Ta có : \(\frac{y-1}{2}=\frac{5-2y}{3}\)
=> \(3\left(y-1\right)=2\left(5-2y\right)\)
=> 3y - 3 = 10 - 4y
=> 3y + 4y = 10 + 3
=> 7y = 13
=> y = 13/7
\(\frac{y-1}{2}=\frac{5-2y}{3}\)
\(\Rightarrow\text{ }3\left(y-1\right)=2\left(5-2y\right)\)
\(3y-3=10-4y\)
\(3y+4y=10+3\)
\(7y=13\)
\(y=\frac{13}{7}\)
\(\frac{y-1}{2}=\frac{5-2y}{3}\)
\(\Leftrightarrow\frac{3\cdot\left(y-1\right)}{2\cdot3}=\frac{2\cdot\left(5-2y\right)}{3\cdot2}\)
\(\Leftrightarrow\frac{3y-3}{6}=\frac{10-4y}{6}\)
\(\Leftrightarrow3y-3=10-4y\)
\(\Leftrightarrow3y+4y=10+3\)
\(\Leftrightarrow7y=13\)
\(\Leftrightarrow y=\frac{13}{7}\)
Vậy \(y=\frac{13}{7}\)
\(\frac{y-1}{2}=\frac{5-2y}{3}\)
\(\Rightarrow\text{ }3\left(y-1\right)=2\left(5-2y\right)\)
\(3y-3=10-4y\)
\(3y+4y=10+3\)
\(7y=13\)
\(y=\frac{13}{7}\)
\(\frac{y-1}{2}=\frac{5-2y}{3}\)
\(\Rightarrow3y-3=10-4y\)
\(3y+4y=10+3\Rightarrow7y=13\Rightarrow y=\frac{13}{7}\)
\(\frac{y-1}{2}=\frac{5-2y}{3}\)
\(\Rightarrow\text{ }3\left(y-1\right)=2\left(5-2y\right)\)
\(3y-3=10-4y\)
\(3y+4y=10+3\)
\(7y=13\)
\(y=\frac{13}{7}\)
\(\frac{y-1}{2}=\frac{5-2y}{3}\)
\(\Rightarrow\text{ }3\left(y-1\right)=2\left(5-2y\right)\)
\(3y-3=10-4y\)
\(3y+4y=10+3\)
\(7y=13\)
\(y=\frac{13}{7}\)
\(\frac{y-1}{2}=\frac{5-2y}{3}\)
\(\Rightarrow\text{ }3\left(y-1\right)=2\left(5-2y\right)\)
\(3y-3=10-4y\)
\(3y+4y=10+3\)
\(7y=13\)
\(y=\frac{13}{7}\)
\(\frac{y-1}{2}=\frac{5-2y}{3}\)
\(\Rightarrow\text{ }3\left(y-1\right)=2\left(5-2y\right)\)
\(3y-3=10-4y\)
\(3y+4y=10+3\)
\(7y=13\)
\(y=\frac{13}{7}\)
