SOLUTION: four numbers have a total of 92. The first number is twice the second number and the third number is 4 times more than the sum of the first two. The fourth number is 3 less than

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Question 18985: four numbers have a total of 92. The first number is twice the second number and the third number is 4 times more than the sum of the first two. The fourth number is 3 less than the difference of the first two numbers. What is the difference?
I have access to the answer and after I have that I can do it...but I can not for the life of me figure how to get the fist number. I am doing a series of these problems and Im almost in tears...please help! Thank you!!!!!
Answer by askmemath(368) About Me  (Show Source):
You can put this solution on YOUR website!
Let the Second number = X
Now, the first number = 2X (since first number is twice the second number)
The third number = (X+2X)+ 4 = 3X +4 (third number is 4 times more than the sum of the first two)
The fourth number = 2X-X -3 = x-3(since it is 3 less than the difference of the first two numbers)

And finally we have the sum of all the numbers = 92
i.e. X+2X+3X +4 +X-3 =92
7x = 92-1
7x = 91
OR X = 13
Can you find out the rest of the numbers? Simply substitute 13 in the rest of the numbers to get their values.