Find three mixed numbers so that the sum is 18 and the difference between the greatest number and the least number is 5 1/5
let the 3 numbers be x, y, and z from smallest to largest
x+y+z = 18
z-x = 26/5
z = 26/5 + x
into the first:
x + y + 26/5 + x = 18
2x+y = 64/5
y = 64/5 - 2x
= (64-10x)/5
but x, y, and z are to be mixed numbers, that is, they must all be greater than 1
so to find y, 64-10x > 5
-10x > -59
x < 5.9
so we need 1 < x < 5.9
so pick value of x within that domain, say
x = 4
then y = (64-40)/5 = 24/5
and z = 26/5 + 24/5 = 10
one possible answer is
x= 4/1, y=24/5, z=10/1
another:
let x = 5/3
y = (64-50/3)/5 = 142/15
z = 26/5 + 5/3 = 103/15
check:
sum = 5/3 + 142/15 + 103/15 = 18
difference 103/15-5/3 = 26/5 = 5 1/5
so you can pick any x you want within the given domain, and find the corresponding triple.
There will be an infinite number of such possible triples
To solve this problem, we need to set up a system of equations. Let's assume that the three mixed numbers are a, b, and c, with a being the least, b being the middle, and c being the greatest number.
First, we know that the sum of the three mixed numbers is 18, so we have the equation:
a + b + c = 18 -- equation (1)
Second, we know that the difference between the greatest number and the least number is 5 1/5, which can be written as a mixed number 5 + 1/5 = 26/5. So, we have the equation:
c - a = 26/5 -- equation (2)
We now have a system of equations consisting of equations (1) and (2). To solve this system, we can use the method of substitution or elimination.
Let's solve this system using the method of substitution. Solving equation (2) for c, we get:
c = a + 26/5
Now, substitute this value of c into equation (1):
a + b + (a + 26/5) = 18
Combine like terms:
2a + b + 26/5 = 18
Multiply through by 5 to eliminate fractions:
10a + 5b + 26 = 90
Move the constant term to the right side:
10a + 5b = 90 - 26
10a + 5b = 64 -- equation (3)
At this point, we have two equations:
2a + b + 26/5 = 18 -- equation (1)
10a + 5b = 64 -- equation (3)
Now, we can solve this system of equations using various methods, such as substitution or elimination, to find the values of a, b, and c.