Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ The box office at a theater is selling tickets for a series of rock concerts. So far, they have sold $94$ balcony tickets and $99$ general admission floor tickets for Friday's show, for a total of $\$7,660$ in receipts. For Saturday's show, $61$ balcony tickets and $59$ general admission floor tickets have been sold, equaling $\$4,824$ in receipts. How much does each ticket cost? $\newline$ A balcony seat ticket costs _, and a general admission floor ticket costs $\$$ _.

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Solve a system of equations using elimination: word problems

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.

\newline

The box office at a theater is selling tickets for a series of rock concerts. So far, they have sold

94

balcony tickets and

99

general admission floor tickets for Friday's show, for a total of

\$7,660

in receipts. For Saturday's show,

61

balcony tickets and

59

general admission floor tickets have been sold, equaling

\$4,824

in receipts. How much does each ticket cost?

\newline

A balcony seat ticket costs _____, and a general admission floor ticket costs

\$

_____.

Define Variables: Let's denote the price of a balcony seat ticket as $b$ and the price of a general admission floor ticket as $f$ . We are given that $94$ balcony tickets and $99$ general admission floor tickets for Friday's show have been sold for a total of $\$7,660$ . This gives us the equation $94b + 99f = 7660$ .
Equations for Friday's Show: For Saturday's show, $61$ balcony tickets and $59$ general admission floor tickets have been sold for a total of $\$$ $4$ , $824$ . This gives us the equation $61b + 59f = 4824$ .
Eliminate Variable $f$ : We now have a system of two equations. We need to eliminate one of the variables, $b$ or $f$ . We choose to eliminate $f$ because its coefficients are close in value, which might make the calculations simpler.
New Equations: To eliminate $f$ , we can multiply the second equation by $99$ and the first equation by $59$ , the coefficients of $f$ in the opposite equations. This gives us the new equations $6039b + 5841f = 476376$ and $5546b + 5831f = 451940$ .
Solve for $b$ : We now subtract the second new equation from the first new equation to eliminate $f$ . This gives us $493b = 24436$ .
Substitute $b$ into Equation: We divide both sides of the equation by $493$ to solve for $b$ . This gives us $b = \frac{24436}{493}$ , which simplifies to $b = 49.57$ . Since ticket prices are typically whole numbers, we can round this to $b = 50$ .
Solve for $f$ : We substitute $b = 50$ into the first original equation and solve for $f$ . This gives us $94 \times 50 + 99f = 7660$ , which simplifies to $4700 + 99f = 7660$ .
Solve for f: We substitute $b = 50$ into the first original equation and solve for $f$ . This gives us $94 \times 50 + 99f = 7660$ , which simplifies to $4700 + 99f = 7660$ . We subtract $4700$ from both sides of the equation to solve for $f$ . This gives us $99f = 2960$ , and dividing both sides by $99$ gives us $f = \frac{2960}{99}$ , which simplifies to $f = 29.90$ . Rounding to the nearest whole number, we get $f$ $0$ .