Evaluate $\frac{7!+8!}{6!+7!}$. Express your answer as a common fraction.
I also notice that 7! appears in both the numerator and denominator, so I can factor it out. I get $\frac{7!+8!}{6!+7!}=\frac{7!(1+8)}{7!(\frac{6!}{7!}+1)}$. I can simplify this by canceling out the 7! in the numerator and denominator. I get $\frac{1+8}{\frac{6!}{7!}+1}$. Now I need to evaluate the fraction $\frac{6!}{7!}$. I know that 7!=7*6!, so I can rewrite this as $\frac{6!}{7*6!}$. I can simplify this by canceling out the 6! in the numerator and denominator. I get $\frac{6!}{7!}=\frac{1}{7}$. So I can plug this into the original expression and get $\frac{1+8}{\frac{1}{7}+1}$. I want to express this as a common fraction, so I need to get rid of the fraction in the denominator. I can do this by multiplying the numerator and denominator by 7, which is the same as multiplying by 1. I get $\frac{1+8}{\frac{1}{7}+1}=\frac{7(1+8)}{7(\frac{1}{7}+1)}$. I can simplify this by distributing the 7 in the numerator and eliminating the fraction in the denominator. I get $\frac{7(1+8)}{7(\frac{1}{7}+1)}=\frac{7+56}{1+7}$. I can simplify this further by adding the terms in the numerator and denominator. I get $\frac{7+56}{1+7}=\frac{63}{8}$. This is the final answer, expressed as a common fraction.

