Question

Un calator si-a propus sa parcurga un drum in 3 zile astfel in prima zi 1/6 din drum plus inca 15 km ,a doua zi 1/5 din rest plus inca 17 km ,iar in a treia zi 1/3 din tot drumul plus inca 3 km. Cat a parcurs in fiecare zi ?

Answer 1

x-lungimea drumului

1/6 din x +15= 1/6 * x +15=  x/6 +15 -in prima zi

x-(x/6+15)=  (6x-x-90)/6= (5x-90)/6  -restul

1/5 din (5x-90)/6 + 17= (5x-90)/30 +17=  (5x-90+510)/30= (5x+420)/30 -a 2a zi

1/3 din x +3 = x/3 +3 -in a3a zi

x/6 +15 +(5x+420)/30 +x/3 +3=x  aducem la acelasi numitor

5x +450 +5x+420 +10x +90 =30x

20x +960 =30x

20x-30x = -960

-10x =-960

10x=960
x=96 km

96/6 +15= 16+15= 31 km -in prima zi

(5*96+420)/30= (480+420)/30= 900/30= 30km -in a2a zi

96/3+3= 21+3=24km -in a3a zi

Answer 2

Fie l -lungimea drumului parcurs de calator
AVem 

a) 1/6 l+15= 1/6 l+15=  l/6 +15  ⇔ l-(l/6+15)= (6l-l-90)/6= (5l-90)/6  

b) 1/5  (5l-90)/6 + 17= (5l-90)/30 +17=  (5l-90+510)/30= (5l+420)/30

c) 1/3 l +3 = l/3 +3 

Deci
l/6 +15 +(5l+420)/30 +l/3 +3=l  ⇔ 5l +450 +5l+420 +10l +90 =30l ⇔20l +960 =30l

⇔20l-30l = -960 ⇔-10l=-960 ⇔10l=960 ⇔l=96 km

d) 96/6 +15= 16+15= 31 km I zi 

e) (5*96+420)/30= (480+420)/30= 900/30= 30km II zi

f)96/3+3= 21+3=24km III zi
R/s 1 zi 31 km, 2 zi-30 km, 3 zi 24 km