1. **State the problem:** Pedro can make a total of 140 tacos or burritos combined, and he must sell at least 600 worth of tacos and burritos daily. 2. **Define variables:** Let $x$ = number of tacos sold, $y$ = number of burritos sold. 3. **Write inequalities based on the problem:** - Total items constraint: $x + y \leq 140$ - Revenue constraint: $3x + 6y \geq 600$ 4. **Rewrite inequalities to express $y$ in terms of $x$:** - From $x + y \leq 140$, we get $$y \leq 140 - x$$ - From $3x + 6y \geq 600$, divide both sides by 3: $$\cancel{3}x + \cancel{3}2y \geq \frac{600}{3}$$ $$x + 2y \geq 200$$ Now isolate $y$: $$2y \geq 200 - x$$ $$y \geq \frac{200 - x}{2}$$ 5. **System of inequalities:** $$\begin{cases} y \leq 140 - x \\ y \geq \frac{200 - x}{2} \end{cases}$$ 6. **Find one possible solution:** Try $x=60$: - Check $y \leq 140 - 60 = 80$ - Check $y \geq \frac{200 - 60}{2} = \frac{140}{2} = 70$ So $y$ must satisfy $70 \leq y \leq 80$. Choose $y=75$. 7. **Verify solution:** - Total items: $60 + 75 = 135 \leq 140$ (valid) - Revenue: $3(60) + 6(75) = 180 + 450 = 630 \geq 600$ (valid) **Final answer:** One possible solution is $x=60$ tacos and $y=75$ burritos.