Inequalities - Definition, Properties and Solved Linear Inequalities Examples

Inequalities are mathematical expressions showing the relationship between two values, indicating that one value is greater than, less than, or not equal to another. Understanding inequalities is crucial for solving various mathematical problems, from basic arithmetic to advanced calculus.

In this article, we will cover the concepts of the inequalities. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of one question has been asked on this concept, including one in 2020.

Inequalities

Inequalities are the relationship between two expressions that are not equal to one another. Symbols denoting the inequalities are <, >, ≤, ≥, and ≠.

• x < 4, “is read as x less than 4”, x ≤ 4, “is read as x less than or equal to 4”.
• Similarly x > 4, “is read as x greater than 4” and x ≥ 4, “is read as x greater than or equal to 4”.

The process of solving inequalities is the same as of equality but instead of equality symbol inequality symbol is used throughout the process.

Types of Inequalities

• - Linear Inequalities: Involve linear expressions.
  - Example: $2 x+3 \leq 7$
  - Quadratic Inequalities: Involve quadratic expressions.
  - Example: $x^2-4 x+3>0$
  - Polynomial Inequalities: Involve polynomials of degree greater than two.
  - Example: $x^3-2 x^2+x-5<0$
  - Rational Inequalities: Involve ratios of polynomials.

  $\frac{x+1}{x-3} \geq 2$

  - Absolute Value Inequalities: Involve absolute value expressions.
  - Example: $|x-2| \leq 5$

  A few rules that are different from equality rules

• If we multiply or divide both sides of the inequality by a negative number, then we reverse the inequality (reversing inequality means > gets converted to < and vice versa, and ≥ gets converted to ≤ and vice versa) (eg 4>3 means -4<-3)
• If we cross multiply a negative quantity in an inequality, then we reverse the inequality (eg 3 > -2 means -3/2 < 1)
• If we cancel the minus sign from both sides of an inequality, then we reverse the inequality. (eg -3 > -4 means 3 < 4)
• As we usually do not know the sign of a variable term like x, (x-2), etc, so we do not cross-multiply them, as we cannot decide if we have to reverse the sign of inequality or not.

We get a range of solutions while solving inequality which satisfies the inequality,

for e.g. a > 3 gives us a range of solutions, means a ? (3, ∞)

Graphically inequalities can be shown as a region belonging to one side of the line or between lines, for example, inequality -3< x ≤ 5 can be represented as below, a region belonging to -3 and 5 are the region of possible x including 5 and excluding -3.

Frequently Used Inequalities
1. $(x-a)(x-b)<0 \Rightarrow x \in(a, b)$, where $a<b$
2. $(x-a)(x-b)>0 \Rightarrow x \in(-\infty, a) \cup(b, \infty)$, where $a<b$
3. $x^2 \leq a^2 \Rightarrow x \in[-a, a]$
4. $x^2 \geq a^2 \Rightarrow x \in(-\infty,-a] \cup[a, \infty)$

How to solve inequalities?

To solve inequalities, follow these steps:

1. Rearrange the inequality so that all the unknowns are on one side of the inequality sign.
2. Divide by the coefficient of the variable to isolate the variable.
3. Write your solution using the inequality symbol.
4. Graph the solution set on number line

Solved Examples Based On the Inequalities:

Hence, the answer is the option 1.

Example 1: Consider the two sets: $A=m \in R$ : both the roots of $x^2-(m+1) x+m+4=0$ are real\} and $B=[-3,5]$
Which of the following is not true?
1) $A-B=(-\infty,-3) \cup(5, \infty)$
2) $A \cap B=-3$
3) $B-A=(-3,5)$
4) $A \cup B=R$

Solution:

$\begin{aligned}
& x^2-(m+1) x+m+4=0 \\
& b^2-4 a c \geqslant 0 \\
& \Rightarrow(m+1)^2-4(m+4) \geqslant 0 \\
& \Rightarrow m^2-2 m-15 \geqslant 0 \\
& \Rightarrow(m-5)(m-3) \geqslant 0 \\
& \Rightarrow m \epsilon(-\infty, 3) \cup(5, \infty)
\end{aligned}$

Example 2: Solution of the inequality $(x+1)(x-2)(x+7)<0$ is
1) $x \in(-7,-1) \cup(2, \infty)$
2) $x \in(-7,1) \cup(2, \infty)$
3) $x \in(-\infty,-7) \cup(-1,2)$
4) $x \in(-\infty,-7) \cup(1,2)$

Solution:
we have $(x+1)(x-2)(x+7)<0$
on number line mark $x=-1,2,-7$

When x > 2, all factors, (x + 1), (x - 2) and (x + 7) is positive

Now put positive and negative signs as shown in the figure

Hence answer is $x \in(-\infty,-7) \cup(-1,2)$
Example 3 : Which values of $x$ satisfy the inequality $(x+1)(x-3)<0$ ?
1) $x \in(-1, \infty)$
2) $x \in(-1,3)$
3) $x \in(-3,1)$
4) $x \in(-\infty, 3)$

Solution:
As we have learned in
Frequently Used Inequalities
1. $(x-a)(x-b)<0 \Rightarrow x \in(a, b)$, where $a<b$
2. $(x-a)(x-b)>0 \Rightarrow x \in(-\infty, a) \cup(b, \infty)$, where $a<b$
3. $x^2 \leq a^2 \Rightarrow x \in[-a, a]$
4. $x^2 \geq a^2 \Rightarrow x \in(-\infty,-a] \cup[a, \infty)$

Now,

Given (x + 1)(x - 3) < 0

on number line mark x = -1, 3

from the concept -1 < x < 3

correct option is $x \in(-1,3)$
Example 4: The solution of the inequation $\frac{x}{2}+\frac{3}{4}>\frac{x}{3}-1$ is
1) $\left(-\infty, \frac{21}{2}\right)$
2) $\left(\frac{-21}{2}, \infty\right)$
3) $(-\infty, \infty)$
4) $(2,3)$

Solution:

$\begin{aligned}
& \frac{x}{2}+\frac{3}{4}>\frac{x}{3}-1 \\
\Rightarrow & \frac{x}{2}-\frac{x}{3}>-1-\frac{3}{4} \\
\Rightarrow & \frac{3 x-2 x}{6}>\frac{-4-3}{4} \\
\Rightarrow & \frac{x}{6}>\frac{-7}{4} \\
\Rightarrow & x>-\frac{7}{4} \times 6 \\
\Rightarrow & x>-\frac{21}{2}
\end{aligned}$

Hence, the answer is option (2).

Example 5: Values of x that satisfy $\frac{x-3}{4}+1 \leqslant \frac{2 x-3}{5}$ are.
1) $\left[\frac{17}{3}, \infty\right)$
2) $(-\infty, \infty)$
3) $\left(-\infty, \frac{17}{3}\right]$
4) $\left[-\frac{17}{3}, \frac{17}{3}\right]$

Solution:

$\begin{aligned}
& \frac{x-3}{4}+1 \leqslant \frac{2 x-3}{5} \\
\Rightarrow & \frac{x-3+4}{4} \leq \frac{2 x-3}{5} \\
\Rightarrow & \frac{x+1}{4} \leq \frac{2 x-3}{5} \\
\Rightarrow & 5(x+1) \leq 4(2 x-3) \\
\Rightarrow & 5 x+5 \leq 8 x-12 \\
\Rightarrow & 5+12 \leq 8 x-5 x \\
\Rightarrow & 17 \leq 3 x
\end{aligned}$

$\begin{aligned} & \Rightarrow \frac{17}{3} \leqslant x \\ & \Rightarrow x \in\left[\frac{17}{3}, \infty\right)\end{aligned}$

Hence, the answer is option (1)

Summary

We concluded that inequalities are a fundamental part of mathematics, providing a way to describe and solve problems involving ranges and constraints. Mastery of inequalities is essential for progressing in algebra, calculus, and applied mathematics, offering valuable tools for both theoretical and practical problem-solving.