Question

For what value of K , x=1, y=3, is the solution of (2k+3)x-(k+1)y-3=0 and proof the solution by putting the value of k
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Answer 1

$\large \blue {Given}$

x = 1

y = 3

$\large \blue {To :\ Find}$

Value of k.

$\large \blue {Solution}$

$\implies$ (2k + 3)x - (k + 1)y = 0

$\implies$ 2xk + 3x - yk + y = 0

$\implies$ 2×1×k + 3×1 - 3×k + 3 = 0

$\implies$ 2k + 3 - 3k + 3 = 0

$\implies$ -k + 6 = 0

$\implies$ -k = -6

$\implies$ k = 6 (minus cancelled)

$\large \pink {Proof}$

$\implies$ ( 2k + 3 )x - ( k + 1 )y - 3 = 0

$\implies$ ( 2×6 + 3 )1 - ( 6 + 1)3 - 3 = 0

$\implies$ ( 12 + 3 ) - ( 18 + 3 ) - 3 = 0

$\implies$ ( 12 + 3 - 18 + 3 - 3) = 0

$\implies$ 0 = 0

Hence proved