Question

Exercitiile 3 si 5 ( b) ) . Va rog :)

Answer 1

3) $\frac{a+3b}{2a+5b}= \frac{5}{9}$⇒ 9(a+3b)=5(2a+5b)⇒ 9a+27b=10a+25b |-25b-9a⇒2b=a (1)
$\frac{3b+c}{b+4c} = \frac{10}{7}$⇒ 7(3b+c)=10(b+4c)⇒ 21b+7c=10b+40c |-7c-10b ⇒11b=33c |÷11 ⇒b=3c (2)
Din (1) si (2)⇒ a=6c 
$\frac{a+b-c}{3a-b+c}$=$\frac{6c+3c-c}{18c-3c+c}= \frac{c(6+3-1)}{c(18-3+1}= \frac{8}{16} = \frac{1}{2}$

Pe 5 nu stiu sa-l fac