An inequality is solved by the same operations as an equation, with one exception: multiplying or dividing both sides by a negative number reverses the relation. The answer is a set of numbers rather than a single number, and that changes how it is written and how it is checked.
Core ideas
One thing is genuinely different from solving an equation, and one thing that looks different is not.
3x − 4 < 11 is true at x = 0, at x = 4, at x = 49/10 and at every number below 5; it is false from 5 upward. The answer is x < 5, a description of infinitely many numbers. There is no single value to report and writing x = 5 reports the one number that fails.
You may add or subtract anything at all, including an expression containing x, and the relation is untouched. Multiplication and division are safe only when you know the sign of what you are multiplying by — which is why you may never divide by x.
Multiplying by a negative turns the relation round, and it follows from a < b meaning b − a is positive together with the fact that a positive times a positive is positive. It is proved below in three lines, and knowing the proof is what stops you applying it at the wrong step.
The material
The four relations, the operations that preserve them, and the single case that turns a statement round.
A linear inequality in x is a statement of the form ax + b < c, or the same with ≤, > or ≥ in place of <. Its solution set is the set of all numbers that make the statement true.
Two statements are equivalent when they have the same solution set. Solving means writing a chain of equivalent statements ending in one whose solution set can be read off.
< is strict and ≤ is not. The distinction is not decoration: x ≤ 5 contains 5 and x < 5 does not. It shows up as a filled or a hollow endpoint on the number line, and as a square or a round bracket in “Interval and Set-Builder Notation”.
One example makes the third rule concrete. 3 < 5 is true. Multiply both sides by −1: −3 and −5. On the number line −3 sits to the right of −5, so −3 > −5. The relation had to turn round or the statement would have become false.
If a < b and c < 0, then ac > bc.
The same holds with ≤ and ≥ throughout. Dividing by c is multiplying by 1/c, which has the same sign as c, so division by a negative reverses the relation too.
a < b means b − a is positive. c < 0 means −c is positive.
The product of two positive numbers is positive, so (b − a)(−c) > 0. Expanding the left side gives −bc + ac, which is ac − bc.
So ac − bc > 0, and that is exactly the statement ac > bc.
□
The first route divides by x, whose sign is unknown, and reaches the conclusion that no number works. The second route only subtracts and then divides by 2, and finds that every negative number works. The check settles it. Whenever a step would divide by something containing the variable, collect terms instead.
Both routes are correct and both give x ≤ −2. Route A never divides by a negative, because collecting the variable on the side where its coefficient is larger leaves that coefficient positive. What route A asks instead is that you read −2 ≥ x correctly: turning a statement round left-to-right reverses the relation, exactly as multiplying by −1 does.
2x + 1 > 2x reduces to 1 > 0, which is true whatever x is: every real number is a solution. 2x + 1 < 2x reduces to 1 < 0, which is false whatever x is: no number is a solution. These are “Identities and Contradictions”'s identity and contradiction with a relation in place of the equals sign, and they are read the same way.
Interactive
The same steps, and the one that reverses
Method
Steps 1 to 3 are “Variables on Both Sides” unchanged. Step 4 is what the relation adds.
Multiplying every term by a positive common denominator is safe and leaves the relation alone. A negative common denominator would reverse it, so choose a positive one and the question never arises.
Compare the two coefficients and move the smaller one across. This is optional, and it removes the reversal step entirely, which is worth the two seconds of thought it costs.
Adding and subtracting never reverse. Dividing by a positive never reverses. Only a negative multiplier or divisor does, and when it does, it always does.
A reversed sign is invisible in the algebra and obvious in the test. The inside number must make the original statement true and the outside number must make it false; testing only inside numbers cannot detect an answer that is too large.
(1/2)x + 1 ≤ (1/3)x. The least common denominator of 2 and 3 is 6, and 6 is positive, so clearing it changes nothing about the relation.
Three tests, not one. x = −6 is the boundary and confirms that ≤ rather than < is right. x = −12 is inside the answer and must work. x = 0 is outside it and must fail; without that third test an answer that is too large looks exactly like a correct one.
Multiply the original statement through by −1 and it becomes −(1/2)x − 1 ≥ −(1/3)x, with the relation turned round. Clearing that one by multiplying by −6 turns the relation round a second time and returns 3x + 6 ≤ 2x. Two reversals cancel, which is a useful thing to have seen once.
For a faded solve, start 7 − 4x > −9 from the supplied first line −4x > −16. Before dividing, write the relation you expect afterwards and why. Finish the solve, then test the boundary, one number inside and one outside against the original statement.
Solve 3 − 5x ≤ 18 and 4x + 1 > 2x + 9 without the lab. Point to the step that reverses the first relation and explain why no step reverses the second. For each answer, test its boundary plus one number from either side in the original statement.
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