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?
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:10R+10S=584.01 (from the first day's sales)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:2(11R+5S)=2(584.45)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:10(R+S)=10(58.401)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:(22R+10S)−(10R+10S)=1168.90−584.0122R−10R=1168.90−584.0112R=584.89
Solve for R: Divide both sides by 12 to solve for R:R=12584.89R=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:R+S=58.40148.74+S=58.401
Solve for S: Subtract 48.74 from both sides to solve for S:S=58.401−48.74S=9.661