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

1 answer

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:

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

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

\[ 16x = -10 \]

  1. 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 \( y = -13x - 4 \).