Question

Stim ca x,y,z sunt direct proporționale 2,3,4 si 2x+y+5z=81,să se afle valoarea expresiei E=3x-y+2z

Answer 1

x / 2 = y / 3 = z / 4 = k

x = 2k

y = 3k

z = 4k

2x + y+ 5z = 81

27k = 81

k = 3

x = 2 × 3 = 6

y = 3 × 3 = 9

z = 4k = 4 × 3 = 12

E = 3 × 6 - 9 + 2 × 12

E = 33

Answer 2

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

$\it \frac{x}{2} = \frac{y}{3} = \frac{z}{4} = k = = > \\ \\ \it \Rightarrow \frac{x}{2} = k \: \Rightarrow \: x = 2k \\ \\ \it \frac{y}{3} = k \: \Rightarrow \: y = 3k \\ \\ \it \frac{z}{4} = k \: \Rightarrow \: z = 4k$

$\it 2x + y + 5z = 81 \: = = > \\ 2 \cdot 2k + 3k + 5 \cdot 4k = 81 = = > \\ \it 4k + 3k + 20k = 81 = = > \\ \it 27k = 81 = = > \\ \it k = \frac{81}{27} = = > \\ \it k = 3$
$\it x = 2k ==> x = 2 \cdot 3 ==> x=6 \\ \it y=3k ==> y= 3 \cdot 3 ==> y=9 \\ \it z=4k ==> z= 4 \cdot 3 ==> z=12$


E = 3x - y + 2z
$\it E = 3x - y + 2z = = > \\ \it E = 3 \cdot 6 - 9 + 2 \cdot 12 = = > \\ \it E = 18 - 9 + 24 = = > \\ \it E = 9 + 24 = = > \\ \it E = 33$