A waitress sold $10$ ribeye steak dinners and $10$ grilled salmon dinners, totaling $\$584.01$ on a particular day. Another day she sold $11$ ribeye steak dinners and $5$ grilled salmon dinners, totaling $\$584.45$ . How much did each type of dinner cost?

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Q. A waitress sold

10

ribeye steak dinners and

10

grilled salmon dinners, totaling

\$584.01

on a particular day. Another day she sold

11

ribeye steak dinners and

5

grilled salmon dinners, totaling

\$584.45

. How much did each type of dinner cost?

Set up equations: Let's denote the cost of one ribeye steak dinner as $R$ and the cost of one grilled salmon dinner as $S$ . We can set up two equations based on the information given: $\newline$ $10R + 10S = 584.01$ (from the first day's sales) $\newline$ $11R + 5S = 584.45$ (from the second day's sales)
Simplify first equation: We can simplify the first equation by dividing all terms by $10$ , which will give us: $R + S = 58.401$
Multiply second equation: Now, let's multiply the second equation by $2$ to help us eliminate one of the variables when we subtract the equations: $\newline$ $2(11R + 5S) = 2(584.45)$ $\newline$ $22R + 10S = 1168.90$
Align S terms: Next, we multiply the simplified first equation by $10$ to align the $S$ terms with the second equation: $\newline$ $10(R + S) = 10(58.401)$ $\newline$ $10R + 10S = 584.01$
Eliminate S: Now we subtract the modified first equation from the modified second equation to eliminate S and solve for R: $\newline$ $(22R + 10S) - (10R + 10S) = 1168.90 - 584.01$ $\newline$ $22R - 10R = 1168.90 - 584.01$ $\newline$ $12R = 584.89$
Solve for R: Divide both sides by $12$ to solve for R: $\newline$ $R = \frac{584.89}{12}$ $\newline$ $R = 48.74$
Substitute for S: Now that we have the value for R, we can substitute it back into the simplified first equation to solve for S: $\newline$ $R + S = 58.401$ $\newline$ $48.74 + S = 58.401$
Solve for S: Subtract $48.74$ from both sides to solve for S: $\newline$ $S = 58.401 - 48.74$ $\newline$ $S = 9.661$