To find the solution to the system of equations given by \( y = 3x + 6 \) and \( y = -13x - 4 \), we can set the two equations equal to each other since they both equal \( y \):
\[ 3x + 6 = -13x - 4 \]
Now, we will solve for \( x \):
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Add \( 13x \) to both sides: \[ 3x + 13x + 6 = -4 \] \[ 16x + 6 = -4 \]
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Subtract 6 from both sides: \[ 16x = -4 - 6 \] \[ 16x = -10 \]
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Divide by 16: \[ x = -\frac{10}{16} = -\frac{5}{8} \]
Now that we have \( x \), we can substitute it back into either equation to find \( y \). We will use the first equation \( y = 3x + 6 \):
\[ y = 3\left(-\frac{5}{8}\right) + 6 \] \[ y = -\frac{15}{8} + 6 \] \[ y = -\frac{15}{8} + \frac{48}{8} \] \[ y = \frac{48 - 15}{8} = \frac{33}{8} \]
So the solution to the system of equations is:
\[ \left( -\frac{5}{8}, \frac{33}{8} \right) \]