Question

U JUVVUJI Call20.CUIT - Topic 04: Topic Test Copy 1 Due 09/11/20 11:59 A computer company can choose between two deals on season box seats for basketball games. Deal A costs $2300 plus $45 for each guest at a game. Deal B costs $5000 plus $15 for each guest at a game. How many guests would the computer company need to bring throughout the season for the charges to be the same?

Answer 1

Answer:

The computer company would need 90 guests throughout the season for the charges to be the same.

Explanation:

There are two given deals on season box seats for basketball games.

Let x represent the number of guest guest

Deal A;

costs $2300 plus $45 for each guest at a game

$2300+45x$

Deal B;

costs $5000 plus $15 for each guest at a game

$5000+15x$

For the charges to be the same, Deal A = Deal B;

$2300+45x=5000+15x$

Let's solve for x;

subtract 15x from both sides

$\begin{gathered} 2300+45x-15x=5000+15x-15x \\ 2300+30x=5000 \end{gathered}$

subtract 2300 from both sides;

$\begin{gathered} 2300-2300+30x=5000-2300 \\ 30x=2700 \end{gathered}$

divide both sides by 30;

$\begin{gathered} \frac{30x}{30}=\frac{2700}{30} \\ x=90 \end{gathered}$

Therefore, the computer company would need 90 guests throughout the season for the charges to be the same.