Two-Step Inequalities

Two moves, one flip rule.

Two operations to undo

In 2x + 3 < 11, two things have been done to x: it was multiplied by 2, and then 3 was added. To get x on its own, undo them in the reverse order. Take away the 3 first, then divide by 2. This is exactly how you would solve the equation 2x + 3 = 11.

Solve it like an equation

Take 3 from both sides: 2x + 3 − 3 < 11 − 3, so 2x < 8. Divide both sides by 2: x < 4. Subtracting a number keeps the order, and so does dividing by a positive number, so the sign stays as <.

The equation 2x + 3 = 11 has the solution x = 4, and that is the boundary. At x = 4 the two sides are equal, 2 × 4 + 3 = 11, so 4 is not a solution of the inequality. The sign then tells you which side of 4 the solutions are on.

Check with one value on each side. x = 3 gives 2 × 3 + 3 = 9, and 9 < 11 is true. x = 5 gives 2 × 5 + 3 = 13, and 13 < 11 is false. So the solutions are the numbers less than 4.

012345678x < 4

Every number to the left of 4 solves 2x + 3 < 11. The circle at 4 is hollow, because 2 × 4 + 3 = 11 is not less than 11.

When x is multiplied by a negative number

Solve −2x + 3 < 11. The first step is the same: take 3 from both sides to get −2x < 8. Now divide both sides by −2. That is a division by a negative number, so the sign reverses: x > −4.

Only that one step reverses the sign. Taking 3 away did not change the order, and adding or subtracting never does, whether the number is positive or negative.

Check with x = 0, on the side of the answer: −2 × 0 + 3 = 3, and 3 < 11 is true. Check with x = −5, on the other side: −2 × (−5) + 3 = 13, and 13 < 11 is false.

-8-7-6-5-4-3-2-1012x > −4

The solutions of −2x + 3 < 11 are every number to the right of −4.

Keeping the x term positive

You can keep the x term positive instead. Add 2x to both sides of −2x + 3 < 11: 3 < 11 + 2x. Subtract 11 from both sides: −8 < 2x. Divide by 2, which is positive: −4 < x. Read from the right, −4 < x says that x is greater than −4, the same answer as before.

The usual mistakes

Reversing the sign because a number is negative. In 3x + 10 < 4, taking 10 away gives 3x < −6, and dividing by 3, a positive number, gives x < −2. The −6 does not reverse the sign. Only dividing by a negative number does.

Forgetting to reverse it. From −2x < 8, writing x < −4 gives values such as x = −5, and −2 × (−5) + 3 = 13, which is not less than 11.

Dividing only one side. From 2x < 8, both sides are divided by 2: the answer is x < 4, not x < 8.

Worked example: The Greatest Number of Rides a Budget Allows

Question Wei Ling has $40 to spend at a fair. It costs $8 to enter and $3 for each ride. (a) Find the greatest number of rides she can afford. (b) She decides to keep at least $10 for food. Find the greatest number of rides she can afford now.

  1. 1.Let n be the number of rides. The entry and the rides cost 8 + 3n dollars, and this cannot be more than 40: 8 + 3n ≤ 40.

    n rides: $8 to enter and $3 for each ride8 + 3n≤40
    n rides: $8 to enter and $3 for each ride8 + 3n≤40
    The entry and n rides cost 8 + 3n dollars, which cannot be more than 40: 8 + 3n ≤ 40.

  2. 2.Subtract 8 from both sides: 3n ≤ 32. Divide both sides by 3: n ≤ 1023. Dividing by a positive number keeps the inequality sign as it is.

    n rides: $8 to enter and $3 for each ride8 + 3n≤403n≤32subtract 8 from both sidesn≤10 2/3divide both sides by 3
    n rides: $8 to enter and $3 for each ride8 + 3n≤40subtract 8 from both sides3n≤32divide both sides by 3n≤10 2/3
    Subtract 8 from both sides, then divide both sides by 3: n ≤ 1023.

  3. 3.The number of rides is a whole number, and the greatest whole number that is not more than 1023 is 10. (a) She can afford 10 rides. Check: 8 + 3 × 10 = 38 ≤ 40, but 11 rides cost 8 + 33 = 41 dollars.

    n rides: $8 to enter and $3 for each ride8 + 3n≤403n≤32subtract 8 from both sidesn≤10 2/3divide both sides by 3051010 2/3check: 8 + 3 × 10 = 38, but 11 rides cost 41
    n rides: $8 to enter and $3 for each ride8 + 3n≤40subtract 8 from both sides3n≤32divide both sides by 3n≤10 2/3051010 2/3check: 8 + 3 × 10 = 38, but 11 rides cost 41
    (a) The greatest whole number that is not more than 1023 is 10, so she can afford 10 rides.

  4. 4.Keeping $10 for food adds 10 to what must fit into the $40: 8 + 3n + 10 ≤ 40, which is 3n + 18 ≤ 40.

    with $10 kept for food: 8 + 3n + 10 ≤ 403n + 18≤40
    with $10 kept for food: 8 + 3n + 10 ≤ 403n + 18≤40
    Keeping $10 for food gives 8 + 3n + 10 ≤ 40, which is 3n + 18 ≤ 40.

  5. 5.Subtract 18 from both sides: 3n ≤ 22. Divide both sides by 3: n ≤ 713.

    with $10 kept for food: 8 + 3n + 10 ≤ 403n + 18≤403n≤22subtract 18 from both sidesn≤7 1/3divide both sides by 3
    with $10 kept for food: 8 + 3n + 10 ≤ 403n + 18≤40subtract 18 from both sides3n≤22divide both sides by 3n≤7 1/3
    Subtract 18 from both sides, then divide both sides by 3: n ≤ 713.

  6. 6.(b) She can afford 7 rides. Check: 18 + 3 × 7 = 39 ≤ 40, but 8 rides would need 18 + 24 = 42 dollars.

    with $10 kept for food: 8 + 3n + 10 ≤ 403n + 18≤403n≤22subtract 18 from both sidesn≤7 1/3divide both sides by 3057107 1/3check: 18 + 3 × 7 = 39, but 8 rides need 42
    with $10 kept for food: 8 + 3n + 10 ≤ 403n + 18≤40subtract 18 from both sides3n≤22divide both sides by 3n≤7 1/3057107 1/3check: 18 + 3 × 7 = 39, but 8 rides need 42
    (b) She can afford 7 rides.

Answer: (a) 10 rides; (b) 7 rides

Common mistakes

  • Rounding 1023 up to 11 because it is nearer. Eleven rides cost $41, which is more than she has. The answer is the greatest whole number that satisfies the inequality, so it is always rounded down here.
  • Dividing 40 by 3 and answering 13 rides. The entry charge of $8 is paid first, so only $32 is left for the rides.

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