The equation $x^{2}-7 x+k=0$ has solutions $x=3$ and $x=a$. The equation $x^{2}-8 x+k+1=0$ has solutions $x=b$ and $x=c$. What is the value of $a+b c$ ?
Using that 3 is a solution to $x^{2}-7 x+k=0$ we have that $3^{2}-7(3)+k=0$ or $k=21-9=12$. This means $x^{2}-7 x+k$ is $x^{2}-7 x+12=(x-3)(x-4)$, the roots of which are 3 and 4 , which means $a=4$.

The polynomial $x^{2}-8 x+k+1$ factors as $(x-b)(x-c)$, so upon expanding the latter expression, we have $b c=k+1=12+1=13$.

Therefore, $a+b c=4+13=17$.

ANSWER: 17