Is this correct?
Now we will add the terms of the left side
[tex]\begin{gathered} \frac{5}{3}x+\frac{1}{3}x=13\frac{1}{3}+\frac{8}{3}x \\ \frac{6}{3}x=13\frac{1}{3}+\frac{8}{3}x \end{gathered}[/tex]Now subtract 8/3 x from both sides
[tex]\frac{6}{3}x-\frac{8}{3}x=\frac{40}{3}+\frac{8}{3}x-\frac{8}{3}x[/tex][tex]-\frac{2}{3}x=\frac{40}{3}[/tex]Cancel the denominator 3 from both sides
-2x = 40
Divide two sides by -2
[tex]\frac{-2x}{-2}=\frac{40}{-2}[/tex]x = -20
