Question 42654
Find two consecutive integers such that the sum of 2 times the first integer and 4 times second integer is 46


Let the consecutive integers be 'n' and 'n+1'.
Two times first integer = 2n.
Four times second integer = 4(n+1).
Then their sum = 2n + 4(n+1) = 6n+4.
Given: this sum = 46.
So 6n+4 = 46
or 6n = 46 - 4 = 42
or n = 42/6 = 7


Hence the consecutive numbers are n = 7 and (n+1) = 7+1 = 8.