To find the solution to the system of equations given by \( x - 5y = 7 \) and \( 3x - 2y = -4 \), we need to solve these equations simultaneously.
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First equation: Rearranging \( x - 5y = 7 \): \[ x = 5y + 7 \]
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Substituting in the second equation: Substitute \( x \) into \( 3x - 2y = -4 \): \[ 3(5y + 7) - 2y = -4 \] Simplifying: \[ 15y + 21 - 2y = -4 \] \[ 13y + 21 = -4 \] \[ 13y = -4 - 21 \] \[ 13y = -25 \] \[ y = -\frac{25}{13} \]
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Finding \( x \): Now substitute \( y \) back into the rearranged first equation: \[ x = 5\left(-\frac{25}{13}\right) + 7 \] \[ x = -\frac{125}{13} + \frac{91}{13} \] \[ x = -\frac{125 - 91}{13} = -\frac{34}{13} \]
So the intersection point (the solution to the system of equations) is: \[ \left( -\frac{34}{13}, -\frac{25}{13} \right) \] Thus, the answer to fill in the blank is:
The intersection point is \((- \frac{34}{13}, -\frac{25}{13})\).