A waitress sold $20$ ribeye steak dinners and $12$ grilled salmon dinners, totaling $\$ 573.12$ on a particular day. Another day she sold $26$ ribeye steak dinners and $6$ grilled salmon dinners, totaling $\$ 582.07$ . How much did each type of dinner cost? $\newline$ The cost of ribeye steak dinners is $\$$ $\square$ and the cost of salmon dinners is $\$$ $\square$ . (Simplify your answer. Round to the nearest hundredth as needed.)

Full solution

Q. A waitress sold

20

ribeye steak dinners and

12

grilled salmon dinners, totaling

\$ 573.12

on a particular day. Another day she sold

26

ribeye steak dinners and

6

grilled salmon dinners, totaling

\$ 582.07

. How much did each type of dinner cost?

\newline

The cost of ribeye steak dinners is

\$

\square

and the cost of salmon dinners is

\$

\square

. (Simplify your answer. Round to the nearest hundredth as needed.)

Set up equations: Let $R$ be the cost of a ribeye steak dinner and $S$ be the cost of a grilled salmon dinner. We can set up two equations based on the information given. Equation $1$ : $20R + 12S = 573.12$ . Equation $2$ : $26R + 6S = 582.07$ .
Eliminate variable $S$ : Multiply Equation $1$ by $3$ to eliminate $S$ . So, $60R + 36S = 1719.36$ .
Subtract equations: Multiply Equation $2$ by $6$ to eliminate $S$ . So, $156R + 36S = 3492.42$ .
Solve for R: Subtract the new Equation $2$ from the new Equation $1$ to solve for R. $(156R + 36S) - (60R + 36S) = 3492.42 - 1719.36$ . This simplifies to $96R = 1773.06$ .
Find R: Divide both sides by $96$ to find $R$ . R = rac{1773.06}{96}. $R = 18.47$ .
Solve for S: Substitute $R = 18.47$ back into Equation $1$ to solve for $S$ . $20(18.47) + 12S = 573.12$ . This simplifies to $369.4 + 12S = 573.12$ .
Find S: Subtract $369.4$ from both sides to solve for $S$ . $12S = 573.12 - 369.4$ . $12S = 203.72$ .
Find $S$ : Subtract $369.4$ from both sides to solve for $S$ . $12S = 573.12 - 369.4$ . $12S = 203.72$ . Divide both sides by $12$ to find $S$ . $S = \frac{203.72}{12}$ . $S = 16.98$ .