= Solution
With $\gamma(v)=(1-v^2/c^2)^{-1/2}$,
$$
x'=\gamma(x-vt),
\qquad
t'=\gamma\left(t-\frac{vx}{c^2}\right).
$$
Adding and subtracting $ct'$ gives
$$
x'_+=\gamma(1-v/c)x_+
=\sqrt{\frac{c-v}{c+v}},x_+,
$$
$$
x'_- =\gamma(1+v/c)x_-
=\sqrt{\frac{c+v}{c-v}},x_-.
$$
The product of the two multipliers is one, so $x'_+x'_-=x_+x_-$, which is $x'^2-c^2t'^2=x^2-c^2t^2$.
Successive transformations multiply the $x_+$ factors. Equating
$$
\lambda(v_3)=\lambda(v_2)\lambda(v_1)
$$
and solving gives the relativistic velocity-addition law
$$
\boxed{v_3=\frac{v_1+v_2}{1+v_1v_2/c^2}.}
$$
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