Question

Manipulating proportions.
If we know that a/b = c/d, how can it be (a+c)/(b+d)

Answer 1

im confused with what you are asking

Answer 2

do you mean :
\[\large \dfrac{a}{b} = \dfrac{c}{d} \implies \dfrac{a+b}{a-b} = \dfrac{c+d}{c-d}\]

?

Answer 3

\[\frac{ a }{ b } = \frac{ c }{ d } = \frac{ a + c }{ b + d } = \frac{ a-c }{ b-d }\]

Answer 4

I don't understand how does the author do this

Answer 5

they all are forms of componendo and dividendo rule

Answer 6

lets see if we can derive this particular rule

Answer 7

How can I prove it?

Answer 8

I just saw a way to prove from (a+kc)/(b+kd)...is there any way to prove it in a way that is not backwards?

Answer 9

try adding c/b both sides

Answer 10

Which part? The ones of the beginning?

I would have (a+c)/b = (c/d)+(c/b)

Answer 11

yes simplify right hand side

Answer 12

\[\dfrac{a}{b} = \dfrac{c}{d}\]

add \(\large \dfrac{c}{b}\) both sides :
\[\large \dfrac{a}{b}+\dfrac{c}{b} = \dfrac{c}{d}+\dfrac{c}{b} \implies \dfrac{a+c}{b+d} = \dfrac{c}{d} \tag{1} \]

subtract \(\large \dfrac{c}{b}\) both sides :
\[\large \dfrac{a}{b}-\dfrac{c}{b} = \dfrac{c}{d}-\dfrac{c}{b} \implies \dfrac{a-c}{b-d} = \dfrac{c}{d} \tag{2} \]

Answer 13

the result follolws from (1) & (2)

Answer 14

Or you could simply apply the regular componendo and dividendo to right hand side

Answer 15

add +1 to both sides of right hand side and simplify

Answer 16

Perfect, thanks a lot :9

Answer 17

np :) this is really a very useful property
http://en.wikipedia.org/wiki/Componendo_and_dividendo

Answer 18

:)

Answer 19

I will look it deeper later