Question

find the values of k for which the following pair of linear equations has no solutions 1) (k+1)x+2y=5 and 3x +ky =7
PLS ANSWER
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Answer 1

a1= k+1
b1 = 2
a2=3
b2=k
in no solution case
a1/a2 = b1/b2 not = c1/c2
k+1/3 = 2/k
k(k+1)=3×2
${k}^{2} + k = 6$
or
$so \: {k}^{2} + k - 6 = 0$
or
${k}^{2} + 3k \: - 2k - 6 = 0$
or
k(k+3) -2(k+3)=0
(k-2)(k+3)=0
so k=2 or k=-3
hope it helps you

Answer 2

Hi there !!

(k+1)x+2y=5 and 3x +ky =7

a1= k+1

b1= 2

c1=-5

a2=3

b2=k

c2=-7

For no solution,

a1/a2=b1/b2≠c1/c2

k-1/3 =2/k≠5/7

k²-k=6

k²-k-6=0

k²-3k+2k-6=0

k(k-3)+2(k-3)=0

K=3 and K= -2 (Ignoring -ve value )

and 14/5≠k

Thankyou :)