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Use the Charnes-Cooper transformation on the positive-denominator region:It gives $Ay\le bt$, $d^{T}y+\beta t=1$ and $c^{T}y+\alpha t=(c^{T}x+\alpha )/(d^{T}x+\beta )$.

We must check that allowing $t=0$ introduces no false feasible solution. If $t=0$, then $Ay\le 0$. The recession cone of the nonempty linear polyhedron $P=\{x:Ax\le b\}$ is $\{y:Ay\le 0\}$. Indeed for $x_0\in P$, $x_0+uy\in P$ for every $u\ge 0$. Boundedness of $P$ therefore forces $y=0$, contradicting $d^{T}y=1$. Thus every transformed feasible point has $t>0$.

Conversely set $x=y/t$. The constraints give $x\in P$ and $d^{T}x+\beta =1/t>0$, with the same objective value. The assumed original optimizer with positive denominator supplies a transformed feasible point. Every transformed point corresponds to an original point and has objective at most that optimizer's value. Hence the transformed linear program attains the original optimum, and its optimizer recovers $x=y/t$. Positivity on every point of $P$ is not needed here; the given positive-denominator optimum suffices.

Apply the Charnes-Cooper transformation to any feasible point $x$ of $Q$:Thenso $(y,t)$ is feasible for $R$, andIn particular, the given solution $x^{\ast }$ of $Q$ produces a feasible point of $R$ with the same objective value, so

Moreover, because every $d_i>0$, the equation $d^{T}y=1$ impliesThe linear objective $c^{T}y$ is therefore bounded on the feasible set. The feasible $y$ lie in the compact simplex $\{y\ge 0:d^{T}y=1\}$. Their subset arising in $R$ is closed: if $b\ne 0$, any nonzero component of $b$ determines $t$ continuously from $Ay=bt$, while if $b=0$ the condition is simply $Ay=0$. Hence the feasible $y$ form a compact set, on which the continuous objective $c^{T}y$ attains a finite maximum. Thus $R$ has a finite maximum at least as large as that of $Q$.

The recession cone of a nonempty convex set $C$ consists of directions $d$ for which $x+td\in C$ for every $x\in C$ and $t\ge 0$. For a linear polyhedron $P=\{x:Ax\le b\}$ it is exactly $\{d:Ad\le 0\}$. A nonzero recession direction gives an unbounded ray, so a bounded nonempty polyhedron has recession cone $\{0\}$. This excludes spurious zero-scale feasible points in the Charnes-Cooper transformation.