Let’s solve each equation one by one. We’re finding the value of the letter (variable) that makes each equation true.
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1. 3a = 12
Divide both sides by 3:
a = 12 ÷ 3 →
a = 4
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2. 30 = 5b
Divide both sides by 5:
b = 30 ÷ 5 →
b = 6
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3. 8c = 72
Divide both sides by 8:
c = 72 ÷ 8 →
c = 9
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4. 48 = 12d
Divide both sides by 12:
d = 48 ÷ 12 →
d = 4
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5. 50 = 5g
Divide both sides by 5:
g = 50 ÷ 5 →
g = 10
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6. 6z = 48
Divide both sides by 6:
z = 48 ÷ 6 →
z = 8
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7. 20 = 4h + h
Combine like terms: 4h + h = 5h
So, 20 = 5h
Divide both sides by 5:
h = 20 ÷ 5 →
h = 4
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8. 3b + 5 = 11
Subtract 5 from both sides:
3b = 11 - 5 → 3b = 6
Divide by 3:
b = 6 ÷ 3 →
b = 2
*(Note: This is a different “b” than in problem #2 — variables can be reused with different values in separate problems.)*
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9. 14 = 6j - 4
Add 4 to both sides:
14 + 4 = 6j → 18 = 6j
Divide by 6:
j = 18 ÷ 6 →
j = 3
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10. 2k - 5 = 13
Add 5 to both sides:
2k = 13 + 5 → 2k = 18
Divide by 2:
k = 18 ÷ 2 →
k = 9
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11. 6r + 1 = 19
Subtract 1 from both sides:
6r = 19 - 1 → 6r = 18
Divide by 6:
r = 18 ÷ 6 →
r = 3
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12. 10 = 5s + 0
Adding 0 doesn’t change anything, so:
10 = 5s
Divide by 5:
s = 10 ÷ 5 →
s = 2
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Final Answer:
a = 4
b = 6
c = 9
d = 4
g = 10
z = 8
h = 4
b = 2
j = 3
k = 9
r = 3
s = 2
Parent Tip: Review the logic above to help your child master the concept of year 6 math worksheet.