Question

Find the value of k for which the system 2x+ky=1 and 3x-5y=7
will have (a) a unique solution and(b) no solution.is there a value of k for which the system has infinitely many solutions

Answer 1

SolutioN :

$\tt \dagger \: \: \: \: \: 2x + ky = 1. \: \: \: \: \: - (1)$

$\tt \dagger \: \: \: \: \: 3x - 5y = 7. \: \: \: \: \: - (2)$

$\rule{90}2$

Let's try With Some Case.

✵ Case 1.

• (a) a unique solution.

☛ Condition.

$\tt \dagger \: \: \: \: \: \dfrac{a_1 }{a_2} \neq \dfrac{b_2}{b_2}$

Where as,

• a1 = 2.
• a2 = 3.
• b1 = k.
• b2 = - 5.
• c1 = 1.
• c2 = 7.

$\tt : \implies \dfrac{2 }{3} \neq \dfrac{k}{ - 5}$

$\tt : \implies \dfrac{ - 10 }{3} \neq k$

$\tt : \implies k \neq \dfrac{ - 10 }{3}$

$\rule{90}2$

✵ Case 2.

• No Solution.

☛ Condition.

$\tt \dagger \: \: \: \: \: \dfrac{a_1 }{a_2} = \dfrac{b_2}{b_2} \neq \dfrac{c_1}{c_2}$

$\tt : \implies \dfrac{2 }{3} = \dfrac{k}{ - 5} \neq \dfrac{1}{7}$

$\tt : \implies {k} = - \dfrac{10}{3}$

And,

$\tt : \implies \dfrac{k}{ - 5} \neq \dfrac{1}{7}$

$\tt : \implies k \neq - \dfrac{5}{7}$

$\rule{90}2$

✵ Case 3.

• Many Solution.

☛ Condition.

$\tt \dagger \: \: \: \: \: \dfrac{a_1 }{a_2} = \dfrac{b_2}{b_2} = \dfrac{c_1}{c_2}$

$\tt : \implies \dfrac{2}{3} = \dfrac{k}{ - 5} = \dfrac{1}{7}$

$\tt : \implies {k} = - \dfrac{10}{3}$

And,

$\tt : \implies \dfrac{k}{ - 5} = \dfrac{1}{7}$

$\tt : \implies k = - \dfrac{5}{7}$