# Finding unknown variables

• Aug 15th 2011, 08:07 AM
KayPee
Finding unknown variables
Two numbers are each multiplied by themselves to give two new numbers. The difference between these two new numbers is less than ten; the difference between the two original numbers was one.
The two original numbers added together was more than 7. What is one of the original numbers?

Let x and y represent the first two numbers

Let a and b represent the two new numbers

xy=ab

a-b < 10

x-y = 1

x + y > 7

Am i on the right path?

What do I do next?
• Aug 15th 2011, 08:58 AM
TheChaz
Re: Finding unknown variables
I think the first sentence means that you have two numbers, say x and y, and they are each squared... this gives you x^2 and y^2 (two "new numbers").

Then x^2 - y^2 < 10
x - y = 1
x + y > 7
Multiply the last two together to get
x^2 - y^2 = (x + y)(x - y) > 7 * 1 = 7
• Aug 22nd 2011, 06:25 PM
KayPee
Re: Finding unknown variables
I'm still at a loss.I don't quite get it.

• Aug 22nd 2011, 06:47 PM
Quacky
Re: Finding unknown variables
Are you aware of the difference of two squares? We have two numbers, \$\displaystyle x\$ and \$\displaystyle y\$. We are told they are multiplied by themselves, to give \$\displaystyle x^2\$ and \$\displaystyle y^2\$.
We are told then that the difference of the new numbers is less than \$\displaystyle 10\$, so \$\displaystyle x^2-y^2<10\$
Using the difference of 2 squares rule,
\$\displaystyle (x+y)(x-y)<10\$
We are told that the original difference was \$\displaystyle 1\$, so \$\displaystyle x-y=1\$
This means that \$\displaystyle (x+y)(1)<10\$
\$\displaystyle (x+y)<10\$
We also know that \$\displaystyle x+y>7\$
This means that
\$\displaystyle 7<x+y<10\$

Assuming \$\displaystyle x\$ and \$\displaystyle y\$ are integers, this doesn't leave many options when \$\displaystyle x-y=1\$. What are the options?
• Aug 22nd 2011, 07:17 PM
KayPee
Re: Finding unknown variables
We are looking at 4 and 5 ?
• Aug 22nd 2011, 07:24 PM
Quacky
Re: Finding unknown variables
I believe so.