Question

The value of ‘k’ , so that the system of equation 3x-y-5=0 and 6x-2y-k=0 have infinitely many solutions is

1 point

K=-10

K=10

K=-8

K=8

​

Answer 1

Answer:

to have infinite many solutions for the equation

3x - y - 5 =0 and 6x - 2y - k = 0

we want

$a \frac{}{1} \div a \frac{}{2} = b \frac{}{1} \div b \frac{}{2} = c \frac{}{1} \div c \frac{}{2}$

so, 3 / 6 = -1 / -2 = -5 / -k

1/2 = 1/2 = 5/k

now, 1/2 = 5/ k

cross multiplication

k = 10

Answer 2

a1, b1, c1 and a2, b2, c2 = k

a1/a2 = b1/b2 not =c1/c2

i.e., 3/6 = -1/-2 not= -5/-k

=> -1/-2 not = -5/-k

=> -k = -10 => k=10