Derive the expressions for the velocities of two bodies after a one-dimensional elastic collision.

(N/A) Consider two bodies of masses $m_1$ and $m_2$ moving in a straight line along the $X$-direction. Let their initial velocities be $v_{1i}$ and $v_{2i}$ respectively,with $v_{1i} > v_{2i}$.
After the collision,let their final velocities be $v_{1f}$ and $v_{2f}$. In an elastic collision,both linear momentum and kinetic energy are conserved.
Conservation of linear momentum:
$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$
$m_1(v_{1i} - v_{1f}) = m_2(v_{2f} - v_{2i})$ ---$(1)$
Conservation of kinetic energy:
$\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$
$m_1(v_{1i}^2 - v_{1f}^2) = m_2(v_{2f}^2 - v_{2i}^2)$
$m_1(v_{1i} - v_{1f})(v_{1i} + v_{1f}) = m_2(v_{2f} - v_{2i})(v_{2f} + v_{2i})$ ---$(2)$
Dividing equation $(2)$ by equation $(1)$:
$v_{1i} + v_{1f} = v_{2f} + v_{2i}$
$v_{1f} = v_{2f} + v_{2i} - v_{1i}$ ---$(3)$
Substituting $(3)$ into $(1)$:
$m_1(v_{1i} - (v_{2f} + v_{2i} - v_{1i})) = m_2(v_{2f} - v_{2i})$
$m_1(2v_{1i} - v_{2i} - v_{2f}) = m_2(v_{2f} - v_{2i})$
$2m_1 v_{1i} - m_1 v_{2i} - m_1 v_{2f} = m_2 v_{2f} - m_2 v_{2i}$
$2m_1 v_{1i} + (m_2 - m_1)v_{2i} = (m_1 + m_2)v_{2f}$
$v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i} + \frac{m_2 - m_1}{m_1 + m_2}v_{2i}$
Similarly,for $v_{1f}$:
$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} + \frac{2m_2}{m_1 + m_2}v_{2i}$