`a) ( x + 5 )/4 + ( x + 4 )/5 +  ( x + 3 )/6 = 0 ` `b) ( 2x)/5 + ( x + 1 )/4 = ( 4x )/3 `

#andy

\[\begin{array}{l}
a)\dfrac{{x + 5}}{4} + \dfrac{{x + 4}}{5} + \dfrac{{x + 3}}{6} = 0\\
MC:60\\
 \Rightarrow 15\left( {x + 5} \right) + 12\left( {x + 4} \right) + 10\left( {x + 3} \right) = 0\\
 \Leftrightarrow 15x + 75 + 12x + 48 + 10x + 30 = 0\\
 \Leftrightarrow 15x + 12x + 10x = 0 - 70 - 48 - 30\\
 \Leftrightarrow 37x =  - 153\\
 \Leftrightarrow x =  - \dfrac{{153}}{{37}}\\
 \Rightarrow S = \left\{ { - \dfrac{{153}}{{37}}} \right\}\\
b)\dfrac{{2x}}{5} + \dfrac{{x + 1}}{4} = \dfrac{{4x}}{3}\\
MC:60\\
 \Rightarrow 12.2x + 15\left( {x + 1} \right) = 20.4x\\
 \Leftrightarrow 24x + 15x + 15 = 80x\\
 \Leftrightarrow 24x + 15x - 80x =  - 15\\
 \Leftrightarrow  - 41x =  - 15\\
 \Leftrightarrow x = \dfrac{{15}}{{41}}\\
 \Rightarrow S = \left\{ {\dfrac{{15}}{{41}}} \right\}
\end{array}\]