Two conditions at once
Each inequality in two variables shades one side of its boundary line. shades the line y = x − 1 and everything above it. shades the line y = 5 − x and everything below it.
When a question gives both inequalities together, a point is a solution only if it satisfies both of them. A point that satisfies one and breaks the other does not count.
Both regions on one set of axes. Where the two shadings lie on top of each other, the color is darker.
Where the shadings overlap
The part of the plane that is shaded twice is where both inequalities hold. Here it is a wedge, bounded by the two lines, opening out to the left. That wedge, and nothing else, solves the pair.
Test (1, 2). For the first inequality, x − 1 = 0, and is true. For the second, 5 − x = 4, and is true. Both hold, so (1, 2) is a solution, and it lies inside the wedge.
Only the overlap is shaded now. (1, 2) is inside it, because it satisfies both inequalities.
One out of two is not enough
Test (4, 0). For , 5 − 4 = 1, and is true. For , 4 − 1 = 3, and is false. The point is in the second region but not the first, so it is outside the overlap and is not a solution of the pair.
(4, 0) lies below the line y = x − 1, so it breaks the first inequality and is outside the overlap.
The corner of the region
The two boundary lines meet at the corner of the wedge. At that point both lines give the same y, so x − 1 = 5 − x. Add x to both sides: 2x − 1 = 5. Add 1: 2x = 6, so x = 3. Then y = 3 − 1 = 2, and the corner is (3, 2).
Both signs are and , so both lines are solid and the corner itself is a solution: and are both true, because each side equals 2. If either sign were strict, its line would be dashed and the corner would be left out.
The boundaries meet at (3, 2), the corner of the region.
The usual mistakes
Accepting a point that satisfies only one inequality. (4, 0) satisfies , but it breaks , so it is not a solution of the pair.
Leaving out the boundary. With and , the points on the lines are solutions, including the corner. Only a strict sign leaves its line out.
Worked example: Kites Limited by Paper and by Time at Once
Question Farid makes large kites and small kites for a stall. A large kite uses 3 sheets of paper and takes 1 hour. A small kite uses 2 sheets of paper and takes 2 hours. He has 12 sheets of paper and 8 hours, and he wants to make at least one kite of each size. He makes x large kites and y small kites. (a) In how many different ways can he choose x and y? (b) Which choice gives the greatest total number of kites?
1.The paper used is 3x + 2y sheets, so 3x + 2y ≤ 12. The time taken is x + 2y hours, so x + 2y ≤ 8.
The paper gives 3x + 2y ≤ 12 and the time gives x + 2y ≤ 8. 2.The boundary 3x + 2y = 12 joins (4, 0) and (0, 6). The point (0, 0) gives 0 ≤ 12, which is true, so the region is the side of the line that contains (0, 0).
The boundary 3x + 2y = 12 joins (4, 0) and (0, 6), and its region contains (0, 0). 3.The boundary x + 2y = 8 joins (8, 0) and (0, 4), and its region also contains (0, 0). Subtracting the second equation from the first gives 2x = 4, so the lines meet at (2, 3). The overlap is the region that lies under both lines.
The boundary x + 2y = 8 joins (8, 0) and (0, 4). The lines meet at (2, 3), and the overlap lies under both lines. 4.List the integer points in the overlap with x ≥ 1 and y ≥ 1. For x = 1: y = 1, 2, 3. For x = 2: y = 1, 2, 3. For x = 3: only y = 1, because 9 + 2y ≤ 12. For x = 4 the paper is used up, so y = 0, which is not allowed.
The integer points in the overlap with x ≥ 1 and y ≥ 1 are listed column by column. 5.(a) There are 3 + 3 + 1 = 7 different ways.
(a) There are 3 + 3 + 1 = 7 different ways. 6.(b) The total x + y is 5 at (2, 3) and at most 4 at the other six points, so he should make 2 large kites and 3 small kites, 5 kites in all. Check: 3 × 2 + 2 × 3 = 12 ≤ 12 and 2 + 2 × 3 = 8 ≤ 8.
(b) The total x + y is greatest at (2, 3): 2 large kites and 3 small kites, 5 kites in all.
Answer: (a) 7 ways; (b) 2 large kites and 3 small kites, 5 kites in all
Common mistakes
- Counting the points that satisfy only the paper inequality. The point (1, 4) uses 11 sheets, which is allowed, but it takes 1 + 8 = 9 hours, so it is outside the overlap.
- Leaving out the points on the boundary lines, such as (2, 3). Both signs are ≤, so using exactly 12 sheets or exactly 8 hours is allowed.
More equations and inequalities problems, worked step by step →