A waitress sold 11 ribeye steak dinners and 33 grilled salmon dinners, totaling $580.27 on a particular day. Another day she sold 30 ribeye steak dinners and 11 grilled salmon dinners, totaling $580.58. How much did each type of dinner cost? The cost of ribeye steak dinners is $◻ and the cost of salmon dinners is $◻.

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Set Up Equations: Let's denote the cost of one ribeye steak dinner as R dollars and the cost of one grilled salmon dinner as S dollars. We can set up two equations based on the information given: 11R + 33S = 580.27 (Equation 1) 30R + 11S = 580.58 (Equation 2)Elimination Method: We will solve this system of equations using the method of substitution or elimination. Let's use the elimination method to eliminate one of the variables. We can multiply Equation 1 by 30 and Equation 2 by 11 to make the coefficients of R the same: 30(11R + 33S) = 30(580.27) 11(30R + 11S) = 11(580.58)Multiply Equations: After multiplying, we get: 330R + 990S = 17408.10 (Equation 3) 330R + 121S = 6386.38 (Equation 4)Subtract Equations: Now, we will subtract Equation 4 from Equation 3 to eliminate R: (330R + 990S) - (330R + 121S) = 17408.10 - 6386.38 990S - 121S = 11021.72 869S = 11021.72Solve for S: Divide both sides by 869 to solve for S: S = 11021.72 / 869 S = 12.68Substitute Back: Now that we have the value of S, we can substitute it back into one of the original equations to solve for R. Let's use Equation 1: 11R + 33(12.68) = 580.27 11R + 418.44 = 580.27Solve for R: Subtract 418.44 from both sides to solve for R: 11R = 580.27 - 418.44 11R = 161.83Final Solution: Divide both sides by 11 to find the value of R: R = 161.83 / 11 R = 14.71

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