Question

For the following system of equation determine the value of k for which the given system of equation has infinitely many solutions.
$5x+2y=k$
$10x+4y=3$

Answer 1

The given system of equations is
$5x+2y-k=0$
$10x+4y-3=0$

This system of equation is of the form
$a_{1}x+b_{1}y+c_1=0$
$a_{2}x+b_{2}y+c_2=0$

where $a_1=5,b_1=2,c_1=-k$

and $a_2=10,b_2=4$ and $c_2=-3$

For infinitely many solutions, we must have
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ i.e., $\frac{5}{10}=\frac{2}{4}=\frac{-k}{-3}\Rightarrow \frac{1}{2}=\frac{k}{3}\Rightarrow k=\frac{3}{2}$

Hence, the given system of equations will have infinitely many solutions, if $k=\frac{3}{2}$.