Since both machines maintain their rate, the number of copies made and the elapsed time are proportional. This means that we can write a proportion like
![\text{time}_1 : \text{copies in time}_1 = \text{time}_2 : \text{copies in time}_2](https://img.qammunity.org/2019/formulas/mathematics/high-school/oixa3fdmt8igqb6hd3lz7s7gkg30ebylqa.png)
We know the performance of machine a for 12 minutes, and we want those for 30 minutes: the proportion becomes
![12 : 100 = 30 : x \iff x = (100\cdot 30)/(12) = (3000)/(12) = 250](https://img.qammunity.org/2019/formulas/mathematics/high-school/gmkoofgo25v175p9xmnkftittzv582oqq1.png)
Similarly, we have for machine b
![10 : 150 = 30 : x \iff x = (150\cdot 30)/(10) = (4500)/(10) = 450](https://img.qammunity.org/2019/formulas/mathematics/high-school/xm852sbwotamyn7ct2014uxzozp7le6p6l.png)
So, together, the two machines make
![250+450 = 700](https://img.qammunity.org/2019/formulas/mathematics/high-school/f0dxy05uw82djm1ydqr9lpt8s5td0yb0p8.png)
copies.