Solution 2. We will try all answer choice:
A. \(1\)
This choice gives us \((x+y)^2=1 \implies x+y=1\) or \(x+y=-1\)
With \(x+y=1\) we have
\(\Big\{ \begin {array}{lr}
2x + 9y = 57 \\
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
2x + 9(1-x) = 57 \\
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 48 \\
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 48 \\
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
x = \frac{-48}{7} \\
x+y=1
\end{array}\)
Eliminate this case because \(x\) is not integer.
With \(x+y=-1\) we have
\(\Bigg\{ \begin {array}{lr}
2x + 9y = 57 \\
x+y=-1
\end{array} \implies \Big\{ \begin {array}{lr}
2x + 9(-1-x) = 57 \\
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 66 \\
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 66 \\
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
x = \frac{-66}{7} \\
x+y=1
\end{array}\)
Eliminate this case because \(x\) is not integer.
B. \(9\)
This choice gives us \((x+y)^2=9 \implies x+y=3\) or \(x+y=-3\)
With \(x+y=3\) we have
\(\Big\{ \begin {array}{lr}
2x + 9y = 57 \\
x+y=3
\end{array} \implies \Big\{ \begin {array}{lr}
2x + 9(3-x) = 57 \\
x+y=3
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 30 \\
x+y=3
\end{array} \implies \Big\{ \begin {array}{lr}
7x = -30 \\
x+y=3
\end{array} \implies \Big\{ \begin {array}{lr}
x = \frac{-30}{7} \\
x+y=3
\end{array}\)
Eliminate this case because \(x\) is not integer.
With \(x+y=-3\) we have
\(\Bigg\{ \begin {array}{lr}
2x + 9y = 57 \\
x+y=-3
\end{array} \implies \Big\{ \begin {array}{lr}
2x + 9(-3-x) = 57 \\
x+y=-3
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 84 \\
x+y=-3
\end{array} \implies \Big\{ \begin {array}{lr}
x = -12 \\
y=9
\end{array}\)
This one satisfies the given conditions. Hence this choice is the correct answer.
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