A school spent [tex]$\$[/tex]150[tex]$ on advertising for a breakfast fundraiser. Each plate of food was sold for $[/tex]\[tex]$8.00$[/tex] but cost the school [tex]$\$[/tex]2.00[tex]$ to prepare. After all expenses were paid, the school raised $[/tex]\[tex]$2,400$[/tex] at the fundraiser. Which equation can be used to find [tex]$x$[/tex], the number of plates that were sold?

A. [tex]$6x - 150 = 2,400$[/tex]
B. [tex]$6x + 150 = 2,400$[/tex]
C. [tex]$10x - 150 = 2,400$[/tex]
D. [tex]$10x + 150 = 2,400$[/tex]



Answer :

Let's carefully analyze the problem step-by-step to find the correct equation to determine [tex]\( x \)[/tex], the number of plates sold:

1. Determine Net Profit per Plate:
- Each plate is sold for [tex]\( \$8.00 \)[/tex].
- The cost to prepare each plate is [tex]\( \$2.00 \)[/tex].
- Therefore, the net profit per plate is:
[tex]\[ 8 - 2 = \$6.00 \][/tex]

2. Establish the Fundraising Goal:
- After covering all expenses, the school raised [tex]\( \$2,400 \)[/tex].

3. Consider the Advertising Cost:
- The school spent [tex]\( \$150 \)[/tex] on advertising.

4. Formulate the Net Profit Equation:
- Let [tex]\( x \)[/tex] be the number of plates sold.
- The total profit from selling [tex]\( x \)[/tex] plates at a net gain of [tex]\( \$6 \)[/tex] per plate would be:
[tex]\[ 6x \][/tex]
- However, the school also needs to subtract the [tex]\( \$150 \)[/tex] advertising cost from this profit:
[tex]\[ 6x - 150 \][/tex]
- This net profit after the advertising cost must equal the total amount raised, [tex]\( \$2,400 \)[/tex].

5. Set Up the Equation:
- So, the equation to find the number of plates sold, [tex]\( x \)[/tex], will be:
[tex]\[ 6x - 150 = 2400 \][/tex]

After analyzing each step logically, we conclude that the correct equation is:

[tex]\[ 6x - 150 = 2400 \][/tex]

Therefore, the equation that can be used to find [tex]\( x \)[/tex], the number of plates sold, is:

[tex]\[ 6x - 150 = 2400 \][/tex]

### Correct Answer:
[tex]\[ \boxed{6x - 150 = 2400} \][/tex]