The first step to solving this inequality is to subtract 18 from all three segments of the equation:

8 (- 18)  < -5x + 18 ( - 18) < 103 ( - 18)

-10 < -5x < 85

We then divide the entire equation by -5. According to inequality rules, when we multiply or divide an inequality by a negative number, we must flip the inequality signs.

(-10/-5) > (-5x/-5) > (85/-5)

2 > x > -17 or in conventional layout -17 < x < 2

In order to solve this inequality, we must split it into two inequalities:

 and 

We then solve each, remembering of course that, when multiplying or dividing by a negative number, we must switch the direction of the inequality sign.

and

Therefore:

To solve a rational inequality, move all expressions to the left first:

The boundary points of the solution set will be the points at which:

 - that is, ;

;

; that is, .

None of these values will be included in the solution set, since equality is not allowed by the inequality symbol.

Test the intervals

by choosing a value in each interval and testing the truth of the inequality.

: test 

True; include the interval 

: test 

False; exclude the interval 

: test 

True; include the interval .

: test 

False; exclude the interval .

The solution set of the inequality is .

To solve a quadratic inequality, move all expressions to the left first:

Since the square of any real number is nonnegative, there is no value of  for which this is true. The solution set is the empty set.

To solve a quadratic inequality, move all expressions to the left first:

The only boundary point for the intervals in the solution set is the point at which  - that is, . This will be excluded from the solution set, as the symbol does not allow equality. But for any other value of , the square of the number must be positive. Therefore, the solution set is 

.

To solve a quadratic inequality, move all expressions to the left first

Since the square of any number must be nonnegative, it follows that for any , 

and the solution set is the set of all real numbers, .

First, solve the inequality:

The boundary points of the intervals to be tested are the points at which:

; that is, , or

; that is, .

Therefore, the intervals  should be tested; do so by testing one value in each interval in the inequality and testing its truth:

 - test 

False - reject this interval.

 - test 

True - accept this interval.

 - test 

False - reject this interval.

The solution set is the interval ; therefore,  if and only if .

From Statement 1 alone, we know that ; this provides insufficient information, since, for example,  and  both fall in this range, but only the former is a solution of . For similar reasons, Statement 2 alone provides infufficient information. 

Assume both statements are true. Together, the two statements are euivalent to saying that . Therefore, it holds that , or , and it can be established that .

To solve a quadratic inequality, move all expressions to the left first:

The square of a real number cannot be less than 0, so 

, the only solution.

To solve a quadratic inequality, move all expressions to the left first:

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers 

If , then one of two things is true.

Either

  (note the change in direction of the inequality symbols)

or

(note the change in direction of the inequality symbols)

The set is .