Question

Numerele a,b,c sund direct proportionale cu 3,4,11. Aflați valoarea raportului: 3a+2b-c supra a+2b+3c

Answer 1

   

$\displaystyle\\\frac{a}{3}=\frac{b}{4}=\frac{c}{11}=k\\\\a=3k\\b=4k\\c=11k\\\\\frac{3a+2b-c}{a+2b+3c}=\\\\=\frac{3\times3k+2\times4k-11k}{3k+2\times4k+3\times11k}=\\\\=\frac{9k+8k-11k}{3k+8k+33k}=\\\\=\frac{6k}{44k}=\frac{6}{44}=\boxed{\frac{3}{22}}$