\frac{a+3b}{2a+b}=
\frac{a}{a+b}=1
a=a+b
b=0
\frac{a+3\cdot 0}{2a+0}=\frac{a}{2a}=0,5
zad.2
\frac{2a}{a+3b}=\frac{4}{11}
mnożę “na krzyż”
4(a+3b)=11 \cdot 2a
4a+12b=22a
12b=18a
b=1,5a
podstawiam b:
\frac{a+3b}{2a+b}=\frac{a+5\cdot 1,5a}{2a+1,5a}=\frac{8,5a}{3,5a}=\frac{85}{35}=\frac{17}{7}=2\frac{3}{7}