a, \(x-5=\frac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\frac{3x-15}{3}=\frac{x+2}{3}\Leftrightarrow\frac{3x-15-x-2}{3}=0\)
\(\Leftrightarrow2x-17=0\Leftrightarrow x=\frac{17}{2}\)
b, \(\frac{x}{3}+\frac{x}{4}=\frac{1}{5}-\frac{x}{6}\)
\(\Leftrightarrow\frac{2x}{6}+\frac{x}{6}=\frac{4}{20}-\frac{5x}{20}\Leftrightarrow\frac{x}{2}=\frac{4-5x}{20}\)
\(\Leftrightarrow\frac{10x}{20}-\frac{4-5x}{20}=0\Leftrightarrow15x-4=0\Leftrightarrow x=\frac{4}{15}\)
a, x - 5 = \(\frac{1}{3}\).(x + 2)
<=> x - 5 = \(\frac{1}{3}\)x + \(\frac{2}{3}\)
<=> x - 5 - \(\frac{1}{3}\)x - \(\frac{2}{3}\)= 0
<=>\(\frac{2}{3}\)x - \(\frac{17}{3}\)= 0
<=>x = \(\frac{17}{2}\)
