To solve the system of equations given by \( y = 3x + 6 \) and \( y = -13x - 4 \), we can set the two equations equal to each other since both are equal to \( y \):
\[ 3x + 6 = -13x - 4 \]
Now, we can solve for \( x \):
- Add \( 13x \) to both sides:
\[ 3x + 13x + 6 = -4 \] \[ 16x + 6 = -4 \]
- Subtract 6 from both sides:
\[ 16x = -4 - 6 \] \[ 16x = -10 \]
- Divide both sides by 16:
\[ x = -\frac{10}{16} = -\frac{5}{8} \]
Now that we have \( x = -\frac{5}{8} \), we can substitute this value back into one of the original equations to find \( y \). We'll use the first equation \( y = 3x + 6 \):
\[ y = 3\left(-\frac{5}{8}\right) + 6 \] \[ y = -\frac{15}{8} + 6 \] Convert 6 to a fraction with a denominator of 8:
\[ y = -\frac{15}{8} + \frac{48}{8} = \frac{33}{8} \]
So the solution to the system of equations is:
\[ \left( -\frac{5}{8}, \frac{33}{8} \right) \]
This is the point where the two lines intersect.