Question 1122922
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<U>One line solution</U>



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;# of adult tickets &nbsp;= &nbsp;&nbsp;{{{(2430-600*3)/(10-3)}}} &nbsp;= &nbsp;90.   &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(*)


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;# of children tickets = the rest 600-90 = 510.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>Check</U>.  &nbsp;&nbsp;&nbsp;&nbsp;3*510 + 10*90 = 2430  dollars.    &nbsp;&nbsp;&nbsp;&nbsp;! &nbsp;Correct &nbsp;!



<U>Explanation</U>


<pre> 
Let assume for a minute that all 600 visitors were children.


Then the revenue would be 600*3 = 1800 dollars, making the shortage of  2430 - 600*3 dollars, comparing with the real (given) revenue.


It is because we assumed that all tickets were at 3 dollars.


Now we should replace back some number of $3 tickets by $10 tickets.


At each such replacement, we diminish the shortage by 10-3 = 7 dollars.


So, the number of replacements is expressed exactly by the formula (*), which I used in my one line solution above.


And the number of replacements is nothing else as the number of adult tickets.
</pre>

Solved.


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An alternative way of solution is to reduce the problem to the system of two equations with two unknowns

or to a single equation for one unknown.


Independently of the method, you will get the same result.


Interesting, that if you apply the determinant method of solution to the system of two equations, &nbsp;<U>you will get &nbsp;THE &nbsp;SAME &nbsp;FORMULA &nbsp;(*) &nbsp;&nbsp;(!)</U>


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See the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/coordinate/lessons/Using-systems-of-equations-to-solve-problems-on-Tickets.lesson>Using systems of equations to solve problems on tickets</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/coordinate/lessons/Three-methods-for-solving-standard-typical-problem-on-tickets.lesson>Three methods for solving standard (typical) problems on tickets</A>

in this site.