Question

Biết \(\dfrac{3x-y}{x+y}\)=\(\dfrac{3}{4}\)và x,y khác 0 Vậy \(\dfrac{x}{y}\)=

Answer 1

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)

\(\Rightarrow3\left(x+y\right)=4\left(3x-y\right)\)

\(\Leftrightarrow3x+3y=12x-4y\)

\(\Leftrightarrow3x-12x=-4y-3y\)

\(\Leftrightarrow-9x=-7y\)

\(\Leftrightarrow\dfrac{x}{y}=\dfrac{-7}{-9}=\dfrac{7}{9}\)

Answer 2

Ta có:

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)

\(\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)

hay \(12x-4y=3x+3y\)

\(\Rightarrow12x-3x=3y+4y\) (chuyển vế)

hay \(9x=7y\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

Vậy \(\dfrac{x}{y}=\dfrac{7}{9}\)

Chúc bn học tốt!!!