Chapter P, Prerequisites - Section P.8 - Solving Basic Equations - P.8 Exercises - Page 61: 105

(a) k=3 (b) k=3 (c) k=any real number

(a) $3(0)+k-5=k(0)-k+1$ $k-5=-k+1$ $2k=6$ $k=3$ (b) $3 (1) +k-5=k(1)-k+1$ $3+k-5=k-k+1$ $k-2=1$ $k=3$ (c) $3(2)+k-5=k(2)-k+1$ $6+k-5=2k-k+1$ $k+1=k+1$ $1=1$ Since we get an identity, we know that any value of k will be a solution for $x=2$. Thus k can be any real number for $x=2$.