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Can someone please help me. I'm having a hard time answering this question.

At a country fair, adults’ tickets sold for £5.50, senior citizens’ tickets sold for £4.00, and children’s tickets sold for £1.50. On the opening night, the number of children’s and seniors’ tickets sold was 30 more than the number of adults’ tickets sold. The number of senior citizens’ tickets sold was 6 more than four times the number of children’s tickets. Total receipts from the ticket sales were £14967. How many of each type of tickets were sold?

Yajat Shamji
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Andrue D.
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    How far have you gotten so far? I'm asking because if you have already done the majority of the work, there's no need for us to do all that again. – Matti P. Jan 26 '23 at 11:02
  • I think what I did was wrong im not that confident can someone help me so that I can also use it as my basis for the other word problems that I have – Andrue D. Jan 26 '23 at 11:04
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    Can you show us what you have done so far? It could be that it's (almost) correct! – Matti P. Jan 26 '23 at 11:06
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    I would start by defining $A$ = number of adult tickets sold; $S$ = number of senior tickets sold and $C$ = number of children's tickets sold. With these, we can form a few equations. Have you done something similar to this? – Matti P. Jan 26 '23 at 11:13
  • yes i did that too – Andrue D. Jan 26 '23 at 11:15
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    Excellent, what equations did you write? – Matti P. Jan 26 '23 at 11:15
  • after defining the given what i did was – Andrue D. Jan 26 '23 at 11:18
  • C+S = A + 3= S= 4C + 6 – Andrue D. Jan 26 '23 at 11:18
  • THEN A+ S + C = 14967 – Andrue D. Jan 26 '23 at 11:19
  • You're quite close, you just have to be a bit careful with the equations. Indeed "the number of children’s and seniors’ tickets sold was 30 more than the number of adults’ tickets sold" translates to $$C + S = A +30 $$ and "The number of senior citizens’ tickets sold was 6 more than four times the number of children’s tickets." translates to $$ S = 4C+6 $$ You made a small notation error here. These are two different equations and you cannot say that $A+30 = S$ or something similar. – Matti P. Jan 26 '23 at 11:22
  • I just mistyped it is c+s = a+30 – Andrue D. Jan 26 '23 at 11:22
  • And as for the last equation with the total ticket sales, note that the dollar amount is the sum of the ticket numbers multiplied by their individual prices. So just add the prices per tickets as factors and you're all set :) – Matti P. Jan 26 '23 at 11:23
  • then what I did is substituting th evalus – Andrue D. Jan 26 '23 at 11:24
  • @MattiP. last question how can I identify some of the values of the missing terms by just substituting it then solving? – Andrue D. Jan 26 '23 at 11:26
  • After putting the ticket prices in, and moving the terms around a bit, you have the equations $$ \left{ \begin{array}{cl} -A + S + C &= 30 \ S - 4C &= 6 \ 5.50~A+4.0~S+1.50~C &= 14~967 \end{array} \right. $$ Do you know how to continue from here? – Matti P. Jan 26 '23 at 11:28
  • no im a bit confused can you help me all through out the equation? – Andrue D. Jan 26 '23 at 11:29
  • First of all, is it clear to you how we got the equations in my last comment? Now we have a system of linear equations that we have to solve. – Matti P. Jan 26 '23 at 11:30
  • yes it is clear – Andrue D. Jan 26 '23 at 11:31

1 Answers1

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After putting the ticket prices in, and moving the terms around a bit, you have the equations $$ \left\{ \begin{array}{ccclr} -A + &S &+ C &= 30 & (1)\\ &S &- 4C &= 6 &(2)\\ 5.50~A&+4.0~S&+1.50~C &= 14~967 & (3) \end{array} \right. $$ Now we have to use elimination to solve for the three variables. We can multiply the row (2) by $-1$ and add it to (1) to get $$\tag{4} -A + 5C = 24 $$ Another way to get rid of the $S$ is to multiply row (1) by 4 and then subtract row (3) from it to get $$ \tag{5} 9.5 A -2.5C = 14~847 $$ If you multiply Equation (4) by 9.5 and add that to equation (5), you get $$ 45 C = 15~075 \qquad \Rightarrow \qquad C = 335 $$ You can use this result in Equation (4) to solve for $A$. Can you continue from here?

Matti P.
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