Question

Last weekend, the Strong family went out to dinner at a local steakhouse. They ordered 3 filet mignon dinners and 2 rib eye dinners for a total of $122.40. Two weekends later, they went out to dinner again at the same restaurant, and ordered 2 filet mignon dinners and 3 rib eye dinners for a total of $114. 85. How much more was a filet mignon dinner than a rib eye dinner, to the nearest cent?

Answer 1

$Let\: cost \: of \: each \: filet \:mignon \\dinner = \ x$

$and \: cost \: of \: each\: rib \: eye\: dinner = \ y$

$Total \: cost= \ 122.40$

$\implies 3x + 2y = \ 122.40 \: --(1)$

/* After two weekends later */

$Second \: times\: Total \:cost= \ 114.85$

$\implies 2x + 3y = \ 114.85 \: --(2)$

/* Multiplying equation (1) by 3 and equation (2) by 2 , we get */

$6x + 4y = 244.80\: --(3)$

$6x + 9y = 434.55 \: ---(4)$

/* Subtract equation (3) from equation (4), we get */

$5y = 189.75$

$\implies y = \frac{189.75}{5}$

$\implies y = \ 37.95\: --(5)$

/* Put y value in equation (1) , we get */

$3x + 2 \times 37.95 = 122.40$

$\implies 3x + 75.90 = 122.40$

$\implies 3x = - 75.90 + 122.40$

$\implies 3x = 46.5$

$\implies x = \frac{46.5 }{3}$

$\implies x = \ 15.5$

Therefore.,

$\red{ cost \: of \: each \: filet \:mignon \:dinner}\green { = \ 15.5}$

$\red{cost \: of \: each\: rib \: eye\: dinner }\green{= \ 37.95}$

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