Question

(3,1) is a solution of ( (21 c-3) x+4 k y+9 k+5 )
( S_{0}, ) det's put ( x=3 ) & ( y=1 ) in given equation
( Rightarrow(2 k-3) times 3+4 k times 1=9 k+5 )
( 7 x-9+4 k=9 k+5 )
( Rightarrow quad 10 k-9 quad=9 k+5 )
( j quad 10 k-frac{9 k}{k}=frac{5+1}{14} )

# Q.13 if (3, 1) is a solution of the equation (2k-3) x +4ky = 9k+5 find the value of k. olutie m alina tion Av 12

Solution