SOLUTION: I have chosen three numbers. The second is twice the first, and the third is three times the second. The sum of the first two when multiplied by the sum of the last two happens to

Question 246730: I have chosen three numbers. The second is twice the first, and the third is three times the second. The sum of the first two when multiplied by the sum of the last two happens to be the same as the first number multiplied by the square of the second number. What are the three numbers?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
I have chosen three numbers.
x, y, z
:
The second is twice the first,
y=2x
or
x=.5y
:
the third is three times the second.
z = 3y
:
The sum of the first two when multiplied by the sum of the last two happens to be the same as the first number multiplied by the square of the second number.
(x+y)*(y+z) = x * y^2
Substitute for x and z
(.5y + y)* (y + 3y) = .5y * y^2
1.5y * 4y = .5y^3
6y^2 = .5y^3
Mult by 2
12y^2 = y^3
Divide both sides by y^2
12 = y
then
z = 3(12)
z = 36
and
x = .5(12)
x = 6
:
Check solution in original equation
(x+y)*(y+z) = x * y^2
(6+12)*(12+36) = 6 * 12^2
18 * 48 = 6 * 144
864 = 864
:
What are the three numbers? 6, 12, 36
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