Linear Inequalities

In this chapter: strict and slack inequalities, the rules for solving them, solution sets over N, Z and R, representation on the number line, double inequalities, word problems, modulus inequalities, the sign scheme for products and quotients, and linear inequalities in two variables as half-planes.

Inequalities and their types

Two real numbers or algebraic expressions related by <, >, ≤ or ≥ form an inequality. Inequalities with < or > are strict; those with ≤ or ≥ are slack (they allow equality).

• Linear in one variable: ax + b < 0, ax + b ≥ 0 and so on, with a ≠ 0.
• Linear in two variables: ax + by < c, ax + by ≥ c and so on.
• Quadratic: ax² + bx + c ≤ 0 with a ≠ 0, and similar forms.

A value of the variable that makes the inequality true is a solution; the set of all solutions is the solution set.

Rules for solving

1. The same number may be added to or subtracted from both sides without changing the sign of the inequality.
2. Both sides may be multiplied or divided by the same positive number without changing the sign.
3. When both sides are multiplied or divided by the same negative number, the sign is reversed: < becomes >, ≤ becomes ≥.

Rule 3 is the whole chapter in one line. From 5 > 3 we get −5 < −3 on multiplying by −1. Never multiply both sides by an expression like (x − 2) whose sign you do not know; you would not know whether to reverse the sign.

The solution set depends on where x lives

Take 30x < 200. Dividing by 30 gives x < 20/3, that is, x < 6.67 roughly.

• If x is a natural number, the solution set is {1, 2, 3, 4, 5, 6}.
• If x is an integer, it is {..., −2, −1, 0, 1, ..., 6}.
• If x is real, it is the interval (−∞, 20/3).

Read the question for this. Word problems about numbers of students or items usually need natural numbers or whole numbers.

Worked example: Solve x/3 > x/2 + 1 for real x.
Solution: Multiply both sides by 6 (positive, so no change): 2x > 3x + 6. Subtract 3x: −x > 6. Multiply by −1 and reverse the sign: x < −6. The solution set is (−∞, −6). Check with x = −12: −4 > −6 + 1 = −5, which is true.

Graphing on the number line

Show the solution of a one-variable inequality on the number line. A filled dot means the point is included (≤, ≥, or a square bracket); an open circle means it is excluded (<, >, or a round bracket).

Double inequalities

An inequality like 2 < 3x − 4 ≤ 11 is solved by doing the same operation on all three parts at once. Add 4: 6 < 3x ≤ 15. Divide by 3: 2 < x ≤ 5, the interval (2, 5]. If you divide by a negative number, reverse both signs, and then rewrite the result with the smaller number on the left.

Worked example: Solve −3 ≤ 4 − 7x/2 ≤ 18.
Solution: Multiply all parts by 2: −6 ≤ 8 − 7x ≤ 36. Subtract 8: −14 ≤ −7x ≤ 28. Divide by −7 and reverse both signs: 2 ≥ x ≥ −4. So the solution is −4 ≤ x ≤ 2, the interval [−4, 2].

Word problems

Translate the words "at least" as ≥, "at most" and "not more than" as ≤, and "more than" as >. Suppose a student scored 70 and 75 in two unit tests and needs an average of at least 60 over three tests. With x as the third score, (70 + 75 + x)/3 ≥ 60 gives 145 + x ≥ 180, so x ≥ 35. The student needs a minimum of 35 marks.

Two tools JEE expects: modulus and sign scheme

These go a little beyond the NCERT chapter but you will use them constantly. For a > 0:

So |x − 1| > 2 means x − 1 < −2 or x − 1 > 2, that is, x < −1 or x > 3, as shown on the third number line above.

For a product or quotient of linear factors, mark the critical points (zeros of each factor) on the number line. When every factor is written with a positive coefficient of x, the expression is positive to the right of the largest critical point, and the sign alternates as you cross each simple critical point going left. Pick the intervals with the sign you need.

• (x − 1)(x − 4) < 0: critical points 1 and 4; negative between them, so 1 < x < 4.
• (x − 1)/(x + 2) ≥ 0: critical points −2 and 1; non-negative for x < −2 or x ≥ 1. The point x = −2 is excluded because the denominator vanishes there; x = 1 is included because the fraction is 0.

Linear inequalities in two variables

The line ax + by = c divides the plane into two half-planes. Each of ax + by < c and ax + by > c is one of them. To graph an inequality:

1. Draw the boundary line ax + by = c, solid for ≤ or ≥ (points on the line are solutions) and dashed for < or >.
2. Take a test point not on the line, usually the origin, and substitute it.
3. If the test point satisfies the inequality, shade its side; otherwise shade the other side.

For a system of such inequalities, the solution region is the common part (intersection) of the individual half-planes. With x ≥ 0 and y ≥ 0 added to the example above, the region shrinks to the triangle with vertices (0, 0), (4, 0) and (0, 6). This is exactly the picture you will draw in linear programming in Class 12.

Common mistakes: (1) Forgetting to reverse the sign after multiplying or dividing by a negative number. (2) Multiplying both sides by an expression containing x, such as (x + 2), without knowing its sign. (3) Giving an interval answer when the question restricts x to natural numbers or integers. (4) Including a point where a denominator is zero. (5) Drawing a solid boundary for a strict inequality, or shading the wrong side because the test point was never checked. (6) Writing the final answer of a double inequality as 2 ≥ x ≥ −4 and then misreading it; rewrite it as −4 ≤ x ≤ 2.

Exam focus

• Solving linear and double inequalities cleanly, with correct sign reversal.
• Writing solution sets in interval form and picking out integer or natural number solutions.
• Modulus inequalities of the form |ax + b| < c and |ax + b| > c.
• Sign scheme for products and quotients of linear factors, which feeds straight into domain questions.
• Identifying the region of a linear inequality in two variables, and which points lie in it.

Practice questions