Inequalities on graphs — Notes & Worked Examples Show / Hide
Explanation:
To represent inequalities on a graph, first draw each boundary line by replacing the inequality sign with =. Use a solid line for inclusive inequalities (≤ or ≥) and a dashed line for strict inequalities (< or >).
To decide which side of a line satisfies the inequality, substitute a simple test point not on the line—such as (0, 0) or (1, 1)—into the inequality and check if it is true.
Example 1: Write down the three inequalities that define the region R (containing (2, 0)) bounded by solid lines x = 1 and y = −2, and dashed line x + y = 4.
Solution:
Example 2: Find all integer coordinates (x, y) that satisfy x > 1, y > 2, and 2x + y ≤ 7.
Solution:
x ≤ 2 y ≥ -3 y ≤ 2x + 1 3x + 2y ≤ 6
Label the region R.
x ≤ 2 y ≥ -3 y ≤ 2x + 1 3x + 2y ≤ 6
Label the region R.
Worked Solution & Answer
Shade interior region R. A1
y > 1 x + y < 5 y > 2x
Label the region **R**.
Worked Solution & Answer
shade the region above y = 1, below x + y = 5 and to the left of y = 2x. A1
2y + 4 < x x < 3 y < 6 - 3x
Label the region **R**.
Worked Solution & Answer
Shade the region satisfying all three inequalities. A1
Write down the three inequalities that define R.
Worked Solution & Answer
line through (0, 1) and (1, 0) is x + y = 1 (region below: x + y ≤ 1) M1
given line y = 2x + 2 (region below: y ≤ 2x + 2) A1
x ≤ 5 y ≤ 4 x + y > 6
Label the region R.
Worked Solution & Answer
The region R is the triangle bounded by these lines with vertices at (2, 4), (5, 4), and (5, 1). A1
x > 3 \quadand y > 1 \quadand x + y ≤ 7
Label the region R.
Worked Solution & Answer
2. Draw dashed horizontal line y = 1 (strict inequality).
3. Draw solid line x + y = 7 through (0, 7) and (7, 0). M1
4. Shade the triangular region between x > 3, y > 1, and below/on x + y ≤ 7 with vertices at (3, 1), (6, 1), and (3, 4).
5. Label this region R. A1
Worked Solution & Answer
horizontal line y = 6 (below line: y ⩽ 6) M1
line through (–2, 0) and (0, 6) is y = 3x + 6 (above: y ⩾ 3x + 6) M1
line through (0, 1) and (2, 0) is y = –1/2 x + 1 (above: y ⩾ –1/2 x + 1) A1
x ≥ 0 x ≤ 2 y ≤ x + 3 2x + 3y ≥ 6
Label the region **R**. [4 marks]
Worked Solution & Answer
Label the region **R**. [4 marks]
Worked Solution & Answer
the line y = 3x – 2 does not cut into or bound this region, so y < 3x – 2 can be removed B1