Question

Find the value of k for which the following system of equation has unique solution 4x+ky+8=0,2x+2y+2=0

Answer 1

for unique solution
a1 upon a2 is not equal to b1 uponb2
therefore
$4 \div 2$
is not equal to k÷ 2
8is not equal to 2k
therefore kis not equal to 4

Answer 2

Answer:

Step-by-step explanation:

since the equations has no solutions , the condition is

        a1/a2 not equal to b1/b2

given: a1/a2=4/2

        b1/b2=k/2

or,   2 not equal to k/2

or,k not equal to 4

the value of k can be any number except 4.