A limit on a total
Pens cost $2 each and pads cost $5 each, and you have $30 to spend. Let p be the number of pens and d the number of pads. The pens cost 2p dollars, the pads cost 5d dollars, and together they cost 2p + 5d dollars.
The total may be less than $30, or exactly $30, but not more. So the cost is less than or equal to 30: . This is an inequality in two variables, p and d.
Which sign the words give
The words of the question choose the sign. "At most", "no more than" and "cannot be more than" all set an upper limit that may be reached, so they give . "At least" and "no fewer than" set a lower limit that may be reached, so they give . "Less than" and "under" give <, and "more than" and "over" give >.
To decide between and <, ask whether the limit itself is allowed. At most $30 allows a bill of exactly $30, so the sign is . Under $30 does not allow it, so the sign is <.
At most and at least let the limit itself in. Under and over leave it out.
Many pairs fit
Many pairs of values satisfy the inequality: every pair whose cost stays within the limit. To test a pair, put the numbers into 2p + 5d and compare the cost with 30.
5 pens and 4 pads cost 2 × 5 + 5 × 4 = 10 + 20 = 30. That is exactly the limit, and is true, so the pair is allowed. 3 pens and 3 pads cost 6 + 15 = 21, and , so that pair is allowed too. 8 pens and 4 pads cost 16 + 20 = 36, which is more than 30, so that pair is not allowed.
Three pairs (p, d) tested against . (5, 4) costs exactly $30, which at most allows.
A lower limit
A lower limit is written the same way with . If you must buy at least 10 items altogether, the number of items is p + d, and .
The same two letters can build different totals. p + d counts the items, and 2p + 5d counts the dollars. Read which quantity the limit is on before you choose the expression.
A team needs at least 11 players, made up of x seniors and y juniors. The number of players is x + y, and at least 11 gives . A team of exactly 11 satisfies it.
The usual mistakes
Using an equals sign. 2p + 5d = 30 would force you to spend exactly $30. At most allows less, so the sign is .
Using for an upper limit. At most 30 is an upper limit, and writing would demand that you spend at least $30, which is the opposite.
Counting items when the limit is on money. limits the number of things bought. The $30 limits the cost, so each pen counts as 2 and each pad as 5.
Drawing the pairs on a graph
Each pair can be drawn as a point, with the number of pens across and the number of pads up. The pairs that cost exactly $30 satisfy 2p + 5d = 30, and they lie on a straight line called the boundary. To draw it, find where it meets each axis. With no pads, 2p = 30 and p = 15. With no pens, 5d = 30 and d = 6. Join (15, 0) and (0, 6). The line is solid, because at most $30 lets a bill of exactly $30 in.
The pairs that fit lie on one side of the line, and the pairs that cost too much lie on the other. To find which side, test a point that is not on the line. (0, 0) is the easiest: buying nothing costs $0, and is true, so the side that holds (0, 0) is the side that fits.
The solid line is 2p + 5d = 30. The shaded side holds (0, 0), so it is the side that fits. (8, 4) costs $36 and lies outside it.
Worked example: Chairs and Tables That Must Fit a Floor Area
Question A hall has 60 m2 of floor for furniture. Each chair needs 2 m2 of floor and each table needs 6 m2. There are x chairs and y tables. (a) Write an inequality in x and y for the furniture to fit, and describe the region of the graph that it gives. (b) Use the inequality to decide whether 12 chairs and 5 tables fit, and whether 18 chairs and 5 tables fit.
1.The chairs use 2x m2 and the tables use 6y m2. Together they cannot use more than 60 m2, so 2x + 6y ≤ 60.
The chairs use 2x m2 and the tables use 6y m2, so 2x + 6y ≤ 60. 2.The boundary is the line 2x + 6y = 60. When x = 0, 6y = 60 and y = 10. When y = 0, 2x = 60 and x = 30. The line joins (0, 10) and (30, 0), and it is drawn solid because the sign ≤ includes the points on the line.
The boundary 2x + 6y = 60 joins (0, 10) and (30, 0). It is solid because ≤ includes the line. 3.Test the point (0, 0): 2 × 0 + 6 × 0 = 0, and 0 ≤ 60 is true. (a) The inequality is 2x + 6y ≤ 60. Its region is the side of the line that contains (0, 0), with x and y not negative.
(a) (0, 0) gives 0 ≤ 60, which is true, so the region is the side of the line that contains (0, 0). 4.For 12 chairs and 5 tables, 2 × 12 + 6 × 5 = 24 + 30 = 54, and 54 ≤ 60 is true. The point (12, 5) lies inside the region.
12 chairs and 5 tables use 24 + 30 = 54 m2, so (12, 5) is inside the region. 5.(b) For 18 chairs and 5 tables, 2 × 18 + 6 × 5 = 36 + 30 = 66, and 66 ≤ 60 is false. So 12 chairs and 5 tables fit, using 54 m2, but 18 chairs and 5 tables do not, because they need 66 m2.
(b) 18 chairs and 5 tables need 66 m2, so (18, 5) is outside the region and they do not fit.
Answer: (a) 2x + 6y ≤ 60: the region on and below the line through (0, 10) and (30, 0); (b) 12 chairs and 5 tables fit (54 m2), 18 chairs and 5 tables do not (66 m2)
Common mistakes
- Writing x + y ≤ 60. That counts pieces of furniture. The limit is on floor area, so each chair counts as 2 and each table as 6.
- Shading the side of the line away from (0, 0) without testing a point. An empty hall uses 0 m2, which certainly fits, so (0, 0) must be in the shaded region.
More equations and inequalities problems, worked step by step →