Question

♥ Hello Friends !!! ♥


Answer this question ⤵⤵


$\mathsf {The \: pair \: of \: equations (p^{2} \: - \: 1)x \: + \: (q^{2} \: - \: 1)y \: + \: r \: = \: 0 \: and}$
$\mathsf {(p+1)x + (q-1)y + r = 0 \: have \: infinitely \: many \: solutions,}$
$\textsf {then which of the following is true ?}$

A) pq = 0

B) pq = - 1

C) pq = 1

D) pq = - 2



❎ NO SPAM ❎


Best answer will be marked as BRAINLIEST!

Answer 1

Hi dear !!!

condition for infinity many solutions are :-

a1/ a2 = b1/ b2= C1 / C2

=> p² -1 / p+1 = q²-1 / q -1 = r / r

p²-1 / p+1 = r / r ---------- { comparing }

p²-1 / p+1 = 1

( p +1 )(p -1) / ( p +1 ) = 1

p -1 = 1

P = 2

Similarly

=> q²-1 / q -1 = r / r

(q +1 ) ( q-1 ) / (q-1) = 1

q + 1 = 1

q = 0
P. =2

q = 0

p× q = 0 ×2

pq=0

hence option number is A correct !!!

hope it help you dear !!!

thanks !!!