A waitress sold 18 ribeye steak dinners and 22 grilled salmon dinners, totaling $579.15 on a particular day. Another day she sold 29 ribeye steak dinners and 11 grilled salmon dinners, totaling $583.96. 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 22 grilled salmon dinners, totaling $579.15 on a particular day. Another day she sold 29 ribeye steak dinners and 11 grilled salmon dinners, totaling $583.96. 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: 18R+22S=579.15 and 29R+11S=583.96.
Write first equation: Write the first equation from the given information: 18R+22S=579.15.
Write second equation: Write the second equation from the given information: 29R+11S=583.96.
Eliminate variable: Multiply the first equation by 11 and the second equation by 22 to eliminate S when we subtract the equations. So we get 198R+242S=6370.65 and 638R+242S=12847.12.
Solve for R: Subtract the second equation from the first to find R: 198R+242S−(638R+242S)=6370.65−12847.12. This simplifies to −440R=−6476.47.
Plug in R: Divide both sides by −440 to solve for R: R=−6476.47/−440. This gives us R=14.71925.
Solve for S: Now plug the value of R back into the first equation to solve for S: 18(14.71925)+22S=579.15. This simplifies to 264.9465+22S=579.15.
Subtract values: Subtract 264.9465 from both sides to solve for S: 22S=579.15−264.9465. This gives us 22S=314.2035.
Find S: Divide both sides by 22 to find S: S=22314.2035. This gives us S=14.28197727.
Round to nearest hundredth: Round R and S to the nearest hundredth: R=$14.72 and S=$14.28.