An equation pins x to one point. The inequality opens it up to a whole stretch of the line — every x to the right of 2.
You already know how to solve a balance — sorry, an equation. You shuffle terms around until the unknown sits alone, and out pops a single number. Linear inequalities work almost the same way, with two twists. The answer is not one number but a whole stretch of the number line. And one move — dividing both sides by a negative — secretly turns the whole thing around. Get those two ideas right and the rest is pure equation-solving.
By the end of this lesson you will be able to: recognize a linear inequality, solve one step by step (clearing fractions, transposing, dividing — flipping the symbol exactly when you divide by a negative), draw its solution as a ray on the number line, and reason through an inequality whose coefficient is a letter. Throughout, we keep the same colors: the solution set is green, the sign-flip trap is red, a boundary value is amber, and the number line and the variable are blue.
12.3.1 Meeting the linear inequality
Take any linear equation — say 2x + 1 = 5 — and swap the = for an inequality symbol: <, >, ≤, or ≥. You get a linear inequality in one variable, like 2x + 1 > 5. "Linear" means the unknown shows up only to the first power — no x², no x under a root, no x in a denominator. Just x itself, scaled and shifted.
An equation asks one tight question: for which single value is the two sides exactly equal? An inequality asks a roomier one: for which values is the left side bigger (or smaller)? Usually many values work at once. The collection of all the x that make the inequality true is called its solution set.
You could find the solution set by brute force — plug in numbers and keep the winners. Here 1 and 2 fail; 3 passes. The pattern is "everything bigger than 2." Next section finds that in one clean sweep.
Key idea
A linear inequality is a linear equation with the = replaced by <, >, ≤, or ≥. Its answer is a solution set — usually a whole range of values, not just one.
Slide x to plug it into 2x + 1 > 5. Watch which values pass. The green stretch is the solution set forming itself.
x =1
12.3.2 The steps (just like an equation, with one watch-point)
Here is the whole recipe. It is the same recipe you used for equations in Stage 10, with a single extra rule slipped into the last step.
1. Clear fractions — multiply both sides by a common denominator so nothing is stacked. 2. Remove parentheses — distribute. 3. Transpose — move variable terms to one side, plain numbers to the other (adding or subtracting never changes the direction). 4. Combine like terms. 5. Divide by the coefficient of x — and here is the watch-point: if that coefficient is negative, flip the inequality symbol.
Let's run the simplest case. Solve 2x + 1 > 5. Subtract 1 from both sides — direction untouched — to get 2x > 4. Divide by 2, which is positive, so no flip: x > 2. Done. The solution set is every number bigger than 2, exactly the pattern we noticed by hand.
Now the move everyone fears. Solve 3 − 2x ≥ 7. Subtract 3: −2x ≥ 4. The coefficient of x is −2, a negative. Divide both sides by −2 and the symbol must turn around — ≥ becomes ≤: x ≤ −2. Why does it flip? Because multiplying by a negative is a mirror: it sends every number to the opposite side of zero, so "bigger" and "smaller" trade places. (That is the property you met in 12.2.)
The single red step is the only place the symbol changes. Adding, subtracting, and dividing by a positive all leave it alone.
Finally, fractions. Solve x − 12 < x + 23. The common denominator of 2 and 3 is 6, a positive number, so multiplying both sides by 6 does not flip anything: 3(x − 1) < 2(x + 2). Distribute: 3x − 3 < 2x + 4. Transpose the 2x left and the −3 right: x < 7. No negative divisor ever appeared, so no flip.
Worked — fractions
x − 12 < x + 23 → ×6 (positive) → 3(x − 1) < 2(x + 2) → 3x − 3 < 2x + 4 → x < 7.
Watch out
Two traps. (a) Flip only when you multiply or divide by a negative — never for adding, subtracting, or scaling by a positive. (b) When you clear a fraction, multiply every term by the common denominator, including the ones with no fraction.
🎮 Try it SOLVE-IT STEPPER
Pick an inequality, then reveal it one step at a time. The divide step turns red and flips only when the coefficient is negative. It finishes with the picture.
solve
12.3.3 The solution set and the number line
An equation's answer is a dot; an inequality's answer is a ray — a stretch of line shooting off in one direction. To draw it you need just two decisions: which way does the ray point, and is the endpoint itself in or out?
Take x > 2. Everything to the right of 2 works, so shade rightward. But is 2 itself a solution? No — 2 is not greater than 2 — so we mark the endpoint with an open (hollow) circle to say "right up to here, but not including this point." Strict symbols < and > always get an open dot.
The solution of x > 2: an open dot at 2 (it's excluded), shaded ray to the right.
Now x ≤ −2. Everything to the left of −2 works, so shade leftward. And −2 itself? The symbol is ≤ — "less than or equal to" — so −2 is a solution. We mark it with a filled circle. Inclusive symbols ≤ and ≥ always get a filled dot.
The solution of x ≤ −2: a filled dot at −2 (it's included), shaded ray to the left.
Open or filled?
Hollow dot ⇔ strict < / > (endpoint excluded). Filled dot ⇔ ≤ / ≥ (endpoint included). The arrow points toward the side that makes the inequality true.
Reading it back
A picture and a sentence say the same thing. "x > 2" reads as "every number greater than 2," and it is also written (2, ∞) in interval notation — round bracket because 2 is out. "x ≤ −2" is (−∞, −2] — square bracket because −2 is in.
🎮 Try it SET → PICTURE
Choose a relation, then flip between strict and inclusive to watch the endpoint dot fill and empty. Feel the difference between > and ≥.
direction
endpoint
boundary a2
12.3.4 When the coefficient is a letter
Sometimes the number in front of x is itself unknown — a letter like a. Consider ax > 1. You might want to "divide by a" and be done. But stop: you cannot divide by a until you know its sign, because a negative divisor would flip the symbol and a zero divisor isn't allowed at all. So we split into three honest cases.
Case a > 0. Dividing by a positive keeps the direction: x > 1a. Plain and clean.
Case a < 0. Dividing by a negative flips it: x < 1a. Same arithmetic, mirrored symbol — the watch-point of 12.3.2 returning in disguise.
Case a = 0. Now there is no x left to divide out: the inequality collapses to 0 > 1, which is plainly false. No value of x can rescue it, so the solution set is empty — no solution.
One inequality, three answers, decided entirely by the sign of a. This "ask the sign first" habit returns whenever a coefficient is unknown.
Watch out
Never divide an inequality by a letter whose sign you don't know. A positive keeps the symbol, a negative flips it, and a zero can wipe out the variable entirely — leaving either "always false" (no solution) or, with a different number, "always true" (all reals). Decide the cases first, then divide.
🎮 Try it LETTER-COEFFICIENT EXPLORER
Slide the coefficient a through positive, zero, and negative for a·x > 1. Watch which case fires and how the solution ray — or the empty line — responds.
a =2
★ The big ideas, in one breath
Solving a linear inequality is solving the matching equation — clear fractions, remove parentheses, transpose, combine, divide — with one extra rule: flip the symbol the instant you divide (or multiply) both sides by a negative, and never otherwise. The answer is not a single number but a ray on the number line: arrow toward the true side, with an open dot for strict </> and a filled dot for inclusive ≤/≥. And when the coefficient is a letter, ask its sign first — positive, negative, or zero — before you dare to divide.
What's next
One inequality gives a ray. 12.4 puts several conditions together and keeps only where their rays overlap. Then 12.5 lets the unknown be squared, and you'll read the answer straight off a parabola. The flip rule you just met never goes away — it shows up in every lesson ahead.
✎ Exercises 12.3
Solve and graph: 3x − 4 > 5.
Show answer
Add 4: 3x > 9. Divide by +3 (no flip): x > 3 — open dot at 3, ray right.
Subtract 1: 2x > 4. Divide by +2 (no flip): x > 2.
Solve: 3 − 2x ≥ 7. Where exactly does the flip happen?
Show answer
Subtract 3: −2x ≥ 4. Divide by −2 — the symbol flips, ≥ → ≤: x ≤ −2. The flip happens at the divide-by-negative step, nowhere else. Filled dot at −2, ray left.
Solve: −x < 4.
Show answer
Multiply both sides by −1 (a negative), so flip: x > −4. Open dot at −4, ray right.
Multiply by 6 (positive — no flip): 3(x − 1) < 2(x + 2), i.e. 3x − 3 < 2x + 4. Transpose: x < 7.
A student wrote: "5 − 3x ≤ 8, subtract 5, −3x ≤ 3, divide by −3, x ≤ −1." Find and fix the mistake.
Show answer
Dividing by −3 must flip the symbol. Correct: x ≥ −1 (not x ≤ −1). Filled dot at −1, ray right.
Solve: 2(x − 3) ≥ 5x.
Show answer
Distribute: 2x − 6 ≥ 5x. Transpose 5x left and −6 right: −3x ≥ 6. Divide by −3 (flip): x ≤ −2.
Write the inequality shown by this picture, then say if 2 is a solution: a filled dot at 2 with the ray going left.
Show answer
Filled = inclusive, ray left = "less than." So x ≤ 2. Yes — 2 is a solution, because the dot is filled (≤ includes the boundary).
Solve for x in terms of the letter a: a·x > 1. List all cases.
Show answer
If a > 0: x > 1/a. If a < 0: x < 1/a (flip). If a = 0: becomes 0 > 1, false → no solution.
Challenge. For which values of the letter m does the inequality m·x < 2 have every real number as a solution?
Show answer
Test the cases. If m ≠ 0 the answer is a ray (x < 2/m or x > 2/m), never all reals. If m = 0 it becomes 0 < 2, which is always true. So m = 0 is the only value making every real number a solution.
🎯 Quick check
Six questions to lock it in. Tap the answer you think is right.
§ For teachers and parents
This lesson addresses CCSS 7.EE.B.4b and A-REI.B.3 (solve word problems leading to inequalities of the form px + q > r or px + q < r; solve linear inequalities in one variable and graph the solution set), and it builds on 6.EE.B.8 (writing an inequality and recognizing its solutions as a set on the number line). The unifying message is that solving an inequality reuses every move from solving the matching equation, plus exactly one new rule. The #1 misconception is flipping at the wrong time: students either forget to reverse the symbol when dividing by a negative, or wrongly reverse it when merely subtracting — and many report the answer as a single number instead of a set. The antidote, repeated like a mantra: solve it like an equation, flip only at a divide-by-negative, and always picture the result as a stretch on the line (open dot for strict, filled for inclusive). Ask "would the boundary value itself make it true?" to settle every open-versus-filled question. For more practice, pair this with 12.2 (where the flip rule is introduced) and 12.7 (where these solutions answer real questions).
eastmath.com · Stage 12 · 12.3 Solving Linear Inequalities · Intuition before notation
eastmath.com · 12.3 Solving Linear Inequalities · 12.3.4 When the coefficient is a letter