A linear inequality in the variable x is a comparison of expressions by either “less than” (<), “less than or equal to” (≤ ), “greater than” (>), or “greater than or equal to” (≥) where each expression is of the form , with a and b real numbers.
To solve an inequality containing a variable, find all values of variable that make the inequality true.
Basic Properties
Let a, b and c be real numbers.
1. Transitive Property
If a < b and b < c then a < c
2. Addition Property
If a < b then a + c < b + c
Adding the same quantity to both sides of an inequality produces an equivalent inequality.
3. Subtraction Property
If a < b then a – c < b – c
Subtracting the same quantity from both sides of an inequality produces an equivalent inequality.
Properties of inequalities
4. Multiplication Property
i. If a < b and c is positive then c a < c b
ii. If a < b and c is negative c a > c b
Multiplying both sides of an inequality by a positive number produces an equivalent inequality.
Multiplying both sides of an inequality by a negative number reverses the direction of the inequality.
Note 1. If each inequality sign is reversed in the above properties, we obtain similar properties.
Note 2. If the inequality sign < is replaced by ≤ or the sign > is replaced by ≥ , we also obtain similar properties.
A solution for an inequality in a variable x is a number such that when we substitute that number for x we have a true statement.
Typically an inequality has infinitely many solutions and the solution set is easily described using interval notation.
Example 1. Solve the inequality x + 3 ≤ 10 and graph the solution set.
Solution. Subtract 3 from both sides and simplify ( Property 2 above): x ≤ 7
The solution set is the set of all x ≤ 7.
In interval notation this set is (-∞, 7].
The graph of the solution is sketched below.
The bracket at 7 indicates that 7 is in the solution set.
Observation 1. Solving linear inequalities is very much like solving linear equations, with one important difference:
when you multiply or divide both sides of an inequality by a negative number, the direction of the inequality is reversed.
We can see this, using an inequality with no variables.
Example 2. The inequality 3 < 7 is TRUE.
Multiply both sides of the inequality by the negative number -2:
(3)(-2) < (7)(-2): this is FALSE, because -6 is to the right of -14 on the number line. Hence, -6 > -14.
(3)(-2) > (7)(-2): this is TRUE. So, when we multiply the original inequality by -2, we must reverse the direction to obtain another true statement.
Example 3. Solve the inequality: 7 – 2x < 3
Solution. Subtract 7 to both sides and simplify: – 2x < – 4 Divide both sides by -2 and REVERSE (-2 is negative) the inequality sign and simplify ( Property 3-ii above):
x > 2
The graph of the solution is sketched below.
The number 2 is not in the solution set.
Observation 2. In general we can not multiply or divide both sides of an inequality by an expression with a variable, because some values of the variable could make the expression positive and some could make it negative.
Exercise 1. Solve the inequality: 6x – 6 > 2x + 2
Solution. Add 6 to both sides and simplify: 6x > 2x + 8
Subtract 2x from both sides and simplify: 4x > 8
Multiply both sides by 1/4 (which is a positive number) and simplify: x > 2
The graph of the solution is sketched below.
Exercise 2. Solve the inequality 2(3x + 2) – 20 > 8(x – 3).
Solution. Multiply factors and group like terms
6x + 4 -20 > 8x – 24
6x – 16 > 8x – 24
Add 16 to both sides and simplify: 6x > 8x – 8.
Subtract 8x from both sides and simplify: – 2 x > – 8
Multiply both sides by -1/2 and REVERSE (-1/2 is negative) the inequality sign and simplify: x < 4.
In conclusion, the solution set consists of all real numbers in the interval (- ∞ , 4).
Exercise 3. Solve the inequality:
Solution. Multiply both sides by 5 (the LCD)
and simplify: 10 – 7 x > -5 x + 15.
Subtract 10 from both sides: – 7 x > -5 x + 5
Add 5x to both sides: -2 x > 5
Divide both sides by -2 and reverse inequality:
x < -5 / 2
In conclusion, the solution set to the above inequality consists of all real numbers in the interval (- ∞, -5/2).
A system of inequalities in one variable x is a set of inequalities in the same variable x that must be simultaneously satisfied. In order to satisfy all inequalities, a solution must be in the solution sets of all inequalities.
The technique for solving the system is identical to that for an individual inequality, with one additional step at the end.
Exercise 4. Solve the following system of inequalities:
Solution. Find all numbers x such that -3 < 5 – 2x and 5 – 2x < 9.
The numbers that satisfy both inequalities are the values in the intersection of the two solution sets. Then we solve both inequalities
The graph of the inequalities solutions is sketched below.
The intersection set of the solutions for both inequalities is the interval the set (-2, 4) in interval notation.
Exercise 5. Solve the following system of inequalities:
Solution. The numbers that satisfy both inequalities are the values in the intersection of the two solution sets. Then we solve both inequalities
1. Functions and their graph. One–one and onto functions. Composite and inverse functions.
2. Linear inequalities. Systems of linear inequalities
3. Polynomial inequalities. Rational inequalities. Absolute-value inequalities
6. Differentiation. Derivative rules, the chain rule.
7. Application of differentiation: L'Hospital's Rule
8. Vertical, Horizontal and Slant asymptotes
9. Higher Order Derivatives. Applications of differentiation: local and absolute extremes of a function
10. Curve Sketching