d) $\frac{x-1}{3}$ +2=3-$\frac{2x+5}{4}$
⇔$\frac{4x-4}{12}$ +$\frac{24}{12}$ =$\frac{36}{12}$ -$\frac{6x+15}{12}$
⇔$\frac{4x-4+24}{12}$ =$\frac{36-6x-15}{12}$
⇔$\frac{4x+20}{12}$ =$\frac{21-6x}{12}$
⇔$\frac{4x+20}{12}$ -$\frac{21-6x}{12}$ =0
⇔$\frac{4x+20-21+6x}{12}$=0
⇔$\frac{10x-1}{12}$=0
⇔$10x-1=0$
⇔$10x=1$
⇔$x=1/10$