Question

ही 1 SET-1 11.kके किसी मान के लिए समीकरण निकाय 3x+4y35 तथा 6x +by= 10 का अनगिनत हल होगा? (A)3 (B) 6 (D) 9 (C) 8 +2y=1 तथाxt
$3x + 4y = 5 \: \: a nd \: 6x + ky = 10$
​with formula

Answer 1

Answer:

8

Explanation:

For infinite many solutions, the condition is

$\frac{a1}{b1} = \frac{a2}{b2} = \frac{a3}{b3}$

so putting values

$\frac{3}{6} = \frac{4}{k} = \frac{5}{10}$

on solving,

$k =\frac{4 \times 6}{3} = \frac{24}{3} = 8$

So your answer is 8

HOPE IT HELPS☺️