if the average of 6 x and 8 y is greater than 100 , and x is twice y , what is the least integer value of x ?
"substitution can be used in the following way : always start with the equation : x = 2 y . it is more straight forward to manage as compared to the inequality . substitute y = x / 2 , not the other way because you need to find the minimum value of x . so you can get rid of y . now go on to the inequality . so 8 y = 8 x / 2 = 4 x now average of 6 x and 4 x is greater than 100 . average of 6 x and 4 x is 5 x . so , 5 x > 100 x > 20 answer : b"
multiply(6, const_2)