Question

1. set up but do not solve a system of equations.

a coal mining company owns three different mines. each day, mine a produces 9 tons of anthracite, 4 tons of ordinary coal and 3 tons of bituminous coal. each day, mine b produces 8 tons of anthracite, 5 tons of ordinary coal and 6 tons of bituminous coal. each day, mine c produces 7 tons of anthracite, 4 tons of ordinary coal and 7 tons of bituminous coal. for how many days must each of the three mines be operated to produce exactly 334 tons of anthracite, 184 tons of ordinary coal and 230 tons of bituminous coal?

Explanation:

Step1: Define variables

Let \( x \) be the number of days Mine A operates, \( y \) be the number of days Mine B operates, and \( z \) be the number of days Mine C operates.

Step2: Set up equation for anthracite

The total anthracite produced is the sum of anthracite from each mine. So, \( 9x + 8y + 7z = 334 \).

Step3: Set up equation for ordinary coal

The total ordinary coal produced is the sum of ordinary coal from each mine. So, \( 4x + 5y + 4z = 184 \).

Step4: Set up equation for bituminous coal

The total bituminous coal produced is the sum of bituminous coal from each mine. So, \( 3x + 6y + 7z = 230 \).

Answer:

The system of equations is \(

$$\begin{cases}9x + 8y + 7z = 334\\4x + 5y + 4z = 184\\3x + 6y + 7z = 230\end{cases}$$

\) where \( x \) is the number of days Mine A operates, \( y \) is the number of days Mine B operates, and \( z \) is the number of days Mine C operates.