Question

Find the value of k for which 5x – 2y + 4 = 0 and 3kx + (4 – 2k)y + 12 = 0 has infinitely many solutions​.please answer it as fast as possible

Answer 1

       $\text{\huge \bf \underline{Answer:}}$

       $\text{The two given equations are:}$

       $5x - 2y + 4 = 0$

       $3kx + (4-2k)y + 12 = 0$

       $\text{First, we have to check the values of }\dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}, \dfrac{c_1}{c_2}$

       $\text{In our case;}$

       $a_1 = 5 \: \: , \: \: b_1 = -2\: \: , \: \: c_1 = 4$

       $a_2 = 3k \: \: , \: \: b_2 = 4-2k\: \: , \: \: c_2 = 12$

$\implies \dfrac{a_1}{a_2} = \dfrac{5}{3k}$

$\implies \dfrac{b_1}{b_2} = \dfrac{-2}{4-2k}$

$\implies \dfrac{c_1}{c_2} = \dfrac{4}{12} = \dfrac13$

       $\text{It is given that the pair of equations must have infinitely many solutions}$

$\implies \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

$\implies \dfrac{5}{3k} = \dfrac{-2}{4-2k} = \dfrac13$

       $\text{Taking the first two}$

$\implies \dfrac{5}{3k} = \dfrac{-2}{4-2k}$

$\implies 5(4-2k) = -2(3k)$

$\implies 20 - 10k = -6k$

$\implies 20 - 10k + 6k = 0$

$\implies 20 - 4k = 0$

$\implies 20 = 4k$

$\implies \dfrac{20}{4} = k$

$\implies \boxed{\boxed{5 = k}}$