Depth · one concept from first principles · step 2 of 4
Find the sign of a
In (b) the ratio |b|/a is at least 1/2, so it is positive. Its top |b| is never negative, so the bottom a must be positive (and b ≠ 0).
A quotient is positive only when top and bottom have the same sign.
7.NS.A.2Identify SubproblemsList the choices for a
By (a), a is a positive common divisor of 8 and 12, that is, a divisor of 4.
Only three cases: a = 1, 2, 4.
6.NS.B.4Identify SubproblemsTurn (b) into a range for |b|
Multiply each part of (b) by the positive number a.
Multiplying by a positive number keeps the inequality directions.
6.NS.C.7Make A Systematic ListBecause is positive, condition (b) says exactly .
Why?
must be positive, because the ratio is at least while its top is never negative.
Why?
is how far sits from , so it is never negative.
Why?
A quotient with unlike signs is negative, so a positive quotient whose top is not negative needs a positive bottom.
Why?
Multiplying every part of (b) by the positive number keeps each inequality pointing the same way.
Read off |b| in each case
a = 1: 1/2 ≤ |b| < 2. a = 2: 1 ≤ |b| < 4. a = 4: 2 ≤ |b| < 8.
That is 1, 3 and 6 values of |b|.
6.NS.C.7Make A Systematic ListCount the pairs
Each value of |b| gives b and -b, so double each count and add.
2 + 6 + 12 = 20 pairs.
6.NS.C.7Make A Systematic ListPin down the few possible denominators first, then each case is a simple count.
- Find the sign of a
- List the choices for a
- Turn (b) into a range for |b|
- Read off |b| in each case
- Count the pairs
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