The cost of each ribeye steak dinner is $ 29 and cost of each grilled salmon dinner is $ 20
Solution:
Let "a" be the cost of each ribeye steak dinners
Let "b" be the cost of each grilled salmon dinners
A waitress sold 11 ribeye steak dinners and 12 grilled salmon dinners, totaling $560.63 on a particular day
Thus we frame a equation as:
11 x cost of each ribeye steak dinners + 12 x cost of each grilled salmon dinners = 560.63
![11 * a + 12 * b = 560.63](https://img.qammunity.org/2021/formulas/mathematics/middle-school/xadvw8kicrmanyq6je3c9enlus4tuztsq7.png)
11a + 12b = 560.63 -------- eqn 1
Another day she sold 16 ribeye steak dinners and 6 grilled salmon dinners, totaling $584.46
Thus we frame a equation as:
16 x cost of each ribeye steak dinners + 6 x cost of each grilled salmon dinners = 584.46
![16 * a + 6 * b = 584.46](https://img.qammunity.org/2021/formulas/mathematics/middle-school/nrjrdbw30s84plh7pbjalh0m9k3sx53h0f.png)
16a + 6b = 584.46 ------------ eqn 2
Let us solve eqn 1 and eqn 2
Multiply eqn 2 by 2
32a + 12b = 1168.92 ------- eqn 3
Subtract eqn 1 from eqn 3
32a + 12b = 1168.92
11a + 12b = 560.63
( - ) ----------------
21a = 1168.92 - 560.63
21a = 608.29
Divide both sides by 21
![a = 28.96 \approx 29](https://img.qammunity.org/2021/formulas/mathematics/middle-school/8p2fzqsyikl8zc9yoave0st73rw1wvrdxf.png)
Substitute a = 29 in eqn 1
11(29) + 12b = 560.63
319 + 12b = 560.63
12b = 560.63 - 319
12b = 214.63
Divide both sides by 12
![b = 20.13 \approx 20](https://img.qammunity.org/2021/formulas/mathematics/middle-school/yogsq7tanempww3djr19wua8i3zrts1wfh.png)
Thus cost of each ribeye steak dinner is $ 29 and cost of each grilled salmon dinner is $ 20