## Linear Algebra: A Modern Introduction

No solution in $\mathbb{Z}_{6}$.
The elements of $\;\mathbb{Z}_{6}\;$ are 0,1,2,3,4,5 So, if there is a solution for this equation, then x is one of these numbers. $If \;\;x=0\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4.(0)+5=2 \; {\color{Red}\rightarrow } \;\; 0+5= 2\;\;{\color{Red} \Rightarrow }\;\;5\neq 2\\ so \;\;0 \;is \;not \;solution\\\\ If \;x=1\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;4.(1)+5=2 \; {\color{Red}\rightarrow } \;\; 4+5= 2\;\;{\color{Red} \Rightarrow }\;\;3\neq 2\\ so\; 1\; is\; solution.\\\\ If\;x=2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;4.(2)+5=2 \; {\color{Red}\rightarrow } \;\; 2+5= 2\;\;{\color{Red} \Rightarrow }\;\;1\neq 2\\ so\; 2\; is\;not\; solution\\\\ If \;x=3\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;4.(3)+5=2 \; {\color{Red}\rightarrow } \;\; 0+5= 2\;\;{\color{Red} \Rightarrow }\;\;5\neq 2\\ so\; 3\; is\; not\; solution\\\\ If\; x=4\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4.(4)+5=2 \; {\color{Red}\rightarrow } \;\; 4+5= 2\;\;{\color{Red} \Rightarrow }\;\;3\neq 2\\ so\; 4 \;is\; not\; solution\\\\ If\; x=5\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4.(5)+5=2 \; {\color{Red}\rightarrow } \;\; 2+5= 2\;\;{\color{Red} \Rightarrow }\;\;1\neq 2\\ so\; 5 \;is\; solution\\\\$ No solution in $\mathbb{Z}_{6}$