jmc

Since $h(1)=5$, we have $a\cdot 1+b=5$, so $a+b=5$. Since $h(-1)=1$, we have $a\cdot (-1)+b=1$, so $-a+b=1$. Adding these two equations gives $2b=6$, so $b=3$. From $a+b=5$, we find $a=2$. Therefore, $h(x)=2x+3$, so $h(6)=2\cdot 6+3=\boxed{15}$.