<p>Can someone help me with this problem? An amusement park charges $8 for an adults' ticket, and $6 for a children's ticket. On a certain day, a total of 150 tickets were sold for a total cost of $1020. How many more children's tickets were sold that day than adults' tickets?</p>
<p>Let x equal number of adult tickets sold.
Let y equal number of children tickets sold.</p>
<p>x + y = 150
8x + 6y = 1020</p>
<p>By solving for y in the first equation, we get y = -x + 150. </p>
<p>Substitute into the second equation to get 8x + 6(-x + 150) = 1020. </p>
<p>Distribute 6 to both terms in parentheses, combine like terms, and solve for x.</p>
<p>x = 60
y = 90</p>
<p>So, 30 more children’s tickets were sold than adult tickets.</p>
<p>Elf is absolutely correct, but I’d like to present an alternate method that uses logic and can, for some folks, be solved in your head in seconds rather than on paper, which might take a minute.</p>
<p>Note that the minimum that would be raised selling 150 tickets is $900 (150 x 6) and that adult tickets cost $2 more each.</p>
<p>The total raised was $120 more than $900, divided by $2 or 60, so 60 of the more expensive adult tickets were sold.</p>
<p>150 - 60 = 90 and 90 - 60 = 30 (your answer)</p>