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Question

A waitress sold 13 ribeye steak dinners and  26 grilled salmon dinners, totaling $597.19 on a particular day. Another day she sold  25 ribeye steak dinners and  13 grilled salmon dinners, totaling $581.85. How much did each type of dinner cost?

 

What are the steps to solve this?? Please help

Answer 1

We have 2 unknowns, so that means we need to have 2 equations to find them.

Let r=cost of a ribeye dinner

     s=cost of a salmon dinner

Eq. I:  13r+26s=597.19

Eq. II: 25r+13s=581.85

Use the elimination method to solve for r by multiplying Eq. II by 2 then subtracting Eq. I from Eq. II:

          2 X Eq. II=50r+26s=1163.70               Eq. II multiplied by 2

                       -(13r+26s=  597.19)              Eq. I

                          37r=566.51

                         r=566.51/37                        divide both sides by 37

                         r≅15.31

Use this value for r in either equation to find the value for s.

                        13r+26s=597.19                 Eq. I

                        13(15.31)+26s=597.19        substituting for r

                        199.03+26s=597.19

                        26s=597.19-199.03            subtract 199.03 from both sides

                        26s=398.16                       simplify

                        s=398.16/26                     divide both sides by 26

                        s≅15.31

Therefore, both dinners cost roughly $15.31.

Answer 2

I give you the system of equations, but I will NOT solve it for you because it will be tedious!

 

Let S = # of steaks and F = # of fish dinners

13S + 26F = 597.19

25S + 13F = 581.85

 

The simplest way to solve this system is with determinants to which you have probably not been exposed yet.

 

You can eliminate the F by multiplying the second equation by 2 and subtracting one equation from the other; this will probably be the next simplest method.

 

There is no trick here, just brute force arithmetic!