Step-by-step explanation:
Given system of linear equations are:
[tex] x + 2y = 3\\
\therefore\: x + 2y - 3 = 0......... (1)\\
\& \:5x + ky + 7 = 0........ (2)\\[/tex]
Equating equations (1) & (2) by
[tex] a_1x+b_1y + c_1=0\: \&\:\\ a_2x+b_2y + c_2=0[/tex]
We find:
[tex] a_1=1, \:\:b_1=2, \:\: c_1=-3\: \\\&\:\\ a_2=5, \:\:b_2=k, \:\: c_2=7\\
[/tex]
The condition for system of linear equations having no solution is given as:
[tex]\frac{a_1}{a_2} =\frac{b_1}{b_2} \neq\frac{c_1}{c_2} \\\\
\therefore \frac{1}{5} =\frac{2}{k} \neq\frac{-3}{7} \\\\\therefore \frac{1}{5} =\frac{2}{k} \\\\
\therefore k = 5\times 2\\\\
\huge \purple {\boxed {\therefore k = 10}} [/tex]
Thus, for k = 10 given system of linear equations have no solution.
