A customer paid a total of \( \$ 6.00 \) for 68 copies at a print shop. Some of the copies were black-and-white copies, and the rest were color copies. - Each black-and-white copy cost \( \$ 0.08 \). (1) - Each color copy cost \( \$ 0.15 \). Which system of equations can be used to find \( b \), the number of black-and-white copies, and \( c \), the number of color copies that the customer paid for at the print shop? (c) A. \( b+c=6.00 \) \( 0.08 b+0.15 c=68 \) B. \( b+c=68 \) \( 0.15 b+0.08 c=6.00 \) C. \( b+c=6.00 \) \[ 0.15 b+0.08 c=68 \] D. \[ \begin{array}{l} b+c=68 \\ 0.08 b+0.15 c=6.00 \end{array} \]

To solve this problem, let's take a look at the equations we need. The total number of copies is 68, so one part of our system must reflect that relationship: \( b + c = 68 \). The total cost is $6.00, where black-and-white copies cost $0.08 each and color copies $0.15 each. This gives us the second equation: \( 0.08b + 0.15c = 6.00 \). So, the correct option is D: \[ \begin{array}{l} b+c=68 \\ 0.08 b+0.15 c=6.00 \end{array} \] We need those equations to pin down the exact quantities of black-and-white and color copies. Now, when solving problems like this, one common mistake is to confuse total counts with total costs. Remember to clearly differentiate the elements like counts of copies versus their total price, or you might end up with mixed numbers that just don’t make sense, creating further confusion in your solving process! It’s like trying to mix apples and oranges—just doesn’t work! If you're looking to delve deeper into similar equations and systems, consider checking out resources on linear algebra or systems of equations. They not only help in academic settings but are also useful in everyday calculations like budgeting, planning purchases, or even understanding more complex data relationships. You can easily find online sources or textbooks that cater to all learning levels!