Answer
$t=\dfrac{1}{6}$
Work Step by Step
Using the properties of equality, the solution to the given equation, $
2(3t+1)-5=t-(t+2)
$, is
\begin{array}{l}
6t+2-5=t-t-2
\\\\
6t-3=-2
\\\\
6t=-2+3
\\\\
6t=1
\\\\
t=\dfrac{1}{6}
.\end{array}
CHECKING:
\begin{array}{l}
2\left(3\cdot\dfrac{1}{6}+1 \right)-5=\dfrac{1}{6}-\left( \dfrac{1}{6}+2 \right)
\\\\
2\left(\dfrac{1}{2}+\dfrac{2}{2} \right)-5=\dfrac{1}{6}-\left( \dfrac{1}{6}+\dfrac{12}{6} \right)
\\\\
2\left(\dfrac{3}{2} \right)-5=\dfrac{1}{6}-\left( \dfrac{13}{6} \right)
\\\\
3-5=\dfrac{1}{6}-\dfrac{13}{6}
\\\\
-2=-\dfrac{12}{6}
\\\\
-2=-2
\text{ (TRUE)}
.\end{array}
Hence, the solution is $
t=\dfrac{1}{6}
$.
