Math problem!!!?

Question

Divide 15 into two parts such that when the larger is divided by 3 and the smaller is divided by 2, the quotients are equal.
thx 4 answering!!!

Answer 1

Let one part be "x", then the other part is 15 - x

So,

x/3 = (15 - x)/2 Multiply both side by 6
2x = 3(15 - x)
2x = 45 - 3x
5x = 45
x = 9 and the other part = 15 - 9 = 6

HTH

Charles

Answer 2

Just turn it into an algebraic problem:
(1) Divide 15 into two parts
let one part be x (larger), and the other be (15 -x) (smaller)

larger divided by 3 = smaller divided by 2
therefore
x/3 = (15-x)/2
2x = 3(15 - x)
= 45 - 3x
5x = 45
x = 9

therefore the two parts are 9, (15-9)
giving 9 and 6!

Just to test the values:
9 + 6 = 15 (Correct)
9 / 3 = 3 and 6 / 2 = 3 (Correct)

Answer 3

Unless i'm reading this wrong, this is a system containing the equations...
x+y=15
x/3=y/2

First, you use the second equation to solve for one of the variables. I like x, so...
x=(3/2)y
Then, you substitute this value of x into the first equation...
(3/2)y + y=15
Simplify...
(5/2)y=15 .... 5y=30 ... y=6

Then, you use this value of y along with the second equation to find your x value.
x/3=y/2 ... x=(3/2)y ... x=(3/2)(6) ... x=9.


Pay close attention to the steps I did. If you plan on pursuing higher level math, solving systems of equations is an assumed knowledge.

Answer 4

15 = a + b <==> 15 - a = b
a/3 = q
b/2 = q

a/3 = b/2 <==> 2a = 3b

2a = 3(15 - a)
2a = 45 - 3a
5a = 45
a = 9
2a = 3b ==> b = 2a/3 = 2*9/3 = 2*3 = 6

a = 9, b = 6

Answer 5

There aren't that many ways to do a+b=15 with integral a and b, so a=9 and b=6 by trial-and-error.