Question

1) Daca $\frac{x}{y}$ = $\frac{5}{7}$ calculati $\frac{3x+2y}{x+4y}$

2) Daca $\frac{a}{b}$ = $\frac{9}{10}$ calculati $\frac{7a-b}{a+b}$

3) Daca $\frac{x}{y}$ = 0,(4) calculati $\frac{3x-y}{2x-4y}$

Answer 1

1) x/y=5/7
7x=5y
x=5y/7

(3x+2y)/(x+4y)=[3·(5y/7)+2y[/[(5y/7)+4y]=(15y/7+2y/1)/(5y/7+4y/1)=
=[(15y+2y·7)/7]/[(5y+4y·7)/7]=(29y/7)/(33y/7)=29y/7·7/33y=simplif cu 7=
=29y/33y=simplif cu y=29/33

2)
a/b=9/10
10a=9b
a=9b/10=0,9b
a=0,9b
(7a-b)/(a+b)=(7·0,9b-b)/(0,9b+b)=
(6,3b-b)/1,9b=5,3b/1,9b=5,3/1,9=53/10:19/10=
=53/10·10/19=53/19

3)
x/y=4/9
9x=4y
x=4y/9
(3x-y)/(2x-4y)=[(3·4y/9)-y]/[(2·4y/9)-(4·4y/9)]=
(4y/3-y)/(8y/9-16y/9)=[(4y-3y)/3]/(-8y/9)=(y/3):(-8y/9)=(y/3)·(-9/8y)=simplificam=-3/8

Answer 2

X=5t    y=7t    inlocuim in relatia de mai sus 
15t+14t \ 5t+28t=
29t\33t=(se simplifica ''t'')= 29\33

2)a=9t      b=10 t
63t-10t\9t+10t=
53t\19t=53\19

3)0,(4)=4\9
x=4t   y=9t
12t-9t \8t-36t=
3t\ -28t= 3\-28(asta dc acolo ai 2x-4y)