3a+5c=7b+30
=>3a+5c-7b=30
\(2a=3b=>\frac{a}{3}=\frac{b}{2}=>\frac{a}{3}.\frac{1}{7}=\frac{b}{2}.\frac{1}{7}=>\frac{a}{21}=\frac{b}{14}\)
\(5b=7c=>\frac{b}{7}=\frac{c}{5}=>\frac{b}{7}.\frac{1}{2}=\frac{c}{5}.\frac{1}{2}=>\frac{b}{14}=\frac{c}{10}\)
\(=>\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a+5c-7b}{63+50-98}=\frac{30}{15}=2\)
\(=>\frac{a}{21}=2=>a=21.2=42\)
\(=>\frac{b}{14}=2=>b=14.2=28\)
\(=>\frac{c}{10}=2=>c=10.2=20\)
Vậy a=42,b=28,c=20.