To find the solution to the system of equations given by \( x - 5y = 7 \) and \( 3x - 2y = -4 \), we need to solve them simultaneously.
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First equation: \( x - 5y = 7 \)
We can solve for \( x \): \[ x = 7 + 5y \]
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Second equation: \( 3x - 2y = -4 \)
Now, we substitute \( x \) from the first equation into the second equation: \[ 3(7 + 5y) - 2y = -4 \]
Expanding this: \[ 21 + 15y - 2y = -4 \] This simplifies to: \[ 21 + 13y = -4 \]
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Solving for \( y \): \[ 13y = -4 - 21 \] \[ 13y = -25 \] \[ y = -\frac{25}{13} \]
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Finding \( x \): Now substitute \( y \) back into the expression for \( x \): \[ x = 7 + 5\left(-\frac{25}{13}\right) \] \[ x = 7 - \frac{125}{13} \] We convert 7 into a fraction: \[ 7 = \frac{91}{13} \] So: \[ x = \frac{91}{13} - \frac{125}{13} = \frac{91 - 125}{13} = \frac{-34}{13} \]
Thus, the solution to the system of equations (the intersection point) is: \[ \left( -\frac{34}{13}, -\frac{25}{13} \right) \]
The final answer for the intersection point (to fill in the answer box) is: \[ \left( -\frac{34}{13}, -\frac{25}{13} \right) \]