Question

Se da proportia : a/b = c/d . Aratati ca 5a+2b/13a+7b = 5c+2d/13c+7d este o proportie

Answer 1

Voi incerca sa scot niste relatii care sa ma ajute.
a/b=c/d=>ad=bc si totodata b/a=d/c
a=bc/d
b=c/ad
c=ad/b
d=a/bc
Si avem
5a+2b/13a+7b=5c+2d/13c+7d<=>
5a-5c+7b-7d=2d/13c-2b/13a<=>
5(a-c)+7(b-d)=2/13(d/c-b/a)
Dar d/c=b/a =>
5(a-c)+7(b-d)=0<=>
5(a-c)=-7(b-d)<=>
a-c=(-7/5)(b-d)
Dar c=ad/b=>a-ad/b=(-7/5)(b-d)=>
(ab-ad)/b=(-7/5)(b-d)
De unde b=5 =>
5a-ad=-7(5-d)<=>
5a-ad=-35+7d<=>5a-ad+35-7d=0<=>
a(5-d)+7(5-d)=0<=>
(a+7)(5-d)=0 => a=-7 ,d=5
iar c=ad/b=-35/5=-7
Si avem a=-7,b=5,c=-7,d=5
Acum sa verificam relatiile
a/b=c/d<=> -7/5=-7/5 adevarat
5a+2b/13a+7b=5c+2d/13c+7d
-35-10/91+35=-35-10/91+35<=>-10/91=-10/91 Ceea ce verifica relatia.
Mi-a luat cam mult in calcule sper sa intelegi.

Answer 2

a/b=c/d⇔(5a)/(2b)=(5c)/(2d)⇔(5a+2b)/(2b)=(5c+2d)/(2d)⇔(5a+2b)/(5c+2d)=b/d.


a/b=c/d⇔(13a)/(7b)=(13c)/(7d)⇔(13a)/(13a+7b)=(13c)/(13c+7d)⇔
⇔(13a+7b)/(13c+7d)=a/c=b/d=(5a+2b)/(5c+2d). Egalitatea dintre  prima fractie si ultima este echivalenta cu (5a+2b)/(13a+7b)=(5c+2d)/(13c+7d)