Question

\(\dfrac{2x-1}{-5}=\dfrac{7}{3+4y}=\dfrac{-6z+4}{-9}=-\dfrac{2}{9}\)

Answer 1

\(\dfrac{2x-1}{-5}=\dfrac{7}{3+4y}=\dfrac{-6z+4}{-9}=-\dfrac{2}{9}\)

Ta có: 

`(2x-1)/(-5) = -2/9`

`=> 2x - 1 = 10/9`

`=> 2x = 10/9 + 1`

`=> 2x = 10/9 + 9/9`

`=> 2x = 19/9`

`=> x = 19/9 : 2`

`=> x = 19/9 xx 1/2`

`=> x = 19/9`

\(\dfrac{7}{3+4y}=-\dfrac{2}{9}\)

`=> 3 + 4y = -63/2`

`=> y = -69/8`

\(\dfrac{-6z+4}{-9}=-\dfrac{2}{9}\)

\(\Rightarrow-6z+4=2\)

\(\Rightarrow z=\dfrac{1}{3}\)

Answer 2

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x-1}{-5}=-\dfrac{2}{9}\Rightarrow2x-1=\dfrac{10}{9}\Rightarrow x=\dfrac{\dfrac{10}{9}+1}{2}=\dfrac{19}{18}\\\dfrac{7}{3+4y}=-\dfrac{2}{9}\Rightarrow3+4y=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\Rightarrow y=\dfrac{-\dfrac{63}{2}-3}{4}=-\dfrac{69}{8}\\\dfrac{-6z+4}{-9}=-\dfrac{2}{9}\Rightarrow-6z+4=\dfrac{\left(-2\right)\left(-9\right)}{9}=2\Rightarrow z=\dfrac{2-4}{-6}\end{matrix}\right.=\dfrac{1}{3}}\)

Answer 3

2x−1−5=73+4y=−6z+4−9=−292x−1−5=73+4y=−6z+4−9=−29

Ta có: 

$\frac{2x-1}{-5}=-\frac{2}{9}$

$\Rightarrow 2x-1=\frac{10}{9}$

$\Rightarrow 2x=\frac{10}{9}+1$

$\Rightarrow 2x=\frac{10}{9}+\frac{9}{9}$

$\Rightarrow 2x=\frac{19}{9}$

$\Rightarrow x=\frac{19}{9}:2$

$\Rightarrow x$