[tex]a:b=3:8\\\\\dfrac{a}{b}=\dfrac{3}{8}\qquad\text{multiply both sides by b}\\\\\underline{\underline{a=\dfrac{3}{8}b}}\\\\b:c=6:11\\\\\dfrac{b}{c}=\dfrac{6}{11}\qquad\text{cross multiply}\\\\6c=11b\qquad\text{divide both sides by 6}\\\\\underline{\underline{c=\dfrac{11}{6}b}}\\\\a+b+c\qquad\text{substitute}\\\\\dfrac{3}{8}b+b+\dfrac{11}{6}b=\dfrac{3\cdot3}{8\cdot3}b+b+\dfrac{11\cdot4}{6\cdot4}b=\dfrac{9}{24}b+\dfrac{24}{24}b+\dfrac{44}{24}b=\dfrac{77}{24}b[/tex]
[tex]\text{If a, b, and c are positive integers, then the sum a+ b + c is positive}\\\text{integer}.\\\\\dfrac{77}{24}b\ \text{is positive integer if}\ b=24k\ (k\ \text{is positive integer}).\\\\\text{The smallest possible value of a + b + c is for k = 1}\\\\b=24(1)=24\\\\\dfrac{77}{24}\cdot24=77\\\\Answer:\ \boxed{77}[/tex]
