Lens 09 of 9
The tightest constraint
Look at which condition shrinks the answer most. Let it generate the candidates; let the others filter.
A problem with two conditions is rarely two equations. One condition is usually tight (it leaves four pairs) and the other loose (it would leave hundreds). Lead with the tight one and the loose one becomes a quick check. Get the order backwards and you are solving a quadratic for no reason.
The clues that call for it
- Quadratic Equations
Two conditions on two numbers
'Sum is 8' leaves four pairs; '15 times the sum of the reciprocals is 8' is only a filter.
- Coordinate Geometry
'A point on the x-axis …'
On the x-axis, y = 0: one unknown is gone before any distance formula.
Worked example
Two numbers sum to 8; 15 times the sum of their reciprocals is also 8. Find them.
- 1
Rank the conditions
'Sum to 8' is far tighter than the reciprocal condition: it yields just four pairs.
a + b = 8 ≫ 15(1/a + 1/b) = 8
- 2
Let the tight one generate
Positive whole-number pairs summing to 8.
- 1, 7
- 2, 6
- 3, 5
- 4, 4
- 3
Demote the loose one to a filter
Only one pair survives the reciprocal check.
15 × (1/3 + 1/5) = 15 × 8/15 = 8 ✓
- 1, 7
- 2, 6
- 3, 5
- 4, 4
- 4
Answer
3 and 5. The tight condition generates, the loose one verifies, never the other way round.
Try the lens
Two positive integers sum to 9 and have product 20. Which condition should you lead with?
Problems that use this lens
- 11Sum 8 and reciprocalsThe sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.Lens · The tightest constraintquadratic equationsPractice
- 1A point on the axisFind the point on the x-axis which is equidistant from (2, −5) and (−2, 9).Lens · The tightest constraintcoordinate geometryNCERT Ex 7.1 Q7