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Solve systems of inequalities and draw intervals on a number line

system of inequalities

As you already know, the solution to any linear inequality is some kind of interval. This will also be true for the systems of inequalities, only here you’ll search for the interval that is common to both of the inequalities. The procedure of solving systems of inequalities is the following: first you solve both of your inequalities individually and then you search for the intersection of the solutions that is their common solution (note that the intersection in mathematics is marked with \cap).

Now we can go to the example 1. Solve the system of linear inequalities.

2x + 1 \ge 3

5x - 2 > 6x - 6

1. First we’ll solve inequalities individually.

2x + 1 \ge 3

2x \ge 2

x \ge 1

This solution written in the form of interval is [1, +\infty\rangle. If we present this solution graphically:

graph solution 1 to infinity

For the second inequality:

5x - 2 > 6x - 6

-x > -4

x < 4

or \langle - \infty, 4 \rangle.

infinity to 4 graph

If we combine these two pictures we get their intersection:

union of two intervals

This gives us a pretty clear solution. Their intersection, the solution to this system, is the interval


Example 2. What would happen if we altered this system just a little bit? If we got a system:

2x + 1 \ge 3

5x - 2 > 6x - 2

The first inequality is the same, so is its solution: [1, + \infty \rangle

But, something did change with the second inequality. Let’s find its solution.

5x - 2 > 6x - 2

- x > 0

x < 0

We got interval \langle - \infty, 0 \rangle. Let’s draw these two intervals.

draw excluded intervals

From the picture we can conclude that these two intervals have no intersection which means that this systems of inequalities has no solutions.

Example 3. Solve the inequality:

\frac{4 - x}{x + 2} > 4

Here you have one inequality, but this is specific kind. If your first instinct is to multiply with (x + 2) you are very wrong. Remember that when you multiply inequality with negative number you’d have to change inequality sign? Well, nothing guaranties that expression (x + 2) is a positive number. If x is any number lesser than -2 this whole inequality would change. Since x can be any number, we have to break down this inequality into two cases: first case when x is lesser than -2 and second where x is greater than two (you can’t have lesser or equal or greater or equal to -2 because x cannot be – 2, otherwise we’d have zero in our denominator).

Now we have two cases that give us two different inequalities. The solution of the main inequality is the union of the solutions of these two cases.

1. For x + 2 > 0 (this means that you multiply with a positive number and that the inequality sign won’t change)


\frac{4 - x}{x + 2} > 4

4 - x > 4(x + 2)

4 - x > 4x + 8

- 5x > 4

x < -\frac{4}{5}

We got our solution, but you still have to be careful because you have a condition x + 2 > 0 or x > – 2. These two inequalities make a systems of inequalities. Let’s draw these solutions on a number line:

solution x bigger than 2

This means that the solution of this case is the interval \langle -2, -\frac{4}{4} \rangle.

2. For x + 2 < 0 (this means that you multiply with a negative number and that the inequality sign will change):

\frac{4 - x}{x + 2} > 4

4 - x < 4(x + 2)

4 - x < 4x + 8

- 5x < 4

x > -\frac{4}{5}

You again have a condition: x + 2 < 0 or x < – 2 .

x less than 2

The solution of this case is an empty set or \emptyset.

The solution of our starting inequality is the union of these two intervals:

\langle - 2, -\frac{4}{5} \rangle \cup \emptyset = \langle - 2, -\frac{4}{5} \rangle


Systems of inequalities worksheets

  Standard form (5.7 MiB, 253 hits)

  Slope-intercept form (5.7 MiB, 242 hits)