Math Detours · Grade 7 · Sharpen · 7

At least, at most

"You must be at least 130 cm tall" is not an equation. Write it as an inequality, solve it, and draw every answer at once on a number line.

Name ____________________Date ____________

Tonight's idea

An equation like \(x + 2 = 5\) has one answer. "You must be at least 130 cm tall" has many: 130, 131, 142.5 and more. It is an inequality. With \(h\) for height in cm, it is written \(h \ge 130\).

\(>\) more than  ·  \(\ge\) at least (more than or equal to)  ·  \(\lt\) less than  ·  \(\le\) at most (less than or equal to)

A number line shows every answer at once. Shade the part that works. A filled dot says the end number counts. An open dot says it does not.

h ≥ 130filled dot: 130 counts110120130140150t < 5open dot: 5 does not count34567
"At least 130" includes 130. "Below 5" stops just short of 5.

Solve an inequality with the same undo moves as an equation. Adding or subtracting slides both sides along the line together, so the one on the left stays on the left. Multiplying or dividing by a positive number stretches or shrinks the line, and left still stays left.

Multiplying by a negative number is another matter. That is for next year: Why the inequality flips.

Watch one: the lift

A lift carries at most 600 kg. A trolley weighs 120 kg and each box weighs 40 kg. How many boxes, \(b\), can go up with the trolley?

"At most" means \(\le\):

\[40b + 120 \le 600\]

Subtract 120 from both sides: \(40b \le 480\). Divide both sides by 40: \(b \le 12\).

So any number of boxes from 0 to 12 is safe. Check the edge. 12 boxes give \(480 + 120 = 600\), just allowed. 13 boxes give 640, too heavy.

Your turn

  1. Write each sentence as an inequality. (a) You must be at least 130 cm tall (\(h\)). (b) A bag may weigh at most 23 kg (\(w\)). (c) The temperature stayed below 5 degrees (\(t\)). (d) More than 40 people came (\(p\)).
  2. Solve each one. Then draw the answers on a number line. (a) \(x + 4 > 9\); (b) \(3x \le 18\); (c) \(2x - 5 \ge 7\).
  3. Ada has $50 for the fair. Getting in costs $14 and each ride costs $4.50. Write an inequality for the number of rides, \(r\), and solve it. What is the most rides she can take?
  4. (a) A bus carries at most 48 students, and 200 students are going on a trip. Write an inequality for the number of buses, \(b\), and find the fewest buses. (b) Tickets cost $48 each and you have $200. Write an inequality for the number of tickets, \(n\), and find the most you can buy. (c) Both sums lead to 4.17. Why do you round up in one and down in the other?

Stretch: fenced in on both sides

\(n\) is a whole number. \(3n + 2\) is at least 20 and at most 35. Write this as two inequalities and solve each one. Then list every value \(n\) can take.

Answers

  1. (a) \(h \ge 130\). (b) \(w \le 23\). (c) \(t \lt 5\). (d) \(p > 40\).
  2. (a) \(x > 5\): open dot at 5, shaded to the right. (b) \(x \le 6\): filled dot at 6, shaded to the left. (c) Add 5: \(2x \ge 12\). Halve: \(x \ge 6\). Filled dot at 6, shaded to the right.
  3. \(4.5r + 14 \le 50\). Subtract 14: \(4.5r \le 36\). Divide by 4.5: \(r \le 8\). At most 8 rides, and 8 rides use exactly $50.
  4. (a) \(48b \ge 200\), so \(b \ge 4.17\) (rounded). Fewest: 5 buses. (b) \(48n \le 200\), so \(n \le 4.17\). Most: 4 tickets. (c) The answer must be a whole number on the allowed side. For buses the allowed side is above 4.17, and the first whole number there is 5. For tickets it is below, and the last whole number there is 4.

Stretch. \(3n + 2 \ge 20\) gives \(3n \ge 18\), so \(n \ge 6\). \(3n + 2 \le 35\) gives \(3n \le 33\), so \(n \le 11\). So \(n\) is 6, 7, 8, 9, 10 or 11.

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