Question

If 2x=3y=4z, then (x+y-z)/(x-y+z)=?

@Hero

Answer 1

This is a problem that is made to look difficult but it isn't.

Answer 2

yes?

Answer 3

@kryton1212 The 2x = 3y = 4z implies that each variable can be written in terms of any other variable. Hence you can re-write the expression in such a way that there is only 1 variable instead of 3.

Answer 4

Wow, that was like a hit and run @genius12

Answer 5

@genius12 how?
2/3=y/x
3/4=z/y
?

Answer 6

@Hero Sorry but I didn't understand the metaphorical phrasing there.

Answer 7

You posted your comment then ran away fast.

Answer 8

But im still here..

Answer 9

Weird, but okay

Answer 10

@kryton1212 not quite

use 2x=3y=4z to get something like

2x = 3y which gives y = 2x/3
2x = 4z which gives z = 2x/4

Now you can replace y and z in the equation.

Answer 11

are you kidding?
2x=2x=2x....

Answer 12

Huh? No. I don't know where you got that. If 2x=3y=4z then 2x=4z.

\[\Large \frac{ x+y-z }{x-y+z } = \]

\[\Huge \frac{ x+\frac{ 2x }{3 }-\frac{ x }{ 2 } }{x-\frac{ 2x }{3 }+\frac{ x }{ 2 } } = \]

Answer 13

ok...

Answer 14

Now try simplifying that... just gotta add fractions, notice you can factor out the x\[\Huge \frac{x( 1+\frac{ 2 }{3 }-\frac{ 1 }{ 2 } )}{x(1-\frac{ 2}{3 }+\frac{ 1 }{ 2 } )} = \]

Answer 15

\[\frac{ \frac{ 6x^{2}+x }{ 6x }}{ \frac{ 6x^{2}-6x }{ 6x } }\]

seems complicated...

Answer 16

I don't know how you got there... see my last post.

Answer 17

7/5
i think i am very stupid

Answer 18

Nope, 7/5 is correct :)

Answer 19

thank you