Solve a linear inequality exactly like an equation: simplify, then isolate the variable by adding, subtracting, multiplying, or dividing on both sides. The one exception is that multiplying or dividing both sides by a negative number flips the inequality sign (< becomes > and vice versa). Once solved, show the answer on a number line (open circle for < or >, closed circle for ≤ or ≥), or list the integer solutions if the question asks for them.
Linear inequalities feel like solving linear equations, and for most of the working, they are. That similarity is exactly the trap: students carry the equation habit all the way through, including the one step where it breaks. Divide both sides of −2x > 6 by −2 without flipping the sign, and you get x > −3 instead of the correct x < −3. The whole solution set ends up backwards, and every integer solution listed after that is wrong.
The second common loss is smaller but just as costly: mixing up open and closed circles on the number line, or including a boundary value in a list of integers when the inequality is strict (< or >) and should have excluded it.
Step 1
Simplify both sides (expand brackets, collect like terms)
Step 2
Move terms across, exactly as for an equation
Step 3
Multiplying or dividing by a negative? Flip the sign
Step 4
Show on a number line, or list integer solutions
For a double inequality such as −3 < 2x + 1 ≤ 7, apply every operation to all three parts (left, middle, right) at the same time, in the same order you would for a single inequality. The moving-terms-across step is exactly the same skill used to change the subject of a formula, so if that step feels shaky, revisit it first.
A strict inequality (< or >) gets an open circle at the boundary, because that exact value is not included. A weak inequality (≤ or ≥) gets a closed (filled) circle, because the boundary value is included. Shade or arrow the line in the direction of the values that satisfy the inequality.
Solve −3 < 2x + 1 ≤ 7 and list the integer values of x that satisfy it.
Solution
−3 − 1 < 2x + 1 − 1 ≤ 7 − 1−4 < 2x ≤ 6−2 < x ≤ 3x is greater than −2 (not equal to −2) and up to and including 3.x = −1, 0, 1, 2, 3.Quick check: pick a value from the middle of your solution, such as x = 1, and substitute it back into the original inequality: 2(1) + 1 = 3, and −3 < 3 ≤ 7 is true. Also try a boundary that should fail: x = −2 gives 2(−2) + 1 = −3, and −3 < −3 is false, confirming −2 is correctly excluded.
Solve 3 − 5x < 18 and state the smallest integer value of x that satisfies it.
Solution
3 − 5x − 3 < 18 − 3−5x < 15< to >:x > −3x strictly greater than −3, so −3 itself is not included.x = −2.Quick check: substitute x = −2 into the original: 3 − 5(−2) = 3 + 10 = 13, and 13 < 18 is true. Now try the excluded boundary x = −3: 3 − 5(−3) = 3 + 15 = 18, and 18 < 18 is false, so −3 correctly fails and −2 is indeed the smallest solution.
The step students get wrong
Forgetting to flip the inequality sign when multiplying or dividing by a negative number. It happens most often when the negative sign is attached to the coefficient of x, as in −5x < 15, because the "divide by −5" step looks routine. Before you divide or multiply, check the sign of the number on both sides of the inequality: if it is negative, flip < to >, or ≤ to ≥, in the same step you do the division. Doing it as a separate afterthought is where it gets forgotten.
Do I flip the sign when I add or subtract a negative number?
No. The flip rule only applies to multiplying or dividing both sides by a negative number. Adding or subtracting any number, positive or negative, never changes the direction of the inequality.
How do I know whether to use an open or closed circle on the number line?
Strict inequalities (<, >) use an open circle because the boundary value itself does not satisfy the inequality. Weak inequalities (≤, ≥) use a closed, filled-in circle because the boundary value does satisfy it and is part of the solution.
For a double inequality, what happens to the middle part if I multiply by a negative?
Both inequality signs flip, and the order of the two outer numbers effectively reverses so the smaller value is written on the left. For example, 2 ≤ x ≤ 5 multiplied throughout by −1 becomes −2 ≥ −x ≥ −5, which is normally rewritten as −5 ≤ −x ≤ −2 so the inequality reads left to right in the usual increasing order.
Is "list the integer solutions" the same as "state the smallest (or greatest) integer solution"?
No, and E-Math papers ask both versions. "List the integer solutions" wants every integer in the range written out. "State the smallest (or greatest) integer" wants a single value: the first integer inside the solution set from that end, remembering that a strict inequality excludes its own boundary integer if that boundary happens to be a whole number.
Does this method still work if the inequality has x on both sides?
Yes. Collect all the x terms on one side first, the same way you would when solving simultaneous equations by elimination: subtract the smaller x term from both sides so you are left with a positive coefficient wherever possible, then finish isolating x as usual, checking the sign of whatever you multiply or divide by at the final step.
— Mr Gan Math Tuition
Mr. Gan works with students who want a method they can trust under exam pressure, not one they second-guess at the negative-sign step.
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