Answer:
[tex]\dfrac{3a-3b}{a-b}=\dfrac{3}{1}[/tex]
Step-by-step explanation:
[tex]\text{Given}\\\\\dfrac{a}{3}=\dfrac{b}{2}\qquad|\text{multiply both sides by 3}\\\\3\!\!\!\!\diagup\cdot\dfrac{a}{3\!\!\!\!\diagup}=3\cdot\dfrac{b}{2}\\\\a=\dfrac{3b}{2}[/tex]
[tex]\text{Substitute to}\ \dfrac{3a-3b}{a-b}:\\\\\dfrac{3\cdot\frac{3b}{2}-3b}{\frac{3b}{2}-b}=\left(\dfrac{9b}{2}-3b\right):\left(\dfrac{3b}{2}-\dfrac{2b}{2}\right)=\left(\dfrac{9b}{2}-\dfrac{6b}{2}\right):\left(\dfrac{3b-2b}{2}\right)\\\\=\left(\dfrac{9b-6b}{2}\right):\dfrac{b}{2}=\dfrac{3b\!\!\!\!\diagup}{2\!\!\!\!\diagup}\cdot\dfrac{2\!\!\!\!\diagup}{b\!\!\!\!\diagup}=\dfrac{3}{1}[/tex]
Other method:
[tex]\dfrac{3a-3b}{a-b}=\dfrac{3(a-b)}{a-b}\qquad|\text{cancel}\ (a-b)\\\\=\dfrac{3}{1}[/tex]
