Answer:
Step-by-step explanation:
[tex]\frac{3a -b}{2c}+\frac{3a-c}{c-b}=\frac{3*3-[-4]}{2*(-2)}+\frac{3*3-[-2]}{-2-[-4]}\\\\=\frac{9+4}{-4}+\frac{9+2}{-2+4}\\\\=\frac{13}{-4}+\frac{11}{2}\\\\=\frac{-13}{4}+\frac{11*2}{2*2}\\\\=\frac{-13}{4}+\frac{22}{4}\\\\=\frac{-13+22}{4}\\\\=\frac{9}{4}\\\\=2\frac{1}{4}[/tex]
