Question

A waitress sold 15 ribeye steak dinners and 18 grilled salmon​ dinners, totaling ​$559.81 on a particular day. Another day she sold 19 ribeye steak dinners and 9 grilled salmon​ dinners, totaling ​583.66. How much did each type of dinner​ cost?

Answer 1

Let the steaks = X and the salmon = y.

Set up two equations:

15x + 18y = 559.81

19x + 9y = 583.66

Now using the elimination method:

Multiply the second equation by -2, then add the equations together.

(15x+18y=559.81)

−2(19x+9y=583.66)

Becomes:

15x+18y=559.81

−38x−18y=−1167.32

Add these equations to eliminate y:

−23x=−607.51

Divide both sides by -23 to solve for x:

x= -607.51 = -23 = 26.413478

Now you have the cost for a steak.

To solve for the cost of the salmon, replace x with the value in the first equation and solve for y.

15(26.413478) + 18y = 559.91

396.202174 + 18y = 559.81

Subtract 396.202174 from both sides:

18y = 163.607826

Divide both sides by 18:

y = 163.607826 / 18

y = 9.089324

Round both x and Y to the nearest cent:

X (Steaks) =$26.41

Y (Salmon) = $9.09