Question

Find the value of k for which the system of equation
x + 2y = 5, 3x + ky - 15 = 0​

Answer 1

Answer

The given system of equation is

$\sf x + 2y - 5 = 0$

$\sf 3x + ky - 15 = 0$

These equations are of the form

$\sf \dashrightarrow {a}_{1} x + {b}_{1} y + {c}_{1} = 0$

$\sf \dashrightarrow {a}_{2}x + {b}_{2}y + {c}_{2} = 0$

where,

$\sf \dashrightarrow {a}_{1} = 1, {b}_{1} = 2, {c}_{1} = -5$

$\sf \dashrightarrow {a}_{2} = 3, {b}_{2} = k, {c}_{2} = -15$

$\sf \therefore \dfrac{{a}_{1}}{{a}_{2}} = \dfrac{1}{3}, \dfrac{{b}_{1}}{{b}_{2}} = \dfrac{2}{k} \: \: and \: \: \dfrac{{c}_{1}}{{c}_{2}} = \dfrac{-5}{-15} = \dfrac{1}{3}$

Let the given system of equations have no solution.

Then, $\sf \dashrightarrow \dfrac{{a}_{1}}{{a}_{2}} = \dfrac{{b}_{1}}{{b}_{2}} \neq \dfrac{{c}_{1}}{{c}_{2}}$

$\sf \dashrightarrow \dfrac{1}{3} = \dfrac{2}{k} \neq \dfrac{1}{3}$

$\sf \dashrightarrow \dfrac{1}{3} = \dfrac{2}{k} \: \: and \: \: \dfrac{2}{k} \neq \dfrac{1}{3}$

So, $\sf k = 6 \: \: and \: \: k \neq 6$ which is impossible.

Thus, there is no value of k for which the given system of equations has no solution.