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Sue is 8 years old.

Sue (8) + Leah (14) + John (19) = 41

Solve algebraically by stating all ages in terms of Leah's (L)

Sue S = L - 6 (Leah = Sue + 6)

John J = L + 5 (Leah = John + 8)

S + J + L = 41

(L-6) + (L+5) + L = 41

3L -1 = 41

3L = 42

L = 14

S = 8

J = 19

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Q: If Leah is 6 years older than her sister Sue and John is 5 years older than Leah and the total of their ages is forty one the how old is Sue?
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