Giải:
Ta có: \(\dfrac{y-5}{7-y}=\dfrac{2}{-3}\)
\(\Rightarrow\left(y-5\right).\left(-3\right)=2\left(7-y\right)\)
\(\Rightarrow-3y+15=14-2y\)
\(\Rightarrow-3y+2y=-15+14\)
\(\Rightarrow-1y=-1\)
Vậy y=1
Ta có:y-5/7-y=2/-3
=>(y-5).(-3)=(7-y).2
=>-3y+15=14-2y
=>-3y+2y=14-15
=>-y=-1
=>y=1
\(\dfrac{y-5}{7-y}\)= \(\dfrac{2}{-3}\)
\(\Leftrightarrow\) -3(y-5)= 2(7-y)
\(\Leftrightarrow\) -3y+15 = 14 - 2y
\(\Leftrightarrow\) -3y+2y = 14-15
\(\Leftrightarrow\)-y = -1
\(\Leftrightarrow\) y = 1
\(\dfrac{y-5}{7-y}=\dfrac{-2}{3}\)
\(\Rightarrow3\left(y-5\right)=-2\left(7-y\right)\)
\(\Rightarrow3y-15=7y-49\)
\(\Rightarrow4y=34\)\(\Rightarrow y=\dfrac{17}{2}\)
