How to Solve Inequalities – Quick Review and Practice

Basic linear inequalities have one of the following forms:

ax + b > 0

ax + b < 0

ax + b > 0

ax + b < 0

where a and b are some real numbers. Our solution to any of these inequalities would be some interval. Let’s see one simple example:

2x – 10 > 16

2x > 16 + 10

2x > 26

x > 26/2

x > 13

So, the interval here is: (3, +∞)

If we have a case where –x is lesser or greater than some number, then we multiply the whole inequality by -1, where the sign of inequality also changes:

-3x + 9 < 12

-3x < 12 – 9

-3x < 3

-x < 3/(-1)

x > -3

So, the interval here is: [3, +∞) Notice the difference in the brackets. This is because this interval contains number 3.

Let’s see a little more complex example:

x / (x + 1) > 0 ∞

x is positive on the right of the 0, negative on the left of the 0. x+1 is positive on right of the -1, and negative on the left of the -1. If we multiply the signs, we get the signs for the function. We are interested in the positive sign (because we need it to be greater than 0), so the interval is:

Whenever we have a fraction, we have to make a table:

x is positive on the right of the 0, negative on the left of the 0. x+1 is positive on right of the -1, and negative on the left of the -1. If we multiply the signs, we get the signs for the function. We are interested in the positive sign (because we need it to be greater than 0), so the interval is:

(-∞, -1) U (0, +∞)

1. Solve the inequality:

-7x – 1 ≥ 13

1) [2 +∞)

2) (7, +∞)

3) (-∞, -2]

4) (2, +∞)

2. Solve the inequality:

2x – 1  ≥ x + 10

1) (-∞, 9)

2) (9, +∞)

3) (-∞, -9]

4) [11, +∞)

3. Solve the inequality:

(x – 6)²  ≥ x² + 12

1) [2, +∞)

2) (2, +∞)

3) (-∞, 2]

4) (12, +∞)

1. 3)  (-∞, -2]

-7x – 1 > 13

-7x > 13 + 1

-7x > 14

-x > 2/(-1)

x < -2

2. 4) [11, +∞)

2x – 1 > x + 10

2x – x > 10 + 1

x > 11

3. 3) (-∞, 2]

(x – 6)² > x² + 12

x² – 12x + 36 > x² + 12

-12x > 12 – 36

-12x > -24

-x > -2/(-1)

x < 2