Question

determinati numerele x,y, si z invers porportionale cu numerele 2,3 si 4 ,stiind ca: a) x+y- z=7 b) x+z-y=10 c) y+z=x=4
va rog frumos ! dau coroană​

Answer 1

{x,y,z} i.p. {2,3,4}

2x=3y=4z=k

k-coeficient de proportionalitate

2x=k=>x=k/2

3y=k=>y=k/3

4z=k=>z=k/4

b)x+z-y=10

$\frac{k}{2} + \frac{k}{4} - \frac{k}{3} = 10 \\ numitor \: comun \: 12 \\ \frac{6k}{12} + \frac{3k}{12} - \frac{4k}{12} = 10 \\ \frac{5k}{12} = 10 = > 5k = 120 = > k = 24$

x=k/2=>x=24/2=>x=12

y=k/3=>y=24/3=>y=8

z=k/4=>z=24/4=>z=6

a)x+y-z=7

$\frac{k}{2} + \frac{k}{3} - \frac{k}{4} = 7 \\ aduci \: la \: acelasi \: numitor(12) \\ \frac{6k}{12} + \frac{4k}{12} - \frac{3k}{12} = 7 \\ \frac{6k + 4k - 3k}{12} = 7 \\ \frac{7k}{12} = 7 = > 7k = 7 \times 12 = > \\ 7k = 84 = > k = 12$

x=k/2=>x=12/2=>x=6

y=k/3=>y=12/3=>y=4

z=k/4=>z=12/4=>z=3

c)y+z-x=4

$\frac{k}{3} + \frac{k}{4} - \frac{k}{2} = 4 \\ \frac{4k}{12} + \frac{3k}{12} - \frac{6k}{12} = 4 \\ \frac{k}{12} = 4 = > k = 4 \times 12 = > k = 48$

x=k/2=>x=48/2=>x=24

y=k/3=>y=48/3=>y=16

z=k/4=>z=48/4=>z=12