Find the value of k for which the linear equations x + 2y = 3 and 5x + ky = 7, does not have a unique solution.

Options

A. 5
B. 7
C. 2
D. 10

Answer

To find the value of \( k \) for which the linear equations \( x + 2y = 3 \) and \( 5x + ky = 7 \) do not have a unique solution, we can follow these steps:

### Step 1: Write the equations in standard form
The given equations are:
1. \( x + 2y - 3 = 0 \) (Equation 1)
2. \( 5x + ky - 7 = 0 \) (Equation 2)

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1:
- \( a_1 = 1 \)
- \( b_1 = 2 \)
- \( c_1 = -3 \)

- For Equation 2:
- \( a_2 = 5 \)
- \( b_2 = k \)
- \( c_2 = -7 \)

### Step 3: Use the condition for no unique solution
The condition for the two linear equations to not have a unique solution is given by:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \]
Substituting the values we identified:
\[ \frac{1}{5} = \frac{2}{k} \]

### Step 4: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 1 \cdot k = 5 \cdot 2 \]
\[ k = 10 \]

### Conclusion
The value of \( k \) for which the linear equations do not have a unique solution is \( k = 10 \).

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