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for what value of k,will the following pair of equation have infinitely many solutions:2x+3y=7 and (k+2)x-3 (1-k)y =5k+1


Tehzeeb Jiya 19th Mar, 2021
Answer (1)
Subhrajit Mukherjee 19th Mar, 2021

Given pair of equations are

2x + 3y = 7 and

(k + 2) x – 3 (1 – k) y = 5k + 1

For Infinitely many solutions,

a1/a2 = b1/b2 = c1/c2  -----------(i)

From the given equations,

a1 = 2, b1=3, c1=7

a2 = k+2, b2 = -3(1-k)

c2 = 5k+1

Thus from equation (i) we get,

2/(k+2) = 3/-3(1-k)=7/5k+1

Thus equating we get,

2/(k+2)=1/(k-1)

=> 2k-2=k+2=> k=4

For k=4 you can have infinite number of solutions.

I hope my answer helps. All the very best for your future endeavors!


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