Có 2a=4b => \(\dfrac{a}{4}=\dfrac{b}{2}\)=> \(\dfrac{a}{4.5}=\dfrac{b}{2.5}\)=>\(\dfrac{a}{20}=\dfrac{b}{10}\) (1)
Có 3b=5c => \(\dfrac{b}{5}=\dfrac{c}{3}\)=>\(\dfrac{b}{5.2}=\dfrac{c}{3.2}\)=>\(\dfrac{b}{10}=\dfrac{c}{6}\) (2)
Từ (1) và (2) => \(\dfrac{a}{20}=\dfrac{b}{10}=\dfrac{c}{6}\)
Đặt \(\dfrac{a}{20}=\dfrac{b}{10}=\dfrac{c}{6}\) = k
=> a=20k , b=10k, c=6k
Thay a=20k , b=10k, c=6k vào a+2b-3c=99, ta có :
20k+2.10k-3.6k=99
=> 20k+20k-18k=99
=> k(20+20-18)=99
=> k= 99:22=4,5
=>a=20.4,5=90, b=10.4,5=45, c=6.4,5=27
Vậy a=90, b=45, c=27
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