|
Question 173635: Four numbers have a total of 92. The first number is twice the second number and the third number is 4 more than the sum of the first two. The fourth number is 3 less than the difference of the first two numbers. What are the numbers?
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! let a,b,c,d = the numbers.
-----
a + b + c + d = 92
-----
first number is twice the second number:
a = 2b
-----
third number is 4 more than the sum of the first two:
c = a + b + 4
-----
fourth number is 3 less than the difference of the first two numbers.
d = a - b - 3
since a > b this would be positive.
it works this way.
if we chose b - a, the difference would be negative and the result would not work out.
the assumption here is that we are looking for the absolute value of the difference which makes the difference always positive.
-----
the rest is just a substitution for what we know.
we start with:
a + b + c + d = 92
since a = 2b, we substitute 2b for a to get:
2b + b + c + d = 92
since c = a + b + 4, we substitute a + b + 4 to get:
2b + b + a + b + 4 + d = 92
since a = 2b, we again substitute 2b for a to get:
2b + b + 2b + b + 4 + d = 92
since d = a - b - 3, e substitute a - b - 3 for d to get:
2b + b + 2b + b + 4 + (a - b - 3) = 92
since a = 2b, we again substitute 2b for a to get:
2b + b + 2b + b + 4 + (2b - b - 3) = 92
we remove parentheses to get:
2b + b + 2b + b + 4 + 2b - b - 3 = 92
we combine likes to get:
7b + 1 = 92
we subtract 1 from both sides of the equation to get:
7b = 91
we divide both sides of the equation by 7 to get:
b = 13
-----
since b = 13, then
a = 2b = 26
c = a + b + 4 = 26 + 13 + 4 = 43
d = a - b - 3 = 26 - 13 - 3 = 10
-----
all variables are:
a = 26
b = 13
c = 43
d = 10
-----
sum of these is 92 satisfying the equation a + b + c + d = 92
a = 2b is good since 26 is 2 * 13
c = a + b + 4 = 43 is good since 26 + 13 + 4 = 43
d = a - b - 3 is good since 26 - 13 - 3 = 10
-----
your answer is:
a = 26
b = 13
c = 43
d = 10
|
|
|
| |
