for what values of k will the pair of equations 3 x + 4 y = 12 and ( kx + 12 y ) / 2 = 15 does not have a unique solution ?
"we have 2 equations 1 . 3 x + 4 y = 12 - - > 9 x + 12 y = 36 2 . ( kx + 12 y ) / 2 = 15 - - > kx + 12 y = 30 substract 1 - 2 , we get ( 9 - k ) x = 6 i . e . x = 6 / ( 9 - k ) then , by looking at options , we get some value of x except for b . when we put k = 9 , x becomes 6 / 0 and hence answer is b"
divide(multiply(2, 4), 3)