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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.)

A waitress sold 1818 ribeye steak dinners and 1212 grilled salmon dinners, totaling $563.44 \$ 563.44 on a particular day. Another day she sold 2626 ribeye steak dinners and 66 grilled salmon dinners, totaling $580.72 \$ 580.72 . How much did each type of dinner cost?\newlineThe 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 1818 ribeye steak dinners and 1212 grilled salmon dinners, totaling $563.44 \$ 563.44 on a particular day. Another day she sold 2626 ribeye steak dinners and 66 grilled salmon dinners, totaling $580.72 \$ 580.72 . How much did each type of dinner cost?\newlineThe 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.)
  1. Set up equations: Let RR be the cost of a ribeye steak dinner and SS be the cost of a grilled salmon dinner. We can set up two equations based on the information given. Equation 11: 18R+12S=563.4418R + 12S = 563.44. Equation 22: 26R+6S=580.7226R + 6S = 580.72.
  2. Modify second equation: Multiply the second equation by 22 to make the coefficient of SS the same as in the first equation. So, 2×(26R+6S)=2×580.722 \times (26R + 6S) = 2 \times 580.72, which gives us 52R+12S=1161.4452R + 12S = 1161.44.
  3. 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.
  4. Solve for R: Divide both sides of the equation by 3434 to solve for RR. 34R34=59834\frac{34R}{34} = \frac{598}{34}, which gives us R=17.59R = 17.59.
  5. 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=563.4418(17.59) + 12S = 563.44. This simplifies to 316.62+12S=563.44316.62 + 12S = 563.44.
  6. Solve for S: Subtract 316.62316.62 from both sides to solve for SS. 12S=563.44316.6212S = 563.44 - 316.62, which gives us 12S=246.8212S = 246.82.
  7. Final solution: Divide both sides by 1212 to find the value of SS. 12S12=246.8212\frac{12S}{12} = \frac{246.82}{12}, which gives us S=20.57S = 20.57.

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