Gọi số cần tìm là x
\(\frac{7+x}{31+x}\)\(=\frac{3}{5}\)
\(5\left(7+x\right)\)\(=3\left(31+x\right)\)
\(5\left(7+x\right)\)\(=5x7+5x\)\(=35+5x\)
\(3\left(31+x\right)\)\(=3x31+3x=93+3x\)
\(=>35+5x=93+3x\)
\(5x-3x=93-35\)
\(5x-3x=58\)
\(x\left(5-3\right)=58\)
\(x2=58\)
\(x=58:2=29\)
Vậy muốn \(\frac{7}{31}\)thành \(\frac{3}{5}\)phải thêm \(29\)đơn vị