What is the type of solution the pair of linear equation `x + 3y = 4` and `2x + y = 5` have.

Options

A. unique
B. Infinite
C. No Solution
D. Both (a) and (b)

Answer

To determine the type of solution for the pair of linear equations \(x + 3y = 4\) and \(2x + y = 5\), we will follow these steps:

### Step 1: Write the equations in standard form
The standard form of a linear equation is \(Ax + By + C = 0\).

1. For the first equation \(x + 3y = 4\):
\[ x + 3y - 4 = 0 \]
Here, \(A_1 = 1\), \(B_1 = 3\), and \(C_1 = -4\).

2. For the second equation \(2x + y = 5\):
\[ 2x + y - 5 = 0 \]
Here, \(A_2 = 2\), \(B_2 = 1\), and \(C_2 = -5\).

### Step 2: Find the ratios of the coefficients
We need to calculate the ratios \( \frac{A_1}{A_2} \), \( \frac{B_1}{B_2} \), and \( \frac{C_1}{C_2} \).

1. Calculate \( \frac{A_1}{A_2} \):
\[ \frac{A_1}{A_2} = \frac{1}{2} \]

2. Calculate \( \frac{B_1}{B_2} \):
\[ \frac{B_1}{B_2} = \frac{3}{1} = 3 \]

3. Calculate \( \frac{C_1}{C_2} \):
\[ \frac{C_1}{C_2} = \frac{-4}{-5} = \frac{4}{5} \]

### Step 3: Analyze the ratios
We need to check the conditions for the types of solutions:

1. **Unique Solution**: If \( \frac{A_1}{A_2} \neq \frac{B_1}{B_2} \).
2. **Infinite Solutions**: If \( \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \).
3. **No Solution**: If \( \frac{A_1}{A_2} = \frac{B_1}{B_2} \) but \( \frac{C_1}{C_2} \neq \).

### Step 4: Compare the ratios
From our calculations:
- \( \frac{A_1}{A_2} = \frac{1}{2} \)
- \( \frac{B_1}{B_2} = 3 \)
- \( \frac{C_1}{C_2} = \frac{4}{5} \)

Since \( \frac{1}{2} \neq 3 \) and \( \frac{1}{2} \neq \frac{4}{5} \), we conclude that:

- The condition for a unique solution is satisfied.

### Conclusion
The pair of linear equations \(x + 3y = 4\) and \(2x + y = 5\) has a **unique solution**.

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