Associated Topics || Dr. Math Home || Search Dr. Math

### Solving an Algebraic Equation

```
Date: 12/21/95 at 6:57:2
From: Anonymous

If x = 2y = 3z =1, then what is x(y-z)squared + y(z-x)squared
+ z(x-y)squared?

The answer I ended up was 7/18, But my teacher said it was 1/3.
I don't get how it is done.
```

```
Date: 12/23/95 at 18:16:38
From: Doctor Elise

Hi!

I'm going to write "^2" instead of "squared" to save space.  Here we go:

>If x=2y=3z=1, then what is x(y-z)square + y(z-x)square + z(x-y)square??

x  = 1
2y = 1, so y = 1/2
3z = 1, so z = 1/3

So, x(y-z)^2 = 1(1/2 -1/3)^2 = (1/6)^2 = 1/36, and

y(z-x)^2 = 1/2(1/3 - 1)^2 = 1/2(-2/3)^2 = 1/2(4/9) = 4/18 = 2/9

and

z(x-y)^2 = 1/3(1-1/2)^2 = 1/3(1/2)^2 = 1/3(1/4) = 1/12

Finally,  1/36 + 2/9 + 1/12 = 1/36 + 8/36 + 3/36 = 12/36 = 1/3

Ta da!

-Doctor Elise,  The Geometry Forum

```
Associated Topics:
Middle School Algebra

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search