Question

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

Answer 1

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

Pe 5 nu il pot face, scuze