Answer:
[tex]$\frac{1}{3}+x+\frac{2}{9}\ge \frac{5}{6}\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\ge \frac{5}{18}\:\\ \:\mathrm{Decimal:}&\:x\ge \:0.27777\dots \\ \:\mathrm{Interval\:Notation:}&\:[\frac{5}{18},\:\infty \:)\end{bmatrix}$[/tex]
Step-by-step explanation:
[tex]$\frac{1}{3}+x+\frac{2}{9}\ge \frac{5}{6}$[/tex]
[tex]\gray{\mathrm{Group\:like\:terms}}[/tex]
[tex]$x+\frac{1}{3}+\frac{2}{9}\ge \frac{5}{6}$[/tex]
[tex]\black{\mathrm{Least\:Common\:Multiplier\:of\:}3,\:9:\quad 9}[/tex]
[tex]\black{\mathrm{Adjust\:Fractions\:based\:on\:the\:LCM}}[/tex]
[tex]$\frac{3}{9}+\frac{2}{9}\ge \frac{5}{6}$[/tex]
[tex]$\gray{\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}}$[/tex]
[tex]$\frac{3+2}{9}\ge \frac{5}{6}$[/tex]
[tex]\gray{\mathrm{Add\:the\:numbers:}\:3+2=5}[/tex]
[tex]$x+\frac{5}{9}\ge \frac{5}{6}$[/tex]
[tex]$\gray{\mathrm{Subtract\:}\frac{5}{9}\mathrm{\:from\:both\:sides}}$[/tex]
[tex]$x+\frac{5}{9}-\frac{5}{9}\ge \frac{5}{6}-\frac{5}{9}$[/tex]
[tex]\gray{\mathrm{Simplify}}[/tex]
[tex]$x\ge \frac{5}{18}$[/tex]
