For all real values of c,

False, the given pair of linear equations

x-2y-8=0

5x-10y=c

Here,   $a_{1}=1, b_{1}=-2, c_{1}=-8$

$a_{2}=5, b_{2}=-10, c_{2}=-c$

Now, $\frac{a_{1}}{a_{2}}=\frac{1}{5}, \frac{b_{1}}{b_{2}}=\frac{-2}{-10}=\frac{1}{5}$

$\frac{c_{1}}{c_{2}}=\frac{-8}{-c}=\frac{8}{c}$

But if $c=40$ (real value), then the ratio $\frac{c_{1}}{c_{2}}$ becomes $\frac{1}{5}$ and then the system of linear

equations has an infinitely many solutions.

Hence, ate = 40, the system of linear equations does not have a unique solution