I let p(x)=ax+b and q(x)=tx+r
I did all the substitution and got eall the terms in terms of b
I could then find q(-1) in terms of b but not as a number on its own.
I do not think that my answer is wrong.
\(a=\frac{3-b}{2}\\ t=\frac{8}{3-b}\\ r=\frac{34-2b}{b-3}\\ \)
etc