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The given equations are: $2x + 5y = 1$ $3x + 2y = 7$ We can write this system in matrix form as $AX = B$, where $A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}$.
First, find the determinant of $A$: $|A| = (2)(2) - (5)(3) = 4 - 15 = -11$ Since $|A| \neq 0$, the inverse exists. The inverse of $A$ is given by: $A^{-1} = \frac{1}{|A|} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix} = \frac{1}{-11} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} -2/11 & 5/11 \\ 3/11 & -2/11 \end{bmatrix}$
We have $AX = B$, so $X = A^{-1}B$. $X = \begin{bmatrix} -2/11 & 5/11 \\ 3/11 & -2/11 \end{bmatrix} \begin{bmatrix} 1 \\ 7 \end{bmatrix} = \begin{bmatrix} (-2/11)(1) + (5/11)(7) \\ (3/11)(1) + (-2/11)(7) \end{bmatrix} = \begin{bmatrix} -2/11 + 35/11 \\ 3/11 - 14/11 \end{bmatrix} = \begin{bmatrix} 33/11 \\ -11/11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$ Therefore, $x = 3$ and $y = -1$.
The point common to the paths of the ants is $(3, -1)$.
Final Answer: (3, -1)
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