A waitress sold 18 ribeye steak dinners and 12 grilled salmon dinners, totaling $563.44 on a particular day. Another day she sold 26 ribeye steak dinners and 6 grilled salmon dinners, totaling $580.72. How much did each type of dinner cost?The cost of ribeye steak dinners is $□ and the cost of salmon dinners is $□ . (Simplify your answer. Round to the nearest hundredth as needed.)
Q. A waitress sold 18 ribeye steak dinners and 12 grilled salmon dinners, totaling $563.44 on a particular day. Another day she sold 26 ribeye steak dinners and 6 grilled salmon dinners, totaling $580.72. How much did each type of dinner cost?The cost of ribeye steak dinners is $□ and the cost of salmon dinners is $□ . (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: 18R+12S=563.44. Equation 2: 26R+6S=580.72.
Modify second equation: Multiply the second equation by 2 to make the coefficient of S the same as in the first equation. So, 2×(26R+6S)=2×580.72, which gives us 52R+12S=1161.44.
Eliminate S: Now subtract the first equation from the modified second equation to eliminate S. 52R + 12S) - (18R + 12S) = 1161.44 - 563.44\. This simplifies to \$34R = 598.
Solve for R: Divide both sides of the equation by 34 to solve for R. 3434R=34598, which gives us R=17.59.
Plug in R: Now plug the value of R back into one of the original equations to solve for S. Using the first equation: 18(17.59)+12S=563.44. This simplifies to 316.62+12S=563.44.
Solve for S: Subtract 316.62 from both sides to solve for S. 12S=563.44−316.62, which gives us 12S=246.82.
Final solution: Divide both sides by 12 to find the value of S. 1212S=12246.82, which gives us S=20.57.