For all real values of c, the pair of equations `x-2y= 8, 5x- 10y= c` have a unique solution. Justify whether it is true or false

Answer

To determine whether the pair of equations \( x - 2y = 8 \) and \( 5x - 10y = c \) has a unique solution for all real values of \( c \), we will analyze the conditions for a unique solution in a system of linear equations.

### Step-by-Step Solution:

1. **Identify the coefficients of the equations**:
The first equation is:
\[ x - 2y = 8 \]
This can be rewritten in the standard form \( Ax + By + C = 0 \):
\[ 1x - 2y - 8 = 0 \]
Here, \( a_1 = 1 \), \( b_1 = -2 \), and \( c_1 = -8 \).

The second equation is:
\[ 5x - 10y = c \]
This can also be rewritten in standard form:
\[ 5x - 10y - c = 0 \]
Here, \( a_2 = 5 \), \( b_2 = -10 \), and \( c_2 = -c \).

2. **Check the condition for a unique solution**:
For a system of linear equations to have a unique solution, the following condition must be satisfied:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]

3. **Calculate \( \frac{a_1}{a_2} \) and \( \frac{b_1}{b_2} \)**:
- Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{1}{5} \]
- Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{-2}{-10} = \frac{2}{10} = \frac{1}{5} \]

4. **Compare the two ratios**:
We found that:
\[ \frac{a_1}{a_2} = \frac{1}{5} \quad \text{and} \quad \frac{b_1}{b_2} = \frac{1}{5} \]
Since both ratios are equal, we have:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \]

5. **Conclusion**:
Since the condition for a unique solution is not satisfied (i.e., \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)), the system of equations does not have a unique solution for any value of \( c \). Therefore, the statement is **false**.