2.1.11 · D5Algebra — Introduction & Intermediate
Question bank — Inequalities — linear, solving, number line representation
This is a reasoning drill, not an arithmetic drill. Every item below targets a misconception or a boundary case that linear inequalities love to hide. Read the question, cover the answer, and force yourself to say why out loud before you reveal. Bare "yes/no" is not an answer here — the reasoning is the point.
Prerequisites this page leans on: the parent topic, Number Line and Real Numbers (direction and ordering), and later ideas from Absolute Value Equations and Inequalities and Systems of Inequalities.
Picture the four building blocks first
Before the traps, look at the picture your eye should reach for. Every linear inequality solution is one of these four shaded rays — an open or closed dot at the boundary, and an arrow showing the direction.
The single most important visual on this page is what a negative multiplier does: it reflects the whole line through zero, so the arrow that pointed left now points right. That reflection is the flip rule.
True or false — justify
Multiplying both sides of by requires flipping the sign.
False — is positive, so order is preserved; you only flip when the multiplier is negative. Result stays .
If then .
False — negating reflects the number line, so the order reverses: . E.g. but .
The solution of is always an interval that stretches to on the left.
False — the direction depends on the sign of . If you get (opens left), but if the flip makes it (opens right).
and describe the same set of solutions.
False — they differ by exactly one point, . The first includes (closed circle), the second excludes it (open circle).
Squaring both sides of gives the equivalent inequality .
False — squaring is not order-preserving across signs. satisfies but is false, so squaring loses solutions.
If , the inequality has no solution because the coefficient of is zero.
False — is true for every , so the solution is all real numbers. A zero coefficient can give "always true" or "always false", never "no info".
Adding the same expression to both sides is always safe, even a negative one like .
True — addition (of any real number, positive or negative) never flips the sign; only multiplying/dividing by a negative flips.
can be rewritten as without changing its meaning.
True — reading the same statement from the other end preserves it; the pointy end still opens toward , so is still the larger side.
The interval and the interval together cover every real number exactly once.
True — the first stops at and includes , the second starts strictly after ; no gap, no overlap, so together they partition .
If two numbers satisfy , their average also satisfies .
True — the region is a single unbroken ray, so any point between two of its members is also inside it.
Spot the error
Where is the mistake? .
Divided by without flipping. Correct: flip the sign to get . Check : ✓, and ✓.
Where is the mistake? ", so I draw an open circle at and shade left."
includes the boundary, so the circle must be closed (filled). Open circles are only for strict or .
Where is the mistake? .
First, the expression only makes sense when , so is excluded from the start (you can't divide by zero). Second, you multiplied both sides by without knowing its sign, so you don't know whether to flip — and you silently threw away the whole negative- region. It needs two cases: Case (multiply, no flip): , combined with gives . Case (multiply, flip): , which contradicts , so nothing there. Correct solution: — note the round brackets exclude both (undefined) and (strict) — see the figure below.
Where is the mistake? ", add : , so ."
Arithmetic slip: , not . Correct is , giving .
Where is the mistake? ", subtract : , divide by : ."
The subtraction is fine, but dividing by must flip the sign: the answer is , not .
Where is the mistake? "The solution to is the interval ."
There is no such interval — no number is both greater than and less than . The compound statement is a contradiction, so the solution set is empty ().
Where is the mistake? " has interval notation ."
Strict excludes the endpoint, so it must be a round bracket: . Also, always takes a round bracket since it is not a reachable value.
Why questions
Why does multiplying by a negative flip the inequality but multiplying by a positive does not?
A positive scale stretches the number line without reordering it; a negative scale reflects it through zero, so what was to the left is now to the right — the order literally reverses.
Why can't we ever write a closed bracket next to , as in 's left side?
is not a real number you can reach or include — it is a direction, not a point. There is no "largest" or "smallest" value to close onto, so it always gets a round bracket.
Why is the solution of a linear inequality a whole interval, while a linear equation usually gives just one number?
An equation demands exact balance (), satisfied at one crossing point; an inequality accepts everything on one side of that crossing point, which is a continuous range.
Why do we say adding to both sides is not "multiplying by a negative"?
Adding shifts every point on the line by the same amount, keeping their relative order intact. Multiplying by a negative scales and reflects — it's the reflection, not the negative number itself, that flips the sign.
Why does checking a single test point confirm your whole solution set is correct?
Because the boundary splits the line into two rays, and every point within one ray behaves identically. One correct point inside plus one correct point outside pins down which ray is the answer.
Why is solved as rather than ?
Subtract to get ; dividing by flips the strict inequality, giving . The flip is the whole story.
Two-sided inequalities
How do you solve correctly in one pass?
Treat it as one object with three parts and do every operation to all three at once. Subtract : . Divide by (positive, no flip): . Solution: — open at , closed at .
In , what happens to both inequality signs when you divide by ?
Both flip at the same time, and the ends swap roles. Dividing all three parts by : , i.e. , the interval .
Where is the mistake? ", so gives and I stop there."
You solved only the left half and forgot the right half . A two-sided inequality needs both bounds; the answer is the overlap , not the ray .
Why must always have to have any solutions?
has to be simultaneously above and below ; if those two demands never overlap, so the solution is — exactly the "" trap.
Edge cases
What is the solution of ?
is true regardless of , so every real number works: the solution is , all of .
What is the solution of ?
is false and has no to fix it, so no value works: the solution set is empty, .
What is the solution of ?
Every number is less than or equal to itself, so it is true for all real : the solution is .
What is the solution of ?
No number is strictly less than itself, so this is never true: the solution set is empty ().
Solve .
Both sides are identical, so it reduces to "true always"; the solution is all real numbers — no need to move any term.
If the coefficient in equals zero, is it still a linear inequality?
No — the definition requires . With there is no variable term to solve; it collapses to a constant comparison that is either always true or always false.
What does the number line for and (both at once) look like?
Nothing is shaded — the two rays open in opposite directions and share only the single point , which each one excludes, so the combined solution is empty ().
What is the solution of ?
Multiply by (flipping) gives , which is always true; equivalently the original is an equality-inclusive statement true for every real .
Connections
- Parent: Linear Inequalities — the rules these traps are built on.
- Number Line and Real Numbers — why "direction" and "reflection" explain the flip.
- Absolute Value Equations and Inequalities — where hidden sign cases (like the trap) become routine.
- Systems of Inequalities — the "empty set" and "all reals" edge cases reappear when combining constraints.
- Quadratic Inequalities — where the "squaring is not order-preserving" trap matters most.
Recall Quick self-audit before any inequality problem
- Did I multiply or divide by anything negative? → flip (and in a two-sided inequality, flip both signs).
- Is my endpoint included ( → closed dot, square bracket) or excluded ( → open dot, round bracket)?
- Am I dividing by a variable whose sign I don't know (or that could be )? → split into cases and exclude any value that makes a denominator zero.
- If the -term vanished, is the leftover comparison always true (all ) or always false ()?
- Pick two test points on purpose: choose one number clearly inside your claimed answer (e.g. for pick ) and one clearly outside (e.g. ). Substitute both into the original inequality — the inside point must make it true, the outside point must make it false. For a two-sided answer like , also test a value just past each end (say and ) to confirm both boundaries.