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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?

A waitress sold 1010 ribeye steak dinners and 1010 grilled salmon dinners, totaling $584.01\$584.01 on a particular day. Another day she sold 1111 ribeye steak dinners and 55 grilled salmon dinners, totaling $584.45\$584.45. How much did each type of dinner cost?

Full solution

Q. A waitress sold 1010 ribeye steak dinners and 1010 grilled salmon dinners, totaling $584.01\$584.01 on a particular day. Another day she sold 1111 ribeye steak dinners and 55 grilled salmon dinners, totaling $584.45\$584.45. How much did each type of dinner cost?
  1. Set up equations: Let's denote the cost of one ribeye steak dinner as RR and the cost of one grilled salmon dinner as SS. We can set up two equations based on the information given:\newline10R+10S=584.0110R + 10S = 584.01 (from the first day's sales)\newline11R+5S=584.4511R + 5S = 584.45 (from the second day's sales)
  2. Simplify first equation: We can simplify the first equation by dividing all terms by 1010, which will give us: R+S=58.401R + S = 58.401
  3. Multiply second equation: Now, let's multiply the second equation by 22 to help us eliminate one of the variables when we subtract the equations:\newline2(11R+5S)=2(584.45)2(11R + 5S) = 2(584.45)\newline22R+10S=1168.9022R + 10S = 1168.90
  4. Align S terms: Next, we multiply the simplified first equation by 1010 to align the SS terms with the second equation:\newline10(R+S)=10(58.401)10(R + S) = 10(58.401)\newline10R+10S=584.0110R + 10S = 584.01
  5. Eliminate S: Now we subtract the modified first equation from the modified second equation to eliminate S and solve for R:\newline(22R+10S)(10R+10S)=1168.90584.01(22R + 10S) - (10R + 10S) = 1168.90 - 584.01\newline22R10R=1168.90584.0122R - 10R = 1168.90 - 584.01\newline12R=584.8912R = 584.89
  6. Solve for R: Divide both sides by 1212 to solve for R:\newlineR=584.8912R = \frac{584.89}{12}\newlineR=48.74R = 48.74
  7. 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:\newlineR+S=58.401R + S = 58.401\newline48.74+S=58.40148.74 + S = 58.401
  8. Solve for S: Subtract 48.7448.74 from both sides to solve for S:\newlineS=58.40148.74S = 58.401 - 48.74\newlineS=9.661S = 9.661

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