Math 1 · Unit 5 · Lesson 2 · part 3 of 6
Two worked examples, one with a flip
Already solved for y.
Two worked examples, one with a flip
Worked: y ≤ 12x − 1
Already solved for y. Boundary y = 12x − 1: b = −1, m = 12. Or-equal: solid. y less than or equal: shade below. Test (0, 0): is 0 ≤ −1? No, so (0, 0)'s side is not shaded, and (0, 0) is above the line. Below is right.
Worked, with a flip: −2y < 4x − 6
Solve for y: multiply both sides by −12, a negative number, so flip: y > −2x + 3. Boundary y = −2x + 3: b = 3, m = −2. Strict: dashed. y greater: shade above. Test (0, 0) in the original: is −2 × 0 < 4 × 0 − 6, that is 0 < −6? No, so shade the side away from (0, 0), which is above the line. It agrees.
Lead with above and below. Confirm with the test point. If the two disagree, look for a missed flip.