Answer: 3/5
Step-by-step explanation:
[tex]\dfrac{a}{a+2}+\dfrac{3}{20}=\dfrac{a+3}{(a+2)+3}\\\\\\\dfrac{a}{a+2}\bigg(\dfrac{20}{20}\bigg)+\dfrac{3}{20}\bigg(\dfrac{a+2}{a+2}\bigg)=\dfrac{a+3}{a+5}\\\\\\\dfrac{23a+6}{20(a+2)}=\dfrac{a+3}{a+5}\\\\\\(23a+6)(a+5)=20(a+2)(a+3)\\\\\\23a^2+121a+30=20a^2+100a+120\\\\\\3a^2+21a-90=0\\\\\\a^2+7a-30=0\\\\\\(a+10)(a-3)=0\\\\\\a=-10\quad a=3[/tex]
Since "a" is a positive integer, disregard a = -10
So the only valid answer is a=3 → a+2=5
[tex]\dfrac{a}{a+2}=\large\boxed{\dfrac{3}{5}}[/tex]
