Question

How to solve a system of linear equations with the same coefficients but different constants?

Answer 1

To solve a system of linear equations with the same coefficients but different constants, you can follow these steps:

### Step 1: Write the System of Equations
Assume you have a system of two linear equations:
\[
a_1 x + b_1 y = c_1
\]
\[
a_2 x + b_2 y = c_2
\]
If the coefficients are the same, it could look like:
\[
ax + by = c_1
\]
\[
ax + by = c_2
\]

### Step 2: Analyze the System
When the coefficients of \(x\) and \(y\) are identical, the two equations represent parallel lines unless \(c_1 = c_2\). Here's how to analyze different scenarios:

1. **If \(c_1 \neq c_2\)**:
- The lines are parallel; thus, there are **no solutions**. This means that the system is inconsistent.

2. **If \(c_1 = c_2\)**:
- The two equations are actually the same line, which means there are **infinitely many solutions**. In this case, any point that lies on the line represented by the equation is a solution.

### Step 3: Conclusion
Based on the values of \(c_1\) and \(c_2\), you can conclude:

- **No solution**: When \(c_1 \neq c_2\) (parallel lines).
- **Infinitely many solutions**: When \(c_1 = c_2\) (identical lines).

### Example
Consider the following system:
1. \(2x + 3y = 5\)
2. \(2x + 3y = 7\)

Here, since \(5 \neq 7\), the lines are parallel. Therefore, there are no solutions to this system.

Now consider:
1. \(2x + 3y = 5\)
2. \(2x + 3y = 5\)

Since both equations are identical, there are infinitely many solutions. Any \((x, y)\) that satisfies \(2x + 3y = 5\) is a solution.

### Alternative Methods
If you wish to investigate this further mathematically, you can use substitution or elimination methods to see that the equations lead to contradictions or repeated solutions.

This process allows you to determine the nature of systems of equations with the same coefficients efficiently.