1. **State the problem:** We want to find the basic reproduction number $R_0$ for the system given by: $$\frac{dS}{dt}=B-(a+b+h)S$$ $$\frac{dI_1}{dt}=aS-(c+h)I_1$$ $$\frac{dI_2}{dt}=bS-(d+h)I_2$$ $$\frac{dT}{dt}=cI_1+dI_2-(h+l)T$$ where $S$, $I_1$, $I_2$, and $T$ are population compartments, and parameters $a,b,c,d,h,l,B$ are positive constants as defined. 2. **Find disease-free equilibrium (DFE):** At DFE, no infected individuals, so $I_1=I_2=T=0$. Setting derivatives zero, from first equation: $$0 = B - (a+b+h)S_0 \implies S_0 = \frac{B}{a+b+h}$$ 3. **Define $R_0$ heuristically without NGM:** Since $R_0$ measures average new infections produced by one infected individual in a fully susceptible population, consider infections from $I_1$ and $I_2$ compartments separately. 4. **New infections produced by $I_1$:** Each infected from unprotected sexual contact infects susceptible at rate $a$, and leaves $I_1$ at rate $c+h$. So expected infectious period in $I_1$ is $\frac{1}{c+h}$. Number of new infections by one $I_1$ individual over infection period: $$R_{0,1} = a \times S_0 \times \frac{1}{c + h} = a \times \frac{B}{a+b+h} \times \frac{1}{c + h}$$ 5. **New infections produced by $I_2$: ** Similarly, transmission rate $b$, removal rate $d+h$, so infectious period $\frac{1}{d+h}$. Number of new infections by one $I_2$: $$R_{0,2} = b \times S_0 \times \frac{1}{d + h} = b \times \frac{B}{a+b+h} \times \frac{1}{d + h}$$ 6. **Combine contributions:** Total $R_0$ is sum of both pathways: $$R_0 = R_{0,1} + R_{0,2} = \frac{aB}{(a+b+h)(c+h)} + \frac{bB}{(a+b+h)(d+h)}$$ 7. **Interpretation:** $R_0$ expresses the expected number of secondary infections caused by one infected individual in an otherwise susceptible population, accounting for the two infection routes and their treatment/removal rates. **Final answer:** $$\boxed{R_0 = \frac{aB}{(a+b+h)(c+h)} + \frac{bB}{(a+b+h)(d+h)}}$$