Worked examples — Inequalities — linear, solving, number line representation

This page is the firing range. The parent note taught you the three rules (add/subtract freely, multiply/divide by positive keeps the sign, multiply/divide by negative flips it). Here we hunt down every type of situation those rules can produce, so no exam question can surprise you.

Before any solving, let's list the enemies.

The scenario matrix

Every linear inequality you will ever meet lives in one of these cells. Read this table like a checklist — by the end, every row will have a worked example with its name attached.

Here is a map of what "flipping" actually does to the number line — the geometric heart of cells B, C and I.

The examples

Cell A — Positive coefficient

Forecast: dividing by $+5$ — do you expect the sign to flip? (No — hold that thought.)

1. Subtract 4 from both sides → $5x\ge 15$. Why this step? Adding or subtracting the same amount slides both sides equally along the number line, so order is preserved (no flip ever from $\pm$).
2. Divide both sides by $5$ → $x\ge 3$. Why this step? $5$ is positive, so scaling keeps left-of/right-of relationships intact. Sign stays $\ge$.
3. Draw it: closed circle at $3$ (because $\ge$ includes the edge), arrow to the right.

Verify: test $x=3$: $5(3)+4=19\ge 19$ ✓ (boundary included). Test $x=4$: $24\ge 19$ ✓. Test $x=0$: $4\ge 19$ is false, correctly excluded. Answer: $x\in [3,\infty )$.

Cell B — Negative coefficient (the flip)

Forecast: we'll divide by $-7$. Predict the final symbol before reading on.

1. Subtract 2 → $-7x<28$. Why this step? Pure subtraction, order preserved.
2. Divide by $-7$ → FLIP → $x>-4$. Why this step? Dividing by a negative reflects the whole number line about $0$ (see figure s01): what was "less than" becomes "greater than".
3. Draw it: open circle at $-4$ (strict $>$), arrow to the right.

Verify: pick $x=0$ (should work since $0>-4$): $-7(0)+2=2<30$ ✓. Pick $x=-5$ (should fail): $-7(-5)+2=37<30$ is false ✓ correctly excluded. Answer: $x\in (-4,\infty )$.

Cell C — Variable on both sides

Forecast: after we gather the $x$'s, will the coefficient be positive or negative? That decides everything.

1. Subtract $9x$ from both sides → $-5x-3\ge 12$. Why this step? Move all $x$-terms to one side; subtraction never flips.
2. Add 3 → $-5x\ge 15$. Why this step? Isolate the $x$-term.
3. Divide by $-5$ → FLIP → $x\le -3$. Why this step? The collected coefficient turned out negative, so the flip fires — this is why we cared about the sign back in the forecast.
4. Draw it: closed circle at $-3$, arrow left.

Verify: test $x=-3$: LHS $4(-3)-3=-15$, RHS $9(-3)+12=-15$, and $-15\ge -15$ ✓ (edge holds). Test $x=-4$: LHS $-19$, RHS $-24$, $-19\ge -24$ ✓. Test $x=0$: LHS $-3$, RHS $12$, $-3\ge 12$ false, correctly out. Answer: $x\in (-\infty ,-3]$.

Cell D — Fraction / denominator

Forecast: the denominator itself is negative. Multiplying both sides by it will flip. Guess the direction now.

1. Multiply both sides by $-3$ → FLIP → $6-2x<-12$. Why this step? We cleared the denominator, but $-3$ is negative, so the $>$ flips to $<$.
2. Subtract 6 → $-2x<-18$. Why this step? Isolate the $x$-term; subtraction is flip-free.
3. Divide by $-2$ → FLIP again → $x>9$. Why this step? A second negative division — flip once more, back to $>$.

Verify: test $x=10$: $\frac{6-20}{-3}=\frac{-14}{-3}\approx 4.67>4$ ✓. Test $x=9$: $\frac{6-18}{-3}=\frac{-12}{-3}=4$, and $4>4$ is false, so $9$ correctly excluded (open boundary). Test $x=0$: $\frac{6}{-3}=-2>4$ false, correctly out. Answer: $x\in (9,\infty )$.

Cell E — Compound "sandwich" inequality

Forecast: this is two inequalities glued together ($-5\le 2x-1$ and $2x-1<7$). Every operation must be applied to all three parts. Guess the final interval.

1. Add 1 to all three parts → $-4\le 2x<8$. Why this step? Whatever we do to the middle we must do to both walls, or the sandwich breaks.
2. Divide all three by $2$ → $-2\le x<4$. Why this step? $2$ is positive, so no flip. (If it were negative, all three would flip and the walls would swap ends.)
3. Draw it: closed circle at $-2$, open circle at $4$, shade between.

Verify: test $x=-2$: $2(-2)-1=-5$, and $-5\le -5<7$ ✓ (left edge in). Test $x=4$: $2(4)-1=7$, and $7<7$ false, so right edge correctly out. Test $x=0$: $-1$, and $-5\le -1<7$ ✓. Answer: $x\in [-2,4)$.

Cell F — Degenerate cases (no solution / all reals)

Forecast: in each, watch what happens to the $x$-terms. Sometimes they cancel entirely — then the truth depends only on the leftover numbers.

(a)

1. Subtract $3x$ → $5<1$. Why this step? Collect the $x$'s; they annihilate.
2. Read it: $5<1$ is false for every $x$. No value of $x$ can rescue it.

Answer (a): no solution, $x\in \varnothing$ (empty set).

(b)

1. Expand → $2x+8\ge 2x+1$. Why this step? Distribute before collecting.
2. Subtract $2x$ → $8\ge 1$. Why this step? $x$-terms cancel again.
3. Read it: $8\ge 1$ is true for every $x$.

Answer (b): all real numbers, $x\in (-\infty ,\infty )$.

Verify: (a) try $x=100$: $305<301$ false; try $x=-100$: $-295<-299$ false — nothing works ✓. (b) try $x=100$: $208\ge 201$ ✓; try $x=-100$: $-192\ge -199$ ✓ — everything works ✓.

Cell G — Boundary / limiting value

Forecast: the answer is an interval with one wall. Guess whether the wall is a filled or hollow dot.

1. Add 6 → $2x\le 6$. Why this step? Isolate the $x$-term.
2. Divide by $+2$ → $x\le 3$. Why this step? Positive divisor, no flip.
3. Interpret the limit: because the symbol is $\le$ (not $<$), the boundary $3$ is attained — closed circle. The largest allowed value is exactly $3$.

Verify: at $x=3$: $2(3)-6=0\le 0$ ✓ (edge included). At $x=3.0001$: $0.0002\le 0$ false — just past the edge it dies, confirming $3$ is the exact ceiling. Answer: $x\in (-\infty ,3]$; maximum value $=3$, reached.

Cell H — Real-world word problem

Forecast: translate first. Earnings $=120+18n$. Guess the smallest whole $n$.

1. Set up the inequality → $120+18n>300$. Why this step? "More than ₹300" is strict $>$; her earnings must exceed it.
2. Subtract 120 → $18n>180$. Why this step? Isolate the parcel term.
3. Divide by $+18$ → $n>10$. Why this step? Positive divisor, no flip.
4. Interpret units: $n$ is a whole number of parcels and must be strictly greater than 10, so the smallest valid integer is $n=11$. Why this step? $n=10$ gives exactly ₹300, which is not "more than", so we step up to the next whole parcel.

Verify: $n=11$: $120+18(11)=120+198=318>300$ ✓. $n=10$: $120+180=300$, and $300>300$ is false ✓ (correctly excluded). Answer: at least 11 parcels.

Cell I — Exam twist: coefficient of unknown sign

Forecast: you cannot just "divide by $k$" — because you don't know if $k$ is positive or negative, you don't know whether to flip! This is exactly the forehead question from the intro.

1. Case $k>0$: divide by $k$ (positive) → $x>\frac{4}{k}$. No flip. Why this step? Positive divisor preserves order.
2. Case $k<0$: divide by $k$ (negative) → $x<\frac{4}{k}$. Flip. Why this step? Negative divisor reflects the number line.
3. Case $k=0$: excluded by "$k\ne 0$"; note that $0\cdot x=0>4$ is never true, so it would give no solution — that's why the problem forbade it.

Verify: take $k=2$ (so case 1, answer $x>2$): test $x=3$ → $2(3)=6>4$ ✓; test $x=1$ → $2>4$ false ✓. Take $k=-2$ (case 2, answer $x<-2$): test $x=-3$ → $-2(-3)=6>4$ ✓; test $x=0$ → $0>4$ false ✓. Answer: $x>\frac{4}{k}$ if $k>0$; $x<\frac{4}{k}$ if $k<0$.

Recall checkpoint

Connections

• Parent topic — the three core rules these examples exercise
• Linear Equations — solve the "equality version" first, then decide the direction
• Absolute Value Equations and Inequalities — next step: sandwich cases (Cell E) appear constantly there
• Systems of Inequalities — combine several single-variable answers into a region
• Quadratic Inequalities — same sign-flip logic, but the boundary is curved
• Number Line and Real Numbers — the geometry behind open/closed circles and the reflection-flip