Question

Find the value of k for which each of the following systems of equations have infinitely many solution:

(i) 2x - 3y = 7
(k+2)x - (2k+1)y = 3(2k-1)....

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Answer 1

Step-by-step explanation:

if

$\frac{2}{k + 2} = \frac{3}{2k + 1} = \frac{7}{2k - 1}$

then it has infinitely many solutions

4k+2 = 3k+6

k=4

but we see that for this value of k 7/2k-1 is not equal to the remaining ones

so for no value of k this equation has infinitely many solutions

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Answer 2

Answer:

Hello ! guys here's Ur ans...

I) 2x - 3y = 7. ........(1)

( k + 2) x - ( 2k+1)=3(2k-1).....(2)..

Now, A1 /A2 = 2x / (k+2)x

b1 / b2 = -3y/ -(2k+1)

C1 / c2 = 7/3(2k-1)..

As , it has infinity solutions...(coincidence lines)

so, A1/ A2 = b1 / b2 = C1/ c2

= 2x / (k +2)x = 7/ 3(2k+1)

= 7kx + 14x = 2x ( 6k + 3)

= 7kx +14x= 12kx + 6x

= 7kx - 12kx = 6x - 14x

= -5kx = - 8x

= 5k = 8

= k = 8/5(ans)..

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