Algebraic solution of linear inequalities in one variable

In this subtopic we apply the manipulation rules systematically to inequalities of the form $ax + b < c$, $ax + b \ge cx + d$, and combinations. The goal: a clean interval description of the solution set.

The basic recipe

To solve a linear inequality in one variable:

1. Eliminate denominators by multiplying both sides by the LCM (track signs).
2. Expand brackets.
3. Collect terms with $x$ on one side, constants on the other.
4. Divide by the coefficient of $x$ (sign reversal if negative).
5. Write the solution as an interval, mark on a number line.

Compound inequalities

Inequalities like $a < f(x) < b$ are two inequalities chained: $a < f(x)$ and $f(x) < b$. Solve each, then take the intersection of the solution sets.

For "or" statements (e.g. $x < 1$ or $x > 3$), take the union.

Number-line representation

Mark the boundaries on a number line. Use an open circle ($\circ$) for strict $<$ or $>$ (endpoint excluded), and a closed circle ($\bullet$) for $\le$ or $\ge$ (endpoint included). Shade the region(s) of $x$ that solve the inequality.

Word-problem translation

Many real-world conditions translate into linear inequalities. Standard phrases:

Read the question carefully , many exam problems pivot on the difference between at least (closed) and more than (open).

Worked examples

Example 1. Solve $5x + 3 \le 2x - 9$.

Subtract $2x$: $3x + 3 \le -9$. Subtract $3$: $3x \le -12$. Divide by $3$: $x \le -4$. Solution: $(-\infty, -4]$.

Example 2. Solve $\dfrac{3x - 2}{5} > \dfrac{x + 1}{2} - 1$.

Multiply by $10$: $2(3x - 2) > 5(x + 1) - 10$. Expand: $6x - 4 > 5x + 5 - 10$. Simplify: $6x - 4 > 5x - 5$. Subtract $5x$: $x - 4 > -5$. Add $4$: $x > -1$. Solution: $(-1, \infty)$.

Example 3. Find all integer solutions of $-3 < 2x - 1 \le 5$.

Add $1$: $-2 < 2x \le 6$. Divide by $2$: $-1 < x \le 3$. Integers in this range: $0, 1, 2, 3$.

Example 4. A solution is to be made by mixing $30$ litres of $30\%$ acid solution with another solution to obtain a $25\%$ acid solution. Find the volume of the second solution if it contains $15\%$ acid, and express the constraint as an inequality if the final volume must be at most $80$ litres.

Let $V$ litres of $15\%$ solution be added. Total acid: $0.30 \cdot 30 + 0.15 V = 9 + 0.15 V$. Total solution: $30 + V$. For $25\%$ concentration: $\dfrac{9 + 0.15 V}{30 + V} = 0.25$, giving $9 + 0.15 V = 7.5 + 0.25 V$, so $V = 15$ litres.

If the constraint is "final volume at most $80$ litres", then $30 + V \le 80$, i.e. $V \le 50$, satisfied by $V = 15$.

Example 5 (harder). A boy needs to score at least $80\%$ in five tests of $100$ marks each. If his scores in the first four tests are $84, 76, 90$ and $73$, find the minimum he needs in the fifth.

Total required: $80\% \cdot 500 = 400$. Already scored: $84 + 76 + 90 + 73 = 323$. Need: $400 - 323 = 77$.

As an inequality with $x =$ score in fifth test: $84 + 76 + 90 + 73 + x \ge 400$, i.e. $x \ge 77$. He needs at least $77$.

Try it yourself

1. Solve $4x + 7 > 19$.
2. Solve $\dfrac{3x + 2}{4} - \dfrac{x - 1}{2} \le 1$.
3. Solve $-7 < 3x + 5 < 8$ and write the integer solutions.
4. Solve $2(x - 3) + 5 \ge 7 - 3(x + 2)$.
5. Solve $\dfrac{5x - 1}{3} > \dfrac{3x + 2}{4}$.
6. A man wants to spend at most $\\₹2000$ on books costing $\\₹150$ each. Find an inequality for the maximum number of books, and solve.
7. The longest side of a triangle is twice the shortest, and the third side is $3$ cm longer than the shortest. If the perimeter is at least $59$ cm, find the minimum length of the shortest side.
8. Solve $\dfrac{x}{2} + \dfrac{x}{3} + \dfrac{x}{6} \ge 5$.
9. Solve $3(2 - x) \ge 2(1 - x)$.
10. Solve $\dfrac{2x + 3}{5} - 1 < \dfrac{x - 4}{3} + 2$.
11. Find all integers satisfying $\dfrac{x - 1}{2} + 3 \le \dfrac{x}{3} + 4$.
12. To pass an exam a student must score at least $35\%$. If the exam is out of $400$, find the minimum mark.

Pitfalls / Tricks

• When the inequality has fractions, multiply through by the positive LCM. If a denominator could be negative, split into cases.
• "At least" includes equality (closed bracket). "More than" excludes (open bracket).
• For compound inequalities like $a < x < b$, both ends must be respected at every step.
• Insight. Once you write the manipulation rules on the back of every page, solving linear inequalities is mechanical , no clever trick needed.