Answer
See the detailed answer below.
Work Step by Step
Let's start with the Lorentz transformation equation for space in $x$-direction.
$$x'=\gamma(x-vt)\tag 1$$
and
$$x=\gamma(x'+vt')\tag 2 $$
Plug $x'$ from (1) into (2),
$$x=\gamma(\gamma \;x-\gamma\; vt+vt') $$
$$x=\gamma^2 \;x-\gamma^2\; vt+\gamma \;vt' $$
$$\dfrac{x}{\gamma^2}= x- vt+\dfrac{vt'}{\gamma} $$
where $\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}$, and hence, $\gamma^2=\left[1-\frac{v^2}{c^2}\right]^{-1}=\dfrac{1}{1-\frac{v^2}{c^2}}$;
Thus,
$$x\left[1-\frac{v^2}{c^2}\right]= x- vt+\dfrac{vt'}{\gamma} $$
$$ x-\frac{v^2x}{c^2} - x+ vt=\dfrac{vt'}{\gamma} $$
$$ \color{red}{\bf\not} v\left[ t-\frac{vx}{c^2} \right] =\dfrac{\color{red}{\bf\not} vt'}{\gamma} $$
Therefore,
$$\boxed{t'=\gamma\left[ t-\frac{vx}{c^2} \right] }$$
To find $t$, we need to plug $x$ from (2) into (1),
$$x'=\gamma(\gamma(x'+vt')-vt) =\gamma^2 x'+\gamma^2 vt'-\gamma vt$$
$$\dfrac{x'}{\gamma^2}= x'+ vt'-\dfrac{vt}{\gamma }$$
Plug $\gamma^2$ from above;
$$ x' \left[1-\frac{v^2}{c^2}\right]= x'+ vt'-\dfrac{vt}{\gamma }$$
$$ x'-\frac{v^2x'}{c^2} - x'- vt'=-\dfrac{vt}{\gamma }$$
$$ -\frac{v^2x'}{c^2} - vt'=-\dfrac{vt}{\gamma }$$
Divide by $-v$;
$$ \frac{v x'}{c^2} +t'=\dfrac{t}{\gamma }$$
Therefore,
$$\boxed{t=\gamma\left[ t'+\frac{v x'}{c^2} \right]}$$
