Solve it almost exactly like an equation — then remember the answer is a whole stretch of line.
Point 1 of 4 in this lesson: 12.3.1 Meeting the linear inequality
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.
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.
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.
Linear: 3x − 7 ≤ 0, 5 > 2 − x, x2 + 1 < x. Not linear: x² > 9 (squared — that's 12.5), 1x < 2 (x downstairs).
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.
2x + 1 > 5 → subtract 1 → 2x > 4 → divide by +2 (positive, keep symbol) → x > 2.
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.)
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.
x − 12 < x + 23 → ×6 (positive) → 3(x − 1) < 2(x + 2) → 3x − 3 < 2x + 4 → x < 7.
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.
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.
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.
Hollow dot ⇔ strict < / > (endpoint excluded). Filled dot ⇔ ≤ / ≥ (endpoint included). The arrow points toward the side that makes the inequality true.
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.
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.
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.
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.
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.
Six questions to lock it in. Tap the answer you think is right.
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).