[tex]( \frac{20a}{100}, \frac{25b}{100} , \frac{30c}{100} ) d.p. ( \frac{1}{2} , \frac{35}{48} , 1)[/tex]
-se simplifica 20 cu 100 si ramane [tex]\frac{1a}{5}[/tex]
-se simplifica 25 cu 100 si ramane [tex]\frac{1b}{4}[/tex]
-se siplifica 30 cu 100 si ramane [tex]\frac{3c}{10}[/tex]
[tex] ( \frac{a}{5} , \frac{b}{4} , \frac{3c}{10} ) d.p. ( \frac{1}{2} , \frac{35}{48} , 1)[/tex]
[tex]{\frac{a}{5}}{\frac{1}{2}}={\frac{b}{4}}{\frac{35}{48}}={\frac{3c}{10}}{1}=k =>[/tex]
=> [tex]\frac{a}{5}=\frac{1k}{2}[/tex]
[tex] \frac{b}{4}=\frac{35k}{48}[/tex]
[tex] \frac{3c}{10}=k[/tex]
=> [tex]a=\frac{5k}{2}[/tex]
[tex]b=\frac{35x4k}{48}[/tex]
[tex]c=\frac{10k}{3}[/tex]
a+b+c=21 (inlocuim pe a,b si c)
[tex] \frac{5k}{2}+\frac{140k}{48}+\frac{10k}{3}=21[/tex] (aducem fractiile la acelasi numitor adica 48 si inmultim toata ecuatia cu 48 deci dispar numitorii )
120k+140k+160k=1008
420k=1008
k=1008\420 => k=2.4
=>[tex] a=\fra{5x2.4}{2}=6[/tex]
[tex]b=\frac{140x2.4}{48}=7[/tex]
[tex]c=\frac{10x2.4}{3}=8[/tex]
verificare : 6+7+8=21
21=21
