if the average of 16 x and 8 y is greater than 80 , 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 16 x and 4 x is greater than 80 . average of 16 x and 4 x is 10 x . so , 10 x > 80 x > 8 answer : a"
multiply(16, const_2)