Number of solutions

When there is exactly one solution

The most common case is that a system of equations has exactly one solution. This happens when the two equations together give a specific pair of values for \( \large x \) and \( \large y \).

 

$$ \large x+y=10 $$

$$ \large x-y=2 $$

 

By solving the system (for example with substitution or elimination) we find:

 

$$ \large x=6, \quad y=4 $$

 

The system therefore has exactly one solution.

 

 

When there are many solutions

Some systems of equations will result in infinitely many solutions:

 

$$ \large x+y = 20 $$

$$ \large 4x+4y=80 $$

 

If we use the Method of Equal Coefficients, we must multiply the upper equation by 4. Both equations then become identical:

 

$$ \large \textcolor{red}{4}x+\textcolor{red}{4}y=80 $$

$$ \large 4x+4y=80 $$

 

When we subtract the two equations, we get:

 

$$ \large 0=0 $$

 

This is true, and it means that all values where \( \large x+y=20 \) are solutions.

 

$$ \large x=5, \quad y=15 $$

$$ \large x=\frac{22}{2}, \quad y=\frac{18}{2} $$

 

Thus there are infinitely many solutions.

 

 

When there is no solution

Some systems of equations have no solution at all:

 

$$ \large x+y = 20 $$

$$ \large 4x+4y=60 $$

 

If we use the Method of Equal Coefficients, we must multiply the upper equation by 4:

 

$$ \large \textcolor{red}{4}x+\textcolor{red}{4}y=80 $$

$$ \large 4x+4y=60 $$

 

When we subtract the two equations, we get:

 

$$ \large 0=20 $$

 

This is not true. It means that no matter what numbers are inserted for \( \large x \) and \( \large y \), the system can never be satisfied.

The system therefore has no solution.

 

 

Summary

• One solution: The system gives one specific result for \( \large x \) and \( \large y \).
• Many solutions: Both equations turn out to be the same, so there are infinitely many solutions.
• No solution: The equations contradict each other, so there is no solution at all.