Let the weights of the three children be denoted as \( a \), \( b \), and \( c \), where \( a < b < c \).
From the problem, we have the following equations based on the given conditions:
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The sum of the weights of the two smallest children: \[ a + b = 71 \]
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The sum of the weights of the two largest children: \[ b + c = 96 \]
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The sum of the weights of the smallest and the eldest child: \[ a + c = 87 \]
Our goal is to find the sum \( a + b + c \).
We can express \( a + b + c \) as follows: \[ a + b + c = (a + b) + c = 71 + c \] and also as: \[ a + b + c = a + (b + c) = a + 96 \] and finally, we can also express it as: \[ a + b + c = (a + c) + b = 87 + b \]
Now, we have three expressions for \( a + b + c \):
- \( 71 + c \)
- \( a + 96 \)
- \( 87 + b \)
To find \( c \) from \( a + b = 71 \) and \( b + c = 96 \), we can solve for \( c \) in terms of \( a \): From \( b + c = 96 \): \[ c = 96 - b \]
Next, we can substitute \( b \) from \( a + b = 71 \): \[ b = 71 - a \] Substituting this into the equation for \( c \): \[ c = 96 - (71 - a) = 96 - 71 + a = 25 + a \]
Now we have expressed \( b \) and \( c \) in terms of \( a \):
- \( b = 71 - a \)
- \( c = 25 + a \)
Next, substitute these values into the equation \( a + c = 87 \): \[ a + (25 + a) = 87 \] This simplifies to: \[ 2a + 25 = 87 \] Subtracting 25 from both sides gives: \[ 2a = 62 \] Dividing by 2: \[ a = 31 \]
Now we can find \( b \) and \( c \): \[ b = 71 - a = 71 - 31 = 40 \] \[ c = 25 + a = 25 + 31 = 56 \]
Now we have the weights:
- \( a = 31 \)
- \( b = 40 \)
- \( c = 56 \)
Finally, we can find the sum of the weights: \[ a + b + c = 31 + 40 + 56 = 127 \]
Thus, the sum of the weights of all three children is \(\boxed{127}\) pounds.
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