Consider the following system of ODE's [1]: \begin{align} \dot x & = \lambda - d_1 x - \frac{k_1 x v}{x + v}\\ \dot y & = \frac{k_1 x v}{x + v} - d_2 y - k_2 y \\ \dot s &= k_2 y - d_3s \\ \dot v &= as - d_4 v \end{align} where $x(t)$, $y(t)$, $s(t)$ and $v(t)$ represent the concentrations of uninfected CD4+ T cells, infected CD4+ T cells in latent stage, productively infected CD4+ T cells and free virus (HIV), respectively. Under some condition, this system has unique strictly positive equilibrium $Q_2 = (x^*, y^*, s^*, v^*)$ that is locally asymptotically stable. Assume this condition. Then at $Q_2$, the system becomes: \begin{align} 0 & = \lambda - d_1 x^* - \frac{k_1 x^* v^*}{x^* + v^*}\\ 0 & = \frac{k_1 x^* v^*}{x^* + v^*} - d_2 y^* - k_2 y^* \\ 0 &= k_2 y^* - d_3 s^* \\ 0 &= a s^* - d_4 v^* \end{align}
My question is: what justified the following equality? (cf. left side of p. 99 in [1]) $$ k_1 = \frac{(d_2 + k_2) y^* (x^* + v^*)}{x^* v^*} \times \left[\frac{(d_2 + k_2) y^* (x + v^*)}{k_1 x v^*} \right] = \frac{x^* (x + v^*)}{x (x^* + v^*)}$$
Specifically, how are the following true? $$ \left[\frac{(d_2 + k_2) y^* (x + v^*)}{k_1 x v^*} \right] = 1 $$ and $$ \frac{(d_2 + k_2) y^* (x^* + v^*)}{x^* v^*} \times \left[\frac{(d_2 + k_2) y^* (x + v^*)}{k_1 x v^*} \right] = \frac{x^* (x + v^*)}{x (x^* + v^*)}$$ I understand that the equation $ 0 = \frac{k_1 x^* v^*}{x^* + v^*} - d_2 y^* - k_2 y^* $ is the key, but I cannot figure out how the above two equalities are true.
[1] Q. Sun, L. Min, Y. Kuang: "Global stability of infection-free state and endemic infection state of a modified human immunodeficiency virus infection model", IET Syst. Biol. 9-3 (2015), 95–103 doi:10.1049/iet-syb.2014.0046
Solving
$$ \left\{ \begin{array}{rcl} \lambda - d_1 x^* - (k_1 x^* v^*)/(x^* + v^*) & = & 0\\ k_1 x^* v^*/(x^* + v^*) - d_2 y^* - k_2 y^* & = & 0\\ k_2 y^* - d_3 s^* & = & 0\\ a s^* - d_4 v^* & = & 0 \end{array} \right. $$
for $k_1, d_1, d_3, d_4$ we obtain
$$ \left\{ \begin{array}{rcl} k_1 & = & ((d_2 + k_2) (v^* + x^*) y^*)/(v^* x^*)\\ d_1 & = & (\lambda - (d_2 + k_2) y^*)/x^*\\ d_3 & = & (k_2 y^*)/s^*\\ d_4 & = & a s^*/v^* \end{array} \right. $$
Substituting those values into
$(d_2 + k_2) y^* (x + v^*)/(k_1 x v^*)$ we obtain $((v^* + x) x^*)/(x (v^* + x^*))$
I hope this helps.