Question

Determine os valores de x,y e z sabendo que são inversamente proporcionais aos numeros 2, 3 e 4 ordem, e que x+y+z=26.

Answer 1

$x=\frac{k}{20}\\\\y=\frac{k}{3}\\\\z=\frac{k}{4}$
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$x+y+z=26\\\\\dfrac{k}{2}+\dfrac{k}{3}+\dfrac{k}{4}=26$

Multiplicando todos os membros da equação por 12 (mmc entre 1, 2, 3 e 4):

$12\cdot\dfrac{k}{2}+12\cdot\dfrac{k}{3}+12\cdot\dfrac{k}{4}=12\cdot26\\\\\\6k+4k+3k=12*26\\\\13k=12*26\\\\1k=12*2\\\\k=24$
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$x=\frac{k}{2}=\frac{24}{2}~~~\therefore~~~\boxed{\boxed{x=12}}\\\\y=\frac{k}{3}=\frac{24}{3}~~~\therefore~~~\boxed{\boxed{y=8}}\\\\z=\frac{k}{4}=\frac{24}{4}~~~\therefore~~~\boxed{\boxed{z=6}}$

Answer 2

$\left \{ {{x+y+z=26} \atop { \frac{x}{ \frac{1}{2} }= \frac{y}{ \frac{1}{3} } = \frac{z}{ \frac{1}{4} }$

$\frac{x+y+z}{ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4} } = \frac{26}{ \frac{6+4+3}{12} } = \frac{26}{ \frac{13}{12} } = \frac{26.12}{13}=24$

$\frac{x}{ \frac{1}{2} } =2x$

$2x=24$

$x= \frac{24}{2}$
x=12

$\frac{y}{ \frac{1}{3} } =3y$

$3x=24$

$y= \frac{24}{3}$
y=8

$\frac{z}{ \frac{1}{4} } =4z$

$4z=24$

$z= \frac{24}{4}$
z=6