## What you need to know

Before we get stuck in to this topic, it’s important to understand what an inequality is and how it works. If you’re not sure, click here (https://mathsmadeeasy.co.uk/gcse-maths-revision/inequalities-number-line-solving-inequalities/) for more info. Additionally, you will need to know how to draw straight-line graphs, more on that here (https://mathsmadeeasy.co.uk/gcse-maths-revision/drawing-straight-line-graphs-gcse-maths-revision-worksheets/).

In this topic, we are going to be looking at how to express multiple inequalities on a graph, and then how to read a graph to determine what inequalities it is expressing.

The way to express an inequality on a number line is as follows:

– Firstly, make the subject. Then, treat the inequality as if it were an equation and plot the straight line. You should plot a solid line if the inequality is inclusive (i.e. ) and a dashed line if it is a strict inequality (i.e. ).

– Identify which side of the line the area that satisfies the inequality is on:

– If it is a greater than (or equal to) – – then you want the area above the line,

– If it is a less than (or equal to) – – then you want the area below the line.

– Shade the area that you identified in step 2.

A caveat to this is if you can’t make  the subject, e.g. with an inequality such as . We still plot the line as if it were an equation, but you can’t shade “above” or “below” a vertical line like . Instead, the area you want is to the right of it, where the  values are bigger than 3. By extension, if you were given  then you would shade to the left of the line.

Example: Shade the area that satisfies the inequality  and mark it an A.

We want to pretend this is an equation and plot the line, so we must firstly rearrange the inequality, so it looks like our familiar straight-line equation. Subtracting 1 from both sides, we get

$y \geq 2x - 1$

Now, we must plot $y = 2x - 1$ as a solid line – since it is an inclusive inequality – and then shade the area above it – since it is a “greater than or equal to” – and mark that area with an A. The result looks like the graph on the right.

More often than not, you are presented with multiple inequalities and asked to shade the area that satisfies all of them. In that case, you repeat step 1 and 2 above for each of the inequalities and once you’re done, you shade the portion of the graph that is in all of the satisfied areas. Let’s have a look.

Example: Shade the region of a graph that satisfies the inequalities $y>2, x\geq-1,$ and $y < -x + 5$ and mark it with an A.

So, we’re pretending that all 3 inequalities are equations and plotting them as such. The first one will be a dashed-line plot of $y=2$ since is a strict inequality; the second will be a solid-line plot of $x=-1$ since it is an inclusive inequality; the third will be a dashed-line plot of $y=-x+5$, which is a line with gradient -1 and y-intercept 5.

We want to shade the area that satisfies all 3 inequalities, or in other words the area that is above the line $y=2$, to the right of the line $x=-1$, and below the line $y=-x+5$.

The completed drawing, with the shaded region marked A, looks like the graph on the right.

Note: you don’t have to write the equations on usually, they’re just there to show what’s going on.

The only other thing to consider now is: what happens when you’re given a pre-drawn graph like the one above and have to deduce the inequalities from it? Let’s see.

Example: Determine the 3 inequalities that describe the shaded area on the graph below.

Firstly, we need to find the equations of the 3 lines that have been drawn on this graph. Once this is done, we can consider them as inequalities.

The first two we can read off: the horizontal line is $y=-2$, and the vertical line is $x=2$. Then, we can see that the slanted line has a y-intercept of -1 and (either by doing the triangle method or otherwise) a gradient of 3, so its equation must be

$y=3x-1$

Now we must convert them to the appropriate inequalities.

The line $y=-2$ is dashed and the shaded area is above it, so the inequality must be $y>-2$. The line $x=2$ is solid and the shaded area is to the left of it, so the inequality must be $x\leq 2$. Finally, the line $y=3x-1$ is solid and the shaded area is below it, so the inequality must be $y\leq 3x-1$. So, the 3 inequalities are

$y>-2,\hspace{2mm}x\leq2\text{, and }y\leq 3x-1$

If you’re ever unsure about which way round your signs go, or whether you should be looking above/below/left/right of a line, you can pick a coordinate from the region you think it is and see whether or not it satisfies your inequality.

## Example Questions

#### 1) Shade the region of a graph that satisfies the inequality $x + 2y \leq + 8$ and mark it with an R.

Firstly, rearrange this equation to make $y$ the subject. Subtract $x$ from both sides to get

$2y \leq - x + 4$

Then, divide both sides by 2 to get

$y \leq -\dfrac{x}{2} + 4$

Pretending this is an equation, the graph would be a solid line with gradient $-\frac{1}{2}$ and y-intercept 4. Once drawn, we should shade and mark the region below the line. The result looks like:

#### 2) Shade the region of a graph that is satisfied by the inequalities $y \geq 1, y \leq 3, x>0,$ and $y > x$, and mark it with an R.

We’re going to pretend that the inequalities are equations and plot them as straight lines. The first one will be the solid plot of the line $y=1$, the second will be a solid plot of the line $y=3$, the third will be a dashed plot of the line $x=0$, and the fourth will be a dashed plot of the line $y=x$.

Now, we want to shade the area that is below the line $y=1$, above the line $y=3$, to the right of the line $x=0$, and above the line $y=x$.

The resulting graph looks like:

#### 3) Determine that 3 inequalities that describe the shaded area on the graph below.

Firstly, determine the equations of the 3 lines and then from them, find the inequalities.

The horizontal line is clearly $y = -2$.

The dashed line has its y-intercept at 2 and a gradient of 2 (you can determine gradient using whatever method you like), so it is $y=2x + 2$.

The final line has its y-intercept at 5 and a gradient of -3 (again, up to you how to determine it), so it is $y = -3x + 5$.

Now, the shaded area is above $y=-2$ and the line is solid, so the inequality is

$y\geq -2$

The shaded area is below $y=2x+2$ and the line is dashed, so the inequality is

$y <2x+2$

The shaded area is below $y=-3x+5$ and the line is solid, so the inequality is

$y \leq -3x+5$

Therefore, the shaded area is described by the 3 inequalities

$y\geq -2,\hspace{2mm} y<2x+2,\text{ and }y\leq-3x+5$

## Graphical Inequalities Revision and Worksheets

Inequalities on a graph
Level 4-5
Inequalities on a graph
Level 6-7