Question

Ajutati-ma va rog sa rezolv acest exercitiu: "Aflati 3 numere, stiind ca sunt invers proportionale cu numerele 4, 6 si 8 si ca suma lor este 52

Answer 1

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Answer 2

       
$4x=6y=8z = k \\ x= \frac{k}{4} \\ \\ y= \frac{k}{6}\\ \\ z= \frac{k}{8} \\ x+y+z=52 \\ \frac{k}{4}+ \frac{k}{6}+\frac{k}{8}=52 \\ \\ \frac{6k}{24}+ \frac{4k}{24}+\frac{3k}{24}=52 \\ \\ \frac{6k+4k+3k}{24}=52 \\ \\ \frac{13k}{24}=52 \\ 13k=52*24 \\ k= \frac{52*24}{13} =4*24=96$

$x= \frac{k}{4} = \frac{96}{4}= \boxed{24} \\ \\ y= \frac{k}{6} =\frac{96}{6} = \boxed{16} \\ \\ z= \frac{k}{8} =\frac{96}{8}= \boxed{12}$