Linear Inequalities
An inequality is a mathematical statement that compares two quantities using the symbols \(<,>, \le, \ge\). When such a comparison involves a linear expression in one variable, it is called a linear inequality.
Definition
A linear inequality in one variable can be written in any one of the following forms: \[ ax + b < 0,\quad ax + b> 0,\quad ax + b \le 0,\quad ax + b \ge 0 \] where \(a\) and \(b\) are real numbers and \(a \ne 0\).
The set of all real values of the variable that satisfy a given linear inequality is called its solution set.
Rules for Solving Linear Inequalities
Rule 1 (Addition or Subtraction Rule)
If the same real number is added to or subtracted from both sides of an inequality, the inequality sign remains unchanged. That is, if \(a < b\), then \[ a + c < b + c \] for any real number \(c\).
Proof:
Adding or subtracting the same real number from both sides changes both quantities by an equal amount.
Hence, their relative order on the number line remains the same, and the inequality sign is preserved.
Rule 2 (Multiplication or Division by a Positive Number)
If both sides of an inequality are multiplied or divided by the same positive real number, the inequality sign remains unchanged. If \(a < b\) and \(c> 0\), then \[ ac < bc \]
Proof:
Multiplication or division by a positive number does not alter the order of real numbers on the number
line. Therefore, the inequality sign remains the same.
Rule 3 (Multiplication or Division by a Negative Number):
If both sides of an inequality are multiplied or divided by the same negative real number, the inequality sign is reversed. If \(a < b\) and \(c < 0\), then \[ ac> bc \]
Proof:
Multiplication or division by a negative number reverses the direction of numbers on the real number
line.
Hence, the order of the quantities is reversed and the inequality sign must be changed.
Representation of Solutions on the Number Line
The solution of a linear inequality is represented graphically on the number line. If the inequality involves \(<\) or \(>\), the boundary point is not included in the solution set and is shown by an open circle. If the inequality involves \(\le\) or \(\ge\), the boundary point is included and is shown by a closed circle.
Linear Inequalities in Two Variables
An inequality of the form \[ ax + by < c \] where \(a, b,\) and \(c\) are real numbers, is called a linear inequality in two variables. The solution of such an inequality is represented graphically as a region in the Cartesian plane bounded by the line \(ax + by=c\).
Important Observations
Linear inequalities generally have infinitely many solutions. Care must be taken while multiplying or dividing both sides of an inequality by a negative number, as the direction of the inequality sign changes. Graphical representation helps in visualizing the complete solution set clearly.
Example-1
Solve \(30 x \lt 200\) when
(i) x is a natural number,
(ii) x is an integer.
Solution
$$\begin{aligned}30x \lt 200\\ x \lt\dfrac{200}{30}\\ x \lt \dfrac{20}{3}\end{aligned}$$(i) Solution set for the inequality when \(x\) is a natural Number
\[\{1,\,2,\,3,\,4,\,5,\,6\}\](ii) Solution set for the inequality when \(x\) is an integer
$$\left\{ \ldots ,3,-2,-1,0,,12,3,4,5,6\right\} $$Example-2
Solve \(5x – 3 \lt 3x +1\) when
(i) x is an integer,
(ii) x is a real number.
Solution
$$\begin{aligned}5x-3 \lt3x+1\\ 5x-3x \lt 1+3\\=2x \lt4\\ x \lt\dfrac{4}{2}\\ x \lt2\end{aligned}$$ Solution set for the inequality when x is an integer $$\left\{ \ldots ,-3,-2,0,1\right\} $$ Solution set when x is a real Number $$x\in \left( -\infty ,2\right) $$Example-3
Solve \(4x + 3 \lt 6x +7\).
Solution
$$\begin{aligned}4x+3 &\lt6x+7\\\\ 4x-6x &\lt7-3\\\\ -2x &\lt4\\\\ x&>-\dfrac{4}{2}\\\\ x &>-2\\\\ x\in\left( -2,\infty \right) \end{aligned}$$Example-4
Solve \(\dfrac{5-2x}{3}\leq\dfrac{x}{6}-5\)
Solution
$$\begin{aligned}\dfrac{5-2x}{3}&\leq \dfrac{x}{6}-5\\\\ \dfrac{5-2x}{3}&\leq \dfrac{x-30}{6}\\\\ \dfrac{6\left( 5-2x\right) }{3}&\leq x-30\\\\ 2\left( 5-2x\right) &\leq x-30\\\\ 10-4x-x&\leq -30\\\\ -5x&\leq -30-10\\\\ -5x&\leq -40\\\\ -x&\leq \dfrac{-40}{5}\\\\ x&\geq 8\\\\ x\in [ 8,\infty ) \end{aligned}$$Example-5
Solve \(7x + 3 \lt 5x + 9\).
Solution
$$\begin{aligned}7x+3 &\lt5x+9\\ 7x-5x &\lt9-3\\ 2x &\lt6\\ x &\lt\dfrac{6}{2}\\ x &\lt3\end{aligned}$$The graphical representation of the solutions are given in Fig Fig 5.1
Example-6
Solve \(\dfrac{3x-4}{2}\geq \dfrac{x+1}{4}-1\)
Solution
6. $$\begin{aligned}\dfrac{3x-4}{2}&\geq \dfrac{x+1}{4}-1\\\\ \dfrac{3x-4}{2}&\geq \dfrac{x+1-4}{4}\\\\ 2\left( 3x-4\right) &\geq x-3\\\\ 6x-x-8&\geq -3\\\\ 5x&\geq -3+8\\\\ x&\geq \dfrac{5}{5}\\\\ x&\geq 1\\\\ x\in [ 1,\infty ) \end{aligned}$$The graphical representation of the solutions are given in Fig Fig 5.2
Example-7
The marks obtained by a student of Class XI in first and second terminal examination are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Solution
Let student got \(x\) marks in annual Examination, then their average is \(\left(\dfrac{62+48+x}{3}\geq60\right)\)
$$\begin{aligned}\dfrac{62+48+x}{3}&\geq 60\\ 110+x&\geq 180\\ x&\geq 180-110\\ x&\geq 70\end{aligned}$$Thus student should get at least 70 marks to get an average of 60
Example-8
Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
Solution
Let consecutive pair of odd number be \(x\) and \(x+2\)
$$\begin{aligned} x &>10\quad\text{and}\\\\ x+x+2 &\lt40\\ 2x &\lt40-2\\ x &\lt\dfrac{38}{2}\\ x &\lt19\end{aligned}$$Since x is an odd number, x can take the values 11, 13, 15, and 17
So, the required possible pairs will be (11,13), (13,15), (15,17), (17,19)