kx+3y=k-3 12x+ky=k find value of x and you
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Sol :
If a1 / a2 ≠ b1 / b2 then the pair of linear equations a1x + b1y + c1 = 0
a2x+ b2y + c2 = 0 has a unique solution.
Given pair of equations K x + 3y = K-3, 12x + Ky = K.
Here a1 = k ,b1 = 3 , a2 = 12 , b2 = k .
k / 12 ≠ 3 / k
k2 ≠ 36.
k ≠ ± 6.
If a1 / a2 ≠ b1 / b2 then the pair of linear equations a1x + b1y + c1 = 0
a2x+ b2y + c2 = 0 has a unique solution.
Given pair of equations K x + 3y = K-3, 12x + Ky = K.
Here a1 = k ,b1 = 3 , a2 = 12 , b2 = k .
k / 12 ≠ 3 / k
k2 ≠ 36.
k ≠ ± 6.
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