Question

Find the value of k so that the following system of equations has no solution:

3x – y – 5 = 0, 6x – 2y + k = 0​

Answer 1

$\color{red}\huge{\underline{\underline{GIVEN\dag}}}$

• $3x - y - 5 = 0$
• $6x - 2y + k = 0$
• This system of lines has 'no solution'.

$\color{red}\huge{\underline{\underline{SOLUTION\dag}}}$

We know that;

Condition for no solution for system of lines is:

$\blue{\boxed{\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}}}$

where,

• a,b & c are co-efficeints.

Here,

• $a_{1}=3$ ; $a_{2}=6$
• $b_{1}=-1$ ; $b_{2}=-2$
• $c_{1}=-5$ ; $c_{2}=k$

$\implies\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\implies\frac{3}{6}=\frac{ - 1}{ - 2}=\frac{ - 5}{k}$

$\implies\frac{1}{2}=\frac{ 1}{ 2}=\frac{ - 5}{k}$

$\implies\frac{ 1}{ 2}=\frac{ - 5}{k}$

$\implies \frac{k}{2} = - 5$

$\color{orange} \boxed{\therefore k = - 10}$

HOPE THAT HELPS!!