Question

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

Answer 1

R=cost of rib-eye dinner; S=cost of salmon dinner

24R+10S=$580.66

10S=$580.66-24R

20S=$1161.32-48R Use this to substitute for 20S below

17R+20S=$590.81 . Substitute for 20S.

17R+($1161.32-48R)=$590.81

-31R=-$570.51

R=$18.40

ANSWER 1: Rib-eye dinner costs $18.40.

Explanation:

24R+10S=$580.66

24($18.40)+10S=$580.66

$441.60+10S=$580.66 441.69

10S=$139.06

S=$13.91

ANSWER 2: A salmon dinner costs $13.91.

CHECK:

17R+20S=$590.81

17($18.40)+20($13.91)=$590.81

Answer 2

• ribeye steak: $24.37
• grilled salmon: $9.30

Explanation:

If we let r and g represent the costs of a ribeye steak dinner and grilled salmon dinner, respectively, the sales figures give rise to two equations.

17r +15g = 553.74

22r +5g = 582.65

If we multiply the second equation by 3 and subtract the first, we have ...

3(22r +5g) -(17r +15g) = 3(582.65) -(553.74)

49r = 1194.21

r = 1194.21/49 ≈ 24.37

g = (553.74 -17r)/15 ≈ 9.30

The steak dinner costs $24.37; the salmon dinner costs $9.30.

_____

Check

17(24.37) +15(9.30) = 553.79 . . . not 553.74

22(24.37) +5(9.30) = 582.64 . . . not 582.65

There is no exact solution to this problem.