Question

does y = 3x + 6, y = -13x - 4 have one solution, no solutions, or infinite solitions

Answer 1

To determine whether the equations \( y = 3x + 6 \) and \( y = -13x - 4 \) have one solution, no solutions, or infinite solutions, we can analyze the slopes of the lines represented by these equations.

1. **Slopes and Intercepts**:
- The first equation \( y = 3x + 6 \) has a slope of \( 3 \) and a y-intercept of \( 6 \).
- The second equation \( y = -13x - 4 \) has a slope of \( -13 \) and a y-intercept of \( -4 \).

2. **Comparison of Slopes**:
- Since the slopes \( 3 \) and \( -13 \) are different, the two lines will intersect at exactly one point.

3. **Conclusion**:
- Since the two lines are not parallel and have different slopes, they will intersect at a single point. Therefore, there is **one solution** to the system of equations.

### Verification by Setting the Equations Equal:
To find the point of intersection, set the two equations equal to each other:

\[
3x + 6 = -13x - 4
\]

Now, solve for \( x \):

1. Add \( 13x \) to both sides:

\[
3x + 13x + 6 = -4
\]
\[
16x + 6 = -4
\]

2. Subtract \( 6 \) from both sides:

\[
16x = -10
\]

3. Divide by \( 16 \):

\[
x = -\frac{10}{16} = -\frac{5}{8}
\]

Now substitute \( x = -\frac{5}{8} \) into either equation to get \( y \). Using the first equation:

\[
y = 3\left(-\frac{5}{8}\right) + 6 = -\frac{15}{8} + 6 = -\frac{15}{8} + \frac{48}{8} = \frac{33}{8}
\]

Thus, the point of intersection (solution) is:

\[
\left(-\frac{5}{8}, \frac{33}{8}\right)
\]

In conclusion, there is **one solution** to the system of equations \( y = 3x + 6 \) and \( y = -13x - 4 \).