Let’s solve each equation one by one.
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1. 3|s| = 21
Divide both sides by 3:
|s| = 7
Absolute value means s can be 7 or -7.
→ s = 7 or s = -7
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2. |x - 29| = 9
This means x - 29 = 9 OR x - 29 = -9
Case 1: x - 29 = 9 → x = 38
Case 2: x - 29 = -9 → x = 20
→ x = 38 or x = 20
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3. |-2r| = 10
Absolute value of a number is always non-negative, and |-2r| = |2r| = 2|r|
So: 2|r| = 10 → |r| = 5 → r = 5 or r = -5
Alternatively, directly:
-2r = 10 → r = -5
or
-2r = -10 → r = 5
→ r = 5 or r = -5
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4. |r| + 3 = 17
Subtract 3 from both sides:
|r| = 14
→ r = 14 or r = -14
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5. |8c| = 2
This means 8c = 2 or 8c = -2
Case 1: 8c = 2 → c = 2/8 = 1/4
Case 2: 8c = -2 → c = -2/8 = -1/4
→ c = 1/4 or c = -1/4
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6. |d| - 7 = 24
Add 7 to both sides:
|d| = 31
→ d = 31 or d = -31
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7. 3|-n| = 24
Note: |-n| = |n|, so this is same as 3|n| = 24
Divide by 3: |n| = 8 → n = 8 or n = -8
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8. |f + 24| = 18
This means f + 24 = 18 OR f + 24 = -18
Case 1: f + 24 = 18 → f = -6
Case 2: f + 24 = -18 → f = -42
→ f = -6 or f = -42
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Final Answer:
s = ±7; x = 20, 38; r = ±5; r = ±14; c = ±1/4; d = ±31; n = ±8; f = -6, -42
Parent Tip: Review the logic above to help your child master the concept of absolute value problems.
