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⚔️ Algebra Kingdom · Algebra

Linear Inequalities

Solve inequalities the way you solve equations, with one extra rule: multiplying or dividing by a negative flips the sign.

In short

  • An inequality describes a whole range of numbers, not a single value; that range is the solution set.
  • Adding, subtracting, and multiplying or dividing by a positive number all leave the direction of the sign alone.
  • Multiplying or dividing both sides by a negative number reverses the direction of the sign — this is the only rule that differs from equations.
  • On a number line an open circle excludes the boundary (< and >) and a filled circle includes it (<= and >=).
  • In a word problem, decide whether to round up or down by asking which way keeps the constraint true.

An answer that is a whole range

An equation such as x + 4 = 9 has exactly one answer: 5. An inequality such as x + 4 > 9 has infinitely many. Every number bigger than 5 works — 5.1, 6, 40, a million.

The four symbols are:

  • x > 3 — x is greater than 3 (3 itself does not count)
  • x < 3 — x is less than 3
  • x >= 3 — x is greater than or equal to 3 (3 does count)
  • x <= 3 — x is less than or equal to 3

The set of all the numbers that work is called the solution set. When you solve an inequality you are not looking for a number, you are describing a region of the number line.

Solve it like an equation

An inequality behaves like a balance that is tipped rather than level. Adding the same weight to both pans, or doubling both pans, does not change which side is heavier. So the ordinary rules still work:

  • You may add the same number to both sides.
  • You may subtract the same number from both sides.
  • You may multiply or divide both sides by the same positive number.

Solve 3x + 5 <= 20 exactly as you would solve 3x + 5 = 20. Subtract 5 from both sides to get 3x <= 15, then divide both sides by 3 to get x <= 5. The direction of the sign never moved, because 3 is positive.

Just as with equations, you undo in the reverse order: undo the adding and subtracting first, then the multiplying and dividing.

The one extra rule

Here is the rule that makes inequalities different, and it is the only one.

Multiplying or dividing both sides by a negative number reverses the direction of the sign.

You can see why with numbers you already trust. Start with a true statement:

  • 2 < 6

Now multiply both sides by -1:

  • -2 and -6. Is -2 < -6? No — -2 is to the right of -6 on the number line, so -2 > -6.

Multiplying by a negative reflects both numbers to the other side of zero, and reflecting swaps which one is further right. So the sign has to turn around with them.

So -4x > 12 becomes x < -3, not x > -3. Note that moving a term across (adding or subtracting) never triggers this — only multiplying or dividing by a negative does.

Showing the answer on a number line

A solution set is easiest to see drawn.

  • Mark the boundary number.
  • Use an open circle for < or > , because the boundary itself is not a solution.
  • Use a filled circle for <= or >=, because the boundary is included.
  • Shade to the right for > and >=, and to the left for < and <=.

So x >= -2 is a filled circle at -2 with everything to the right shaded, and x < 4 is an open circle at 4 with everything to the left shaded.

The picture is also a check. Pick any number in the shaded part, put it back into the original inequality, and it must give a true statement. Pick one from the unshaded part and it must give a false one. Two quick tests catch a flipped sign immediately.

Inequalities in words

Most real constraints are inequalities rather than equations, because money, time and space run out.

  • "at most", "no more than", "up to", "can afford" mean <=
  • "at least", "no fewer than", "must reach" mean >=
  • "under", "less than", "fewer than" mean <

There is one more step in a word problem: interpreting the answer. If n <= 7.4 and n counts whole tents, then the greatest possible n is 7 — you round down, because 8 is too many. If n >= 7.4 and n counts whole quests, the least possible n is 8 — you round up, because 7 is not enough.

The rounding direction comes from the situation, not from the decimal. Always ask: does going one step further break the constraint?

Worked examples

Example 1

Solve 5x - 7 >= 23.

  1. Two things were done to x: multiply by 5, then subtract 7. Undo them in reverse order.
  2. Add 7 to both sides: 5x >= 30.
  3. Divide both sides by 5. Since 5 is positive, the direction of the sign stays the same.
  4. x >= 6.
  5. Check: x = 6 gives 5 x 6 - 7 = 23, and 23 >= 23 is true. x = 5 gives 18, and 18 >= 23 is false, so the boundary is in the right place.

Example 2

Solve 4 - 3x < 19.

  1. Get the x term on its own: subtract 4 from both sides, giving -3x < 15.
  2. Now divide both sides by -3. This is the negative-divisor case, so the sign must reverse.
  3. x > -5.
  4. Check with a number in the solution set: x = 0 gives 4 - 0 = 4, and 4 < 19 is true.
  5. Check with a number outside it: x = -10 gives 4 + 30 = 34, and 34 < 19 is false. The reversal was correct.

Example 3

A guild charges 30 gold to join and 8 gold for each training session. Mira has 150 gold. What is the greatest number of sessions she can book?

  1. Let n be the number of sessions. The total cost is 30 + 8n gold.
  2. "Can afford" means the cost is at most what she has: 30 + 8n <= 150.
  3. Subtract the 30 gold joining fee from both sides: 8n <= 120.
  4. Divide by 8: n <= 15.
  5. n counts whole sessions, so the greatest number is 15. Check: 30 + 8 x 15 = 150 gold exactly, and a 16th session would cost 8 more than she has.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Solve for x. x + 6 > 10 Which statement is the solution?

  1. x < 4
  2. x > 4
  3. x > 10
  4. x < 10

Answer: B. x > 4

  1. x + 6 > 10
  2. x + 6 - 6 > 10 - 6 (do the same to both sides)
  3. x > 4
  4. Check: x = 5 gives 5 + 6 = 11, and 11 > 10 is true.

Problem 2

Difficulty 3 of 5

Solve for x. 8x + 14 >= -34 Write your answer like "x > 3".

Answer: x >= -6

  1. 8x + 14 >= -34
  2. 8x >= -48 (subtract 14 on both sides)
  3. x >= -6 (divide both sides by 8; 8 is positive, so the sign stays)
  4. Check: x = -6 gives 8 x (-6) + 14 = -34, and -34 >= -34 is true.

Problem 3

Difficulty 4 of 5

Solve for x. -10x + 21 > -159 Write your answer like "x > 3".

Answer: x < 18

  1. -10x + 21 > -159
  2. -10x > -180 (subtract 21 on both sides)
  3. Divide both sides by -10. Dividing by a negative reverses the direction of the sign.
  4. x < 18
  5. Check: x = 17 gives -10 x 17 + 21 = -149, and -149 > -159 is true.

Common mistakes

  • Forgetting to flip the sign after dividing by a negative: writing -2x > 10 as x > -5 instead of x < -5.
  • Flipping the sign when nothing negative was divided out — moving a term across the inequality never flips it.
  • Dividing before undoing the constant, so 3x + 6 <= 18 turns into x + 6 <= 6.
  • Using a filled circle for < or an open circle for <=, which changes whether the boundary number is a solution.
  • Rounding the wrong way in a word problem, for example claiming 8 tents can be bought when n <= 7.4.

What you should be able to do

  • Solve one- and two-step inequalities.
  • Flip the inequality sign when multiplying or dividing by a negative.
  • Graph a solution set on a number line.
  • Write and solve inequalities from constraint word problems.

Where this fits in the curriculum

Common Core

  • 6.EE.B.8

    Grade 6 — Write an inequality to represent a constraint and graph its solutions on a number line.

  • 7.EE.B.4.B

    Grade 7 — Solve word problems leading to inequalities of the form px + q > r, and graph and interpret the solution set.

  • HSA-REI.B.3

    High school — Solve linear inequalities in one variable, including with letters for coefficients.

Ontario

  • G7.C2.4

    Grade 7 — Solve inequalities that involve multiple terms and whole numbers, and verify and graph the solutions.

  • G8.C2.4

    Grade 8 — Solve inequalities that involve integers, and verify and graph the solutions.

    The Grade 9 de-streamed course has no expectation for solving an inequality — it only graphs regions (C4.2) — so Ontario finishes this skill in Grade 8.

SAT

  • Heart of Algebra

    Solving and interpreting linear inequalities and their solution sets.

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