x is less than x plus one is less than negative two, the graph will enable you to do that.

Pause, have a go at these problems now.

Feedback time, let's see how we did.

Question one part a was solving this double inequality.

We should have solved those two inequalities and found that x is greater than three and also x is less than six and we should have written that as one solution set, x is greater than three and less than six.

Drawing a graph to show why this solution is correct would have looked like so.

That's the linear equations y equals six minus x, y equals 2x minus three and y equals nine.

We can see in that region that 2x minus three is greater than six minus x plus being less than nine.

Question two, we're using the graph to solve these inequalities.

For a, we should have got x is greater than negative three and less than three, that's that region there.

We can see in that region that x plus one is above negative two but below 10 minus 2x.

For part b, we should have got three is less than x is less than six, that's that region there.

In that region, we can see that 10 minus 2x is above negative two but below x plus one.

For part c, I asked you to do something different.

Explain why there's no solution to this inequality.

Lots of ways you could have done this, you might have noticed this; 10 minus 2x is less than x plus one.

Well that would be that region there as demonstrated by the arrow that I've drawn in the graph.

And then you might have said in the region where x plus one is greater than 10 minus 2x, neither will ever have a value less than negative two.

That's why this inequality has no solution.

That's the end of the lesson now, sadly, but we have learned, we've learned that we can solve inequalities algebraically and understand how this relates to the graphical solution.

We've also learned that we can solve inequalities graphically and relate this back to the algebraic solution.

Hope you've enjoyed this lesson as much as I have and I look forward to seeing you again soon for more mathematics.

Goodbye for now.