If the system of equations 2x + 3y = 5, 4x + ky = 10


If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k =

(a) 1

(b) $1 / 2$

(c) 3

(d) 6


The given system of equations

$2 x+3 y=5$

$4 x+k y=10$

$\frac{a_{1}}{a_{2}}=\frac{2}{4}, \frac{b_{1}}{b_{2}}=\frac{3}{k}, \frac{c_{1}}{c_{2}}=\frac{5}{10}$

For the equations to have infinite number of solutions



If we take


$2 k=12$





$30=5 k$



Therefore, the value of k is 6.

Hence, the correct choice is $d$

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