- #1

RChristenk

- 64

- 9

- Homework Statement
- If ##\dfrac{2x-3y}{3z+y}=\dfrac{z-y}{z-x}=\dfrac{x+3z}{2y-3x}##, prove that each of these ratios is equal to ##\dfrac{x}{y}##

- Relevant Equations
- Basic fraction and algebra manipulation

I have the solution:

## \dfrac{2x-3y}{3z+y} = \dfrac{z-y}{z-x} = \dfrac {x+3z}{2y-3x} ## (1)

## \dfrac{2x-3y}{3z+y} = \dfrac{3(z-y)}{3(z-x)} = \dfrac {x+3z}{2y-3x} ## (2)

##= \dfrac{(2x-3y)-3(z-y)+x+3z}{(3z+y)-3(z-x)+(2y-3x)} ## (3)

##=\dfrac{x}{y}##

My question is how is it possible to go from (2) to (3). Because in (2) there are three fractions and two equal signs, yet they are merged in (3). I don't understand how this can be done. Thank you.

## \dfrac{2x-3y}{3z+y} = \dfrac{z-y}{z-x} = \dfrac {x+3z}{2y-3x} ## (1)

## \dfrac{2x-3y}{3z+y} = \dfrac{3(z-y)}{3(z-x)} = \dfrac {x+3z}{2y-3x} ## (2)

##= \dfrac{(2x-3y)-3(z-y)+x+3z}{(3z+y)-3(z-x)+(2y-3x)} ## (3)

##=\dfrac{x}{y}##

My question is how is it possible to go from (2) to (3). Because in (2) there are three fractions and two equal signs, yet they are merged in (3). I don't understand how this can be done. Thank you.