Question

FOR WHAT VALUE OF K, THE FOLLOWING PAIR OF EQUATIONS HAVE INFINITELY MANY SOLUTIONS, 2x + 3y=7, (k+a) x + (2k-1)y =4k+1

Answer 1

HERE IS YOUR ANSWER,

FOR INFINITELY MANY SOLUTIONS

a1÷a2 = b1÷b2 = c1÷c2

a1÷a2= 2÷k+a
b1÷b2= 3÷2k-1
c1÷c2= 7÷4k-1

we can't equate a1÷a2 because it have the alphabet a. we need only k. so we equate b1÷b2 and c1÷c2.

b1÷b2= 3÷2k-1. = c1÷c2= 7÷4k+1

3÷2k-1 = 7÷4k+1
we do cross multiply to get the solution.

3(4k+1) = 7(2k-1)
12k+3 = 14k-7
12k-14k= -7-3
-2k= -10
k= -10÷ -2
minus and minus get cancelled
k=10÷2
so, k= 5

HENCE THE ANSWER IS 5.


HOPE IT HELPS

PLS MARK IT AS THE BRAINLIEST.

Answer 2

Answer:

k= 5

Step-by-step explanation:

a1÷a2 = b1÷b2 = c1÷c2

a1÷a2= 2÷k+a

b1÷b2= 3÷2k-1

c1÷c2= 7÷4k-1

b1÷b2= 3÷2k-1. = c1÷c2= 7÷4k+1

3÷2k-1 = 7÷4k+1

3(4k+1) = 7(2k-1)

12k+3 = 14k-7

12k-14k= -7-3

-2k= -10

k= -10÷ -2

k=10÷2

so,