In the same way we have been working with systems of equations, we can also work with systems of inequalities. When we are solving a system of inequalities, we solve it the same way we would solve a system of equations, by finding what values make it true.

The solutions in the above animation are repeated with a more detailed explanation below.

We have the system of inequalities,

x + y > 5

y ≥ 5x - 3

Determine if the points (2, 7) and (4, 9) are each solution to the system of inequalities above.

So, for the points to be solutions, they will have to make both inequalities in the system true. So, sub each point into each inequality and check if it makes it true.

Sub in (2, 7) into

x + y > 5

2 + 7 > 5

9 > 5

It makes this inequality true.

And for,

y ≥ 5x - 3

7 ≥ 5(2) - 3

7 ≥ 10 - 3

7 ≥ 7

The point makes this inequality true as well. Therefore, it makes both the inequalities true and is therefore a solution.

Now for the point (4, 9)

x + y > 5

4 + 9 > 5

13 > 5,

It makes this inequality true.

Now for y ≥ 5x - 3

9 ≥ 5(4) - 3

9 ≥ 20 - 3

9 ≥ 17

This point makes this inequality false and is therefore not a solution as it needs to make **both** inequalities in the system true.

Graph the system of inequalities

x + y > 5

y ≥ 5x - 3,

For graphing these inequalities, first, think of the inequality symbol as an equals sign and then graph that line.

For > or < we graph a dotted line, for ≥ or ≤ we graph a line that is solid.

For graphing inequalities, we also shade the area above or below the plotted line that includes the values that make the inequality true. For “greater than” and “greater than or equal to” we shade the area above the plotted line. For “less than” or “less than or equal to” we shade the area below the plotted line.

So, remember, the area that we shade has all the points that are a solution for the inequality that we are graphing.

For example, if we graph the line y > x, this will look like the line of y = x, going through the points (0, 0) (1, 1) (2, 2) etc. except for an inequality like this, the solution is the y values **greater** than x. So, we shade the area **above** the line. The points in the shaded area are all the points where y > x. For example (1, 2) as the y-value of 2 is greater than the x-value of 1 or (3, 7) as the y-value of 7 is greater than the x-value of 3, these points would be in the shaded area.

The points that would **not** be in the solution set are the shaded are the points where the x-value is **greater** than the y -value, like (3, 2) as the x-value of 3 is **not** less than the y-value of 2 or (10, 2) as the x-value of 10 is not less than the y-value of 2. These points would **not** be in the shaded area with the solution set.

So, we can first think of x + y > 5 as x + y = 5

We can rearrange this to be

y = 5 - x

Swap around the -x and 5 to make it easier to read

y = - x + 5

Now, remembering our y = mx + b as the format to plotting a line, *m* is the gradient and *b* is the y-intercept.

For y = -x + 5

The gradient, *m* is -1 or -1 ÷ 1 meaning every 1 y-value we go down we go 1 x-value to the right and every 1 y-value we go up we go 1 x-value to the left.

5 is the y-intercept,

Start our graph at 5 on the y-intercept, and plot the points on the **dotted** line.
Now we graph y ≥ 5x - 3

Think of it as an equation like y = 5x - 3

Our gradient, *m*, will be 5, or 5 ÷ 1

For every 5 y-values we go up, we go 1 x-value to the right and for every 1 y-value we down, we go 1 x-value to the left.

Our y-intercept will be -3.

Start our graph at the y-intercept -3, plot our points, then, fill in our solid line.

Now that we have plotted both of our lines, shade the area for all the values that make the lines for that inequality true.

Note, the area that is overlapped by shading from each plotted line is the area that includes values that are solutions to the system of inequalities, as they are the values that make both inequalities true.

The solutions in the above animation are repeated with a more detailed explanation below.

For the system of inequalities below, graph the solutions.

State one point that is in the solution set and one that is not.

y < {-⅔) x + 1

x ≥ -1

So, for y < {-⅔) x + 1, we graph it the same way we did in the previous question,
Imagine y < {-⅔) x + 1 as a line, then we shade in the area **below** the line as the y-values are less than the expression. The line we graph will also be a dotted line as it is a “less than” and not a “less than or equal to”

So, imagining y < {-⅔) x + 1 as y = {-⅔) x + 1

We have the gradient m = -⅔

Meaning for every 2 y-values we go down, we go 3 x-values to the right and for every 2 y-values we go up, we go 3 x-values to the left.

Remember, as it is a negative gradient, we move inversely to the normal one y-value up one x-value right and one y-value down one x-value left.

Our y-intercept, *b*, is 1

So, graphing this, we start at the y-intercept, 1, then plot our points following the gradient.

Once we have plotted the points and drawn the dotted line, shade the area **below** the line, as this is plotting all the values for *y* **less than** {-⅔) x + 1

For x ≥ -1 we treat it slightly differently, we must get a bit more theoretical and really think what this means.

Imagine this as x = -1, meaning a line where all the x-values are -1, this will simply be a straight line running up and down from the x = -1 mark, i.e. every point where x = - 1,

As this is a “greater than or equal to” we draw a **solid** line, and then shade all the values that are **greater** than -1, i.e. all the values to the right of the line,

State one point that is in the solution set and one point that is not.

Now that we have graphed and shaded the true values for each inequality, the solution to the system of inequalities is the area that is shaded by both of our shades.

For our true point in the solution set, simply pick one point that is shaded by both shades,

That can be (0, -1) or (2, -3) or (5, -7)

The false points that we choose can be anywhere that is not shaded by both shades

(3, 0) or (-2, -2) or (-7,8)

We can also have systems that have more than two inequalities or equations.

The solutions in the above animation are repeated with a more detailed explanation below.

We have the system of inequalities:

x < 3

y ≥ -3

y ≤ 3x

Graph the solution set to each of these inequalities.

So, although we now have three inequalities in our system, we still follow the same procedure.

Start by graphing each of inequalities, and then shading the area around that graphed line that has the solutions to that inequality.

Graph x < 3, so plot the line of x = 3 (a straight line running up and down from the x=3 value)

Then, draw a dotted line, and shade the area with all the values **less** than 3, so, the area to the **left** of x = 3

Now we graph y ≥ -3, so first plot the **solid** line for y = -3, this time we have a y-value line, so we plot a **solid** line running horizontally from the point y = -3, (**solid** as it is a greater than or equal to)

From that we shade all the values that are **greater** than -3, so this will be the whole area **above** the line of y = - 3

Now we have our third inequality of the system, y ≤ 3x,

We treat this as it is y = 3x,

The gradient, *m*, will be 3 or alternatively, 3 ÷ 1

Every 3 y-values we go right we go 1 x-value up and every 3 y-values we go down we go 1 x-value down/

Our y-intercept, b, of y = 3x (y = 3x + 0) is 0/

So, start plotting our points at y = 0, then fill in a **solid** line, are we are dealing with a “less than or equal to” inequality,

Once we have plotted our **solid** line, shade the area **below** the line as we are dealing with a **less than or equal to** inequality.

Now, our solution to this system of inequalities is the area that is shaded by all three shades. Any point that we pick that sits in this area will satisfy and make all the inequalities in our system true.