smc

First, the problem states: $x$ cows $x+1$ cans $x+2$ days Multiply the current number of cows by $\frac{x+3}{x}$ to get the number of cows you want, and divide the number of days by the same amount because an increase in cows will cause a decrease in time. $x\cdot \frac{x+3}{x}x+1(x+2)\cdot \frac{x}{x+3}$ Finally, multiply the current amount of cans by $\frac{x+5}{x+1}$ to get the number of cans you want, and multiply the number of days by the same amount because an increase in cans will cause an increase in time. $x\cdot \frac{x+3}{x}(x+1)\cdot \frac{x+5}{x+1}(x+2)\cdot \frac{x}{x+3}\cdot \frac{x+5}{x+1}$ Therefore, our answer is $\boxed{\frac{x(x+2)(x+5)}{(x+1)(x+3)}}$