Question

problem 3: describe how you can tell a system of equations has no solutions.

Explanation:

1. For linear systems solved algebraically: When simplifying, you end up with a contradiction like $0 = c$ where $c$ is a non-zero number. For graphing linear systems: The lines are parallel (same slope, different y-intercepts), so they never intersect.
2. For non-linear systems: After substitution/elimination, the resulting equation has no real solutions (e.g., $x^2 = -5$), or the graphs of the equations never intersect.

Answer:

A system of equations has no solutions if:

1. For linear systems:

• Algebraically: Simplifying leads to a false numerical statement (e.g., $0=5$).
• Graphically: The equations represent parallel, non-overlapping lines (same slope, different intercepts).

1. For non-linear systems:

• Algebraically: Substitution/elimination results in an equation with no real valid solutions (e.g., a squared term equal to a negative number).
• Graphically: The graphs of the equations do not intersect at any point in the real plane.