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Answer: A has 70 oranges and B has 50 oranges.
Let A have \(x\) oranges and B have \(y\) oranges.
A's statement: if B gives A 10 oranges, A has \(x + 10\) and B is left with \(y - 10\):
\[x + 10 = 2(y - 10) \;\Rightarrow\; x - 2y = -30 \quad (1)\]
B's statement: if A gives B 10 oranges, B has \(y + 10\) and A is left with \(x - 10\):
\[y + 10 = x - 10 \;\Rightarrow\; x - y = 20 \quad (2)\]
Subtract (1) from (2): \(y = 50\). Then from (2), \(x = 70\).
Check: \(70 + 10 = 80 = 2 \times 40\), and \(50 + 10 = 60 = 70 - 10\).
Answer: A has 70 oranges and B has 50 oranges.
Let A have x oranges and B have y oranges.
First condition: If B gives 10 oranges to A, A will have \(x + 10\) and B will have \(y - 10\).
\(x + 10 = 2(y - 10)\)
\(x + 10 = 2y - 20\)
\(x - 2y = -30\) .....(1)
Second condition: If A gives 10 oranges to B, B will have \(y + 10\) and A will have \(x - 10\).
\(y + 10 = x - 10\)
\(x - y = 20\) .....(2)
Subtracting (1) from (2):
\((x - y) - (x - 2y) = 20 - (-30)\)
\(y = 50\)
Substituting \(y = 50\) in (2): \(x - 50 = 20\) ⇒ \(x = 70\)
Check: \(70 + 10 = 80 = 2(50 - 10)\) ✓ and \(50 + 10 = 60 = 70 - 10\) ✓
Hence A has 70 oranges and B has 50 oranges.
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