[tex]\left\{\begin{array}{ccc}x+y=1&\to x=1-y\\2x-y+z=1\\x+2y+z=\dfrac{8}{3}\end{array}\right\\\text{substitute x=1-y to the other equations}\\ \left\{\begin{array}{ccc}2(1-y)-y+z=1\\1-y+2y+z=\dfrac{8}{3}\end{array}\right\\ \left\{\begin{array}{ccc}2-2y-y+z=1&|-2\\1+y+z=\dfrac{8}{3}&|-1\end{array}\right\\ \left\{\begin{array}{ccc}-3y+z=-1\\y+z=\dfrac{5}{3}\end{array}\right\\\text{substract both sides of the equations}\\-4y=-\dfrac{8}{3}\ \ \ \ |:(-4)\\y=\dfrac{2}{3}[/tex]
[tex]\text{substitute the value of y to the second equation}\\\\\dfrac{2}{3}+z=\dfrac{5}{3}\ \ \ \ |-\dfrac{2}{3}\\\\z=\dfrac{3}{3}\to z=1\\\\\text{substitute the value of y to the equation}\ x=1-y\\\\x=1-\dfrac{2}{3}\to x=\dfrac{1}{3}\\\\\text{Answer:}\ \left\{\begin{array}{ccc}x=\dfrac{1}{3}\\\\y=\dfrac{2}{3}\\\\z=1\end{array}\right[/tex]
