Question

Determinati numerele x,y si z invers proportionale cu numerele 3 5 9 si 15 stiind ca 3x-2y-z=12​

Answer 1

{x,y,z} i.p. {3,5,9}

3x=5y=9z=k

k-coeficient de proportionalitate

3x=k=>x=k/3

5y=k=>y=k/5

9z=k=>z=k/9

3x-2y-z=12

$k - 2 \times \frac{k}{5} - \frac{k}{9} = 12 \\ k - \frac{2k}{5} - \frac{k}{9} = 12 \\ aduci \: la \: acelasi \: numitor - > 45 \\ \frac{45k}{45} - \frac{18k}{45} - \frac{5k}{45} = 12 \\ \frac{45k - 18k - 5k}{45} = 12 \\ \frac{22k}{45} = 12 = > 22k = 45 \times 12 = > \\ 22k = 540 = > k = \frac{540}{22} = \frac{270}{11}$

$x = \frac{k}{3} = \frac{ \frac{270}{11} }{3} = \frac{270}{11} \times \frac{1}{3} = \frac{270}{33} = \frac{90}{11}$

$y = \frac{k}{5} = \frac{ \frac{270}{11} }{5} = \frac{270}{11} \times \frac{1}{5} = \frac{270}{55} = \frac{54}{11}$

$z = \frac{k}{9} = \frac{ \frac{270}{11} }{9} = \frac{270}{11} \times \frac{1}{9} = \frac{270}{99} = \frac{30}{11}$