Linear and Nonlinear Inequalities

I nequality Relations

The number a is less than b, written a < b, if b - a is positive. Then b is greater than a, written b > a. If a is either less than or equal to b, this is written $ a b $ . Then b is greater than or equal to a, written $ b a $ . Geometrical Interpretation: If a < b, then a is to the left of b on a real number line (Fig. 6-1). If a > b, then a is to the right of b.

EXAMPLE 6.1

In Fig. 6-1, a < d and b > c. Also, a < c and b > d.

Combined Inequalities and Intervals

If a < x and x < b, the two statements are often combined to write: a < x < b. The set of all real numbers x satisfying a < x < b is called an open interval and is written $ (a, b) $ . Similarly, the set of all real numbers x satisfying the combined inequality $ a x b $ is called a closed interval and is written $ [a, b] $ . The following table shows various common inequalities and their interval representations.

Inequality Notation Graph
a<x> <fcel> <nl>
a<x> <fcel> <nl>
a<x> <ecel>
Inequality Notation Graph
x≥a [a,∞)] b

I nequality Statements Involving Variables

An inequality statement involving variables, like an equation, is in general neither true nor false; rather, its truth depends on the value(s) of the variable(s). For inequality statements in one variable, a value of the variable that makes the statement true is a solution to the inequality. The set of all solutions is called the solution set of the inequality.

Equivalent Inequalities

Inequalities are equivalent if they have the same solution sets.

EXAMPLE 6.2 The inequalities x < -5 and $ x + 5 < 0 $ are equivalent. Each has the solution set consisting of all real numbers less than -5, that is, $ (-∞, -5) $ .

The process of solving an inequality consists of transforming it into an equivalent inequality whose solution is obvious. Operations of transforming an inequality into an equivalent inequality include the following:

  1. ADDING OR SUBTRACTING: The inequalities $ a < b $ , $ a + c < b + c $ , and a - c < b - c are equivalent, for c any real number.

  2. MULTIPLYING OR DIVIDING BY A POSITIVE NUMBER: The inequalities $ a < b, ac < bc, $ and $ a/c < b/c $ are equivalent, for c any positive real number.

  3. MULTIPLYING OR DIVIDING BY A NEGATIVE NUMBER: The inequalities $ a < b, ac > bc, $ and $ a/c > b/c $ are equivalent, for c any negative real number. Note that the sense of an inequality reverses upon multiplication or division by a negative number.

  4. Simplifying expressions on either side of an inequality.

Similar rules apply for inequalities of the form a > b and so on.

Linear Inequalities

A linear inequality is one which is in the form $ ax + b < 0 $ , $ ax + b > 0 $ , $ ax + b $ , or $ ax + b $ , or can be transformed into an equivalent inequality in this form. In general, linear inequalities have infinite solution sets in one of the forms shown in the table above. Linear inequalities are solved by isolating the variable in a manner similar to solving equations.

EXAMPLE 6.3 Solve: 5 - 3x > 4.

\[ \begin{aligned}5-3x&>4\\-3x&>-1\end{aligned} \]

\[ x<\frac{1}{3} \]

Note that the sense of the inequality was reversed by dividing both sides by -3.

An inequality that is not linear is called nonlinear.

Solving Nonlinear Inequalities

An inequality for which the left side can be written as a product or quotient of linear factors (or prime quadratic factors) can be solved through a sign diagram. If any such factor is not zero on an interval, then it is either positive on the whole interval or negative on the whole interval. Hence:

  1. Determine the points where each factor is 0. These are called the critical points.

  2. Draw a number line and show the critical points.

  3. Determine the sign of each factor in each interval; then, using laws of multiplication or division, determine the sign of the entire quantity on the left side of the inequality.

  4. Write the solution set.

EXAMPLE 6.4 Solve: $ (x-1)(x+2)>0 $

The critical points are 1 and -2, where, respectively, x - 1 and $ x + 2 $ are zero. Draw a number line showing the critical points (Fig. 6-2). These points divide the real number line into the intervals $ (-∞, -2) $ , $ (-2, 1) $ , and $ (1, ∞) $ . In $ (-∞, -2) $ , x - 1 and $ x + 2 $ are negative; hence the product is positive. In $ (-2, 1) $ , x - 1 is negative and $ x + 2 $ is positive; hence the product is negative. In $ (1, ∞) $ , both factors are positive; hence the product is positive.

The inequality holds when $ (x-1)(x+2) $ is positive. Hence the solution set consists of the intervals: $ (-∞,-2) ∪ (1,∞) $ .

SOLVED PROBLEMS

6.1. Solve: $ 3(y - 5) - 4(y + 6) $

Eliminate parentheses, combine terms, and isolate the variable:

\[ 3(y-5)-4\left(y+6\right)\leq7 \]

\[ \begin{aligned}3y-15-4y-24&\leq7\\-y-39&\leq7\\-y&\leq46\\y&\geq-46\end{aligned} \]

The solution set is \([-46,\infty)\).

6.2. Solve: $ - > 5 - $

Multiply both sides by 24, the LCD of all fractions, then proceed as in the previous problem.

\[ \frac{2x-3}{3}-\frac{5x+4}{6}>5-\frac{3x}{8} \]

\[ 24\cdot\frac{(2x-3)}{3}-24\cdot\frac{(5x+4)}{6}>120-24\cdot\frac{3x}{8} \]

\[ 16x-24-20x-16>120-9x \]

\[ \begin{aligned}-4x-40&>120-9x\\5x&>160\\x&>32\end{aligned} \]

The solution set is $ (32,) $ .

6.3. Solve: $ -8 < 2x - 7 $

A combined inequality of this type can be solved by isolating the variable in the middle.

\[ \begin{aligned}-8&<2x-7\leq5\\-1&<2x\leq12\\-\frac{1}{2}&<x\leq6\end{aligned} \]

The solution set is $ (- , 6] $ .

6.4. Solve: $ 0 < 3 - 5x $

\[ \begin{aligned}0<3-5x\leq10\\-3<-5x\leq7\\\frac{3}{5}>x\geq-\frac{7}{5}\\-\frac{7}{5}\leq x<\frac{3}{5}\end{aligned} \]

The solution set is $ (extract/viewrange_chunk_1_1_5_16ffc1d0/images/page_4_img_in_chart_box_296_1292_846_1480.jpg)

The critical points divide the real number line into the intervals $ (-∞,-2) $ , $ (-2,10) $ , and $ (10,∞) $ . In $ (-∞,-2) $ , x - 10 and $ x + 2 $ are negative, hence the product is positive. In $ (-2,10) $ , x - 10 is negative and $ x + 2 $ is positive; hence the product is negative. In $ (10,∞) $ , both factors are positive; hence the product is positive. The equation part of the inequality is satisfied at both critical points, and the inequality holds when $ (x + 2)(x - 10) $ is negative; hence the solution set is $ [-2,10] $ .

6.7. Solve: $ 2x^{2} + 2 5x $

Get 0 on the right side, put the left side into factored form, then form a sign diagram

\[ \begin{aligned}&2x^{2}-5x+2\geq0\\&(x-2)(2x-1)\geq0\\ \end{aligned} \]

Draw a number line showing the critical points $ $ and 2 (Fig. 6-4).

The critical points divide the real number line into the intervals $ (-∞, ) $ , $ (, 2) $ , and $ (2, ∞) $ . The product has sign, respectively, positive, negative, positive in these intervals. The equation part of the inequality is satisfied at both critical points, and the inequality holds when $ (2x - 1)(x - 2) $ is positive, hence the solution set is $ (-∞, ] ∪ [2, ∞) $ .

6.8. Solve: $ x^{3} < x^{2} + 6x $

Get 0 on the right side, put the left side into factored form, then form a sign diagram.

\[ \begin{aligned}x^{3}-x^{2}-6x&<0\\x(x-3)(x+2)&<0\end{aligned} \]

Draw a number line showing the critical points -2, 0, and 3 (Fig. 6-5).

The critical points divide the real number line (Fig. 6-5) into the intervals $ (-, -2) $ , $ (-2, 0) $ , $ (0, 3) $ , and $ (3, ) $ . The product has sign, respectively, negative, positive, negative, positive in these intervals. The inequality holds when $ x(x - 3)(x + 2) $ is negative, hence the solution set is $ (-, -2) (0, 3) $ .

6.9. Solve: $ $

Draw a number line showing the critical points -5 and 3 (Fig. 6-6).

The critical points divide the real number line into the intervals $ (-, -5) $ , $ (-5, 3) $ , and $ (3, ) $ . The quotient has sign, respectively, positive, negative, positive in these intervals. The equation part of the inequality is satisfied at the critical point -5, but not at the critical point 3, since the expression $ $ is not defined there. The inequality holds when $ $ is negative; hence the solution set is [-5, 3].

6.10. Solve: $ $

The solution of this inequality statement differs from that of the corresponding equation. If both sides were multiplied by the denominator x - 3, it would be necessary to consider separately the cases where this is positive, zero, or negative.

It is preferable to get 0 on the right side and combine the left side into one fraction, then form a sign diagram.

\[ \begin{aligned}\frac{2x}{x-3}-3&\geq0\\\frac{2x}{x-3}-\frac{3(x-3)}{x-3}&\geq0\\\frac{9-x}{x-3}&\geq0\end{aligned} \]

Draw a number line showing the critical points 3 and 9 (Fig. 6-7).

The critical points divide the real number line into the intervals $ (-, 3) $ , $ (3, 9) $ , and $ (9, ) $ . The quotient has sign, respectively, negative, positive, negative in these intervals. (Note the reversal of signs in the chart for $ 9 - x $ .) The equation part of the inequality is satisfied at the critical point 9, but not at the critical point 3, since the expression $ $ is not defined there. The inequality holds when $ $ is positive, hence the solution set is $ (3, 9] $ .

6.11. Solve: $ $

Draw a number line showing the critical points $ -5, - $ , and 2 (Fig. 6-8). Note that the factor $ x^{2} + 4 $ has no critical point; its sign is positive for all real x; hence it has no effect on the sign of the result.

The critical points divide the real number line into the intervals $ (-∞,-5) $ , $ (-5,-) $ , $ (- ,2) $ , and $ (2,∞) $ . The quotient has sign, respectively, positive, negative, negative, positive in these intervals. (Note that the factor $ (2x + 3)^{2} $ is positive except at its critical point.) The equation part of the inequality is satisfied at the critical points $ - $ and 2, but not at the critical point -5. The inequality holds when the expression under consideration is positive; hence the solution set is $ (-∞, -5) ∪ {-} ∪ [2, ∞) $ .

6.12. For what values of x does the expression $ $ represent a real number?

The expression represents a real number when the quantity $ 9 - x^{2} $ is nonnegative. Solve the inequality statement $ 9 - x^{2} $ , or $ (3 - x)(3 + x) $ , by drawing a number line showing the critical points 3 and -3 (Fig. 6-9).

The critical points divide the real number line into the intervals $ (-∞, -3) $ , $ (-3, 3) $ , and $ (3, ∞) $ . The product has sign, respectively, negative, positive, negative in these intervals. The equation part of the inequality is satisfied at the critical points, and the inequality holds when $ 9 - x^{2} $ is positive, hence the expression $ $ represents a real number when x is in $ [-3, 3] $ .

6.13. For what values of x does the expression $ $ represent a real number?

The expression represents a real number when the quantity under the radical is nonnegative. Solve the inequality $ $ by drawing a number line showing the critical points -5, 0, and 2 (Fig. 6-10).

The critical points divide the real number line into the intervals $ (-∞, -5) $ , $ (-5, 0) $ , $ (0, 2) $ , and $ (2, ∞) $ . The quotient has sign, respectively, positive, negative, positive, negative in these intervals. The equation part of the inequality is satisfied only at the critical point 0, and the inequality holds when the quantity under the radical is positive; hence the entire expression represents a real number when x is in $ (-∞, -5) ∪ [0, 2) $ .

SUPPLEMENTARY PROBLEMS

6.14. Solve: (a) $ < $ ; (b) $ 0.05(2x-3)+0.02x>15 $ ; (c) $ 4(5x-6)-3(6x-3)>2x+1 $

Ans. (a) $ (, ) $ ; (b) $ (126.25, ) $ ; (c) No solution

6.15. Solve: (a) -0.01 < x - 5 < 0.01; (b) $ < 7 $ ; (c) -6 < 3 - 7x $ $

Ans. (a) (4.99, 5.01); (b) $ $ ; (c) $ (-, -1) (3, ) $

6.17. Solve: (a) $ x^{2} $ ; (b) $ x^{2} + 1 < 1 $ ; (c) $ < 1 $ ; (d) $ $ ; (e) $ $

Ans. (a) [-1,1]; (b) no solution; (c) \((-\infty,0) \cup (1,\infty)\); (d) \((-\infty,-1] \cup [1,\infty)\); (e) \((-\infty,-1) \cup \{0\} \cup (1,\infty)\)

6.18. Solve: (a) $ 5 > $ ; (b) $ $ ; (c) $ $

Ans. (a) $ (-,0) (,) $ ; (b) $ (-,-3) {0} (3,) $ ; (c) $ (-,-2) :::