Question 1094912
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A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30. 
The team has sold 3235 tickets overall. It has sold 182 more $20 tickets than $10 tickets. 
The total sales are $62,170. How many tickets of each kind have been sold? 
How many $10 dollar tickets were sold?
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<pre>
Let T be the number of $10-tickets sold.

Then the number of $20-tickets is (T+182), according to the condition,

and the number of $30-tickets is (3235 - T - (T+182)) = 3053-2T.


Hence, your "money equation" is 

10T + 20*(T+182) + 30*(3053-2T) = 62170.


Simplify and solve for T:

(10T + 20T - 60T) = 62170 - 20*182 - 30*3053,

-30T = -33060  =====>  T = {{{(-33060)/(-30)}}} = 1102.


Thus,  1102 of the $10-tickets, 1102 + 182 = 1284 of the $20-tickets  and  3235-1102- 1284 = 849 of the $30-tickets were sold.


<U>Check</U>.  $10*1102 + $20*1284 + $30*849 = 62170.   ! Correct !
</pre>


<U>Lesson to learn from this solution</U>


<pre>
    This problem is for one unknown, not for three.

    Your major task, when you start solve it, is to select the key unknown,
    and then to express other quantities via that unknown.

    Then write the basic governing equation and solve it accurately.
</pre>


The way @josgarithmetic tries to teach you &nbsp;<U>is NOT THE WAY for solving such problems</U>.


It is the way to &nbsp;<U>NOWHERE</U>.



To see more similar solved problems, &nbsp;look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/equations/More-solved-problems-on-a-single-linear-equation.lesson>More solved word problems on a single linear equation</A>

in this site.



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Dear tutor @greenestamps !


This problem is designed and assigned to 6-th or 7-th grade students, who did not study yet systems of equations and 
DEFINITELY did not study systems of equations in three unknowns.