Number Line Graphs, Double Inequalities and Systems

Drawing the Answer: Number Line Graphs

A solution set over the reals is a piece of the number line — so draw it. The standard convention:

• Strict inequality ($<$ or $>$): put an open circle on the boundary number — it is not a solution — and darken the line on the side where solutions lie.
• Slack inequality ($\le$ or $\ge$): put a filled (dark) circle on the boundary — it is a solution — and darken the correct side.

Worked instance: $7x + 3 < 5x + 9$ reduces to $2x < 6$, i.e. $x < 3$ — open circle at 3, dark line to the left. And: $\frac{3x - 4}{2} \ge \frac{x + 1}{4} - 1$ multiplies up to $2(3x - 4) \ge x - 3$, i.e. $6x - 8 \ge x - 3$, so $5x \ge 5$, giving $x \ge 1$ — filled circle at 1, dark line to the right.

[Board Tip] When a question says "show the graph of the solutions," the open-vs-filled circle carries a mark. Match it to the sign before drawing anything.

Double Inequalities: Operate on All Three Parts

A condition like $-8 \le 5x - 3 < 7$ is two inequalities sharing a middle. Solve them together by doing each move to all three parts:

Worked instance: $-8 \le 5x - 3 < 7$. Add 3 throughout: $-5 \le 5x < 10$. Divide by 5 (positive): $-1 \le x < 2$ — the interval $[-1, 2)$.

A negative-divisor instance: $-5 \le \frac{5 - 3x}{2} \le 8$. Multiply by 2: $-10 \le 5 - 3x \le 16$. Subtract 5: $-15 \le -3x \le 11$. Divide by $-3$ — negative, so both signs flip and the chain reverses: $5 \ge x \ge -\frac{11}{3}$, i.e. $-\frac{11}{3} \le x \le 5$.

Key Point: Whatever you do to the middle, do to both ends. Dividing all three parts by a negative flips both inequality signs — then rewrite the chain smallest-to-largest.

[JEE Tip] Keep the variable only in the middle. If $x$ appears in an end as well, split the chain into its two separate inequalities and solve as a system instead.

Systems of Inequalities: Intersect the Solutions

When two (or more) inequalities must hold simultaneously, solve each on its own, then keep the common part — the intersection of the solution sets.

Worked instance: solve $3x - 7 < 5 + x$ and $11 - 5x \le 1$. The first gives $2x < 12$, i.e. $x < 6$. The second gives $-5x \le -10$, flip: $x \ge 2$. Drawing both on one number line, the overlap is $2 \le x < 6$ — the interval $[2, 6)$.

Key Point: System = intersection. Draw each solution set on the same number line; the answer is where all the dark lines overlap. If they never overlap, the system has no solution.

A double inequality is just a system in disguise ($-8 \le 5x - 3$ and $5x - 3 < 7$) — chains are the compact way to write systems whose middle expression is shared.

[Board Tip] Report the final answer with correct brackets: the boundary that came from a slack inequality gets a square bracket, the strict one a round bracket — exactly as in $[2, 6)$ above.

Solved Examples

Example 1: Graph included

Solve $7x + 3 < 5x + 9$ and show the graph of the solutions.

Solution:

Step 1 — Collect. Subtract $5x$ and 3 from both sides: $2x < 6$, so $x < 3$.

Step 2 — Choose the circle. Strict sign — 3 is NOT a solution: OPEN circle at 3.

Step 3 — Darken the correct side. Solutions lie below 3: dark line to the left, i.e. $(-\infty, 3)$.

Takeaway: Sign type decides the circle; direction of the inequality decides the darkened side.

Example 2: Fractions then graph

Solve $\frac{3x - 4}{2} \ge \frac{x + 1}{4} - 1$ and graph the solutions.

Solution:

Step 1 — Simplify the right side first. $\frac{x + 1}{4} - 1 = \frac{x + 1 - 4}{4} = \frac{x - 3}{4}$.

Step 2 — Clear denominators with the positive 4. $2(3x - 4) \ge x - 3$.

Step 3 — Expand and collect. $6x - 8 \ge x - 3$ gives $5x \ge 5$, so $x \ge 1$.

Step 4 — Graph. Filled circle at 1 (slack — included), dark line rightwards: $[1, \infty)$.

Takeaway: Combine each side into one fraction before clearing denominators — fewer terms to track.

Example 3: The basic chain

Solve $-8 \le 5x - 3 < 7$.

Solution:

Step 1 — Add 3 to ALL THREE parts. $-5 \le 5x < 10$.

Step 2 — Divide all parts by the positive 5. $-1 \le x < 2$, i.e. $x \in [-1, 2)$.

Step 3 — Graph. Filled circle at $-1$ (slack end), open circle at 2 (strict end), dark line between.

Takeaway: Whatever you do to the middle, do to both ends — the chain moves as one object.

Example 4: Chain with a negative divisor

Solve $-5 \le \frac{5 - 3x}{2} \le 8$.

Solution:

Step 1 — Multiply all parts by 2. $-10 \le 5 - 3x \le 16$.

Step 2 — Subtract 5 throughout. $-15 \le -3x \le 11$.

Step 3 — Divide by $-3$, flipping BOTH signs. $5 \ge x \ge -\frac{11}{3}$.

Step 4 — Rewrite smallest-to-largest. $-\frac{11}{3} \le x \le 5$, i.e. $x \in \left[-\frac{11}{3}, 5\right]$.

Takeaway: A negative division flips both signs AND forces a reorder — do the two moves together, deliberately.

Example 5: A system with a graph

Solve the system $3x - 7 < 5 + x$ and $11 - 5x \le 1$, and represent the solutions on the number line.

Solution:

Step 1 — Solve the first inequality. $3x - x < 5 + 7$ gives $2x < 12$, so $x < 6$.

Step 2 — Solve the second. $-5x \le 1 - 11 = -10$; divide by $-5$ and FLIP: $x \ge 2$.

Step 3 — Intersect on one number line. The overlap of $x < 6$ and $x \ge 2$ is $2 \le x < 6$, i.e. $[2, 6)$ — filled at 2, open at 6.

Takeaway: Solve each inequality separately; the system's answer is where the dark lines overlap.

Example 6: Both strict

Solve $5x + 1 > -24$ and $5x - 1 < 24$.

Solution:

Step 1 — First inequality. $5x > -25$, so $x > -5$.

Step 2 — Second. $5x < 25$, so $x < 5$.

Step 3 — Intersect. $-5 < x < 5$: the open interval $(-5, 5)$.

Takeaway: Two mirror-image strict conditions produce a symmetric open interval.

Example 7: Flip inside a system

Solve $3x - 7 > 2(x - 6)$ and $6 - x > 11 - 2x$.

Solution:

Step 1 — First inequality. $3x - 7 > 2x - 12$ gives $x > -5$.

Step 2 — Second. $-x + 2x > 11 - 6$ gives $x > 5$.

Step 3 — Intersect. Both point the same way; the STRONGER condition $x > 5$ wins: solution $(5, \infty)$.

Takeaway: When both conditions face the same direction, the intersection is simply the tighter one.

Example 8: Chain with negative multiplier

Solve $6 \le -3(2x - 4) < 12$.

Solution:

Step 1 — Divide all parts by $-3$, flipping both signs. $-2 \ge 2x - 4 > -4$.

Step 2 — Add 4 throughout. $2 \ge 2x > 0$.

Step 3 — Divide by the positive 2. $1 \ge x > 0$.

Step 4 — Reorder. $0 < x \le 1$, i.e. $(0, 1]$ — note the slack end moved to 1 with the flip.

Takeaway: After a flip, track WHICH end carries the slack sign — it travels with the boundary, not with the position in the chain.

Example 9: A heavier system

Solve $5(2x - 7) - 3(2x + 3) \le 0$ and $2x + 19 \le 6x + 47$.

Solution:

Step 1 — Expand and solve the first. $10x - 35 - 6x - 9 \le 0$ gives $4x \le 44$, so $x \le 11$.

Step 2 — Solve the second. $19 - 47 \le 6x - 2x$ gives $-28 \le 4x$, so $x \ge -7$.

Step 3 — Intersect. $-7 \le x \le 11$, i.e. $[-7, 11]$ — both ends slack, both included.

Takeaway: Opposite-facing conditions trap x in a closed interval; brackets record the two slack signs.

Example 10: An empty intersection

Solve the system $x + 3 < 2$ and $2x - 1 > 5$.

Solution:

Step 1 — Solve each. First: $x < -1$. Second: $2x > 6$, so $x > 3$.

Step 2 — Attempt the intersection. No real number is simultaneously below $-1$ and above 3 — the dark lines never overlap.

Step 3 — State the conclusion. Solution set: $\varnothing$.

Takeaway: A perfectly legitimate answer; state it explicitly rather than manufacturing a fake interval.