Solved on Feb 13, 2024

Sell 21 tickets for a school play, with adult tickets at $6 and child tickets at$ 4. Solve for the number of adult and child tickets sold.
$a$ = number of adult tickets, $c$ = number of child tickets $a + c = 21$ $6a + 4c = 104$

STEP 1

Assumptions
1. Let $a$ be the number of adult tickets sold.
2. Let $c$ be the number of child tickets sold.
3. The total revenue from selling tickets is $104.<br />4. The price of an adult ticket is$ 6.
5. The price of a child ticket is $4.
6. The total number of tickets sold is 21.

STEP 2

We are given the system of equations:
$\begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array}$
We will solve this system to find the values of $a$ and $c$ .

STEP 3

First, we will express $c$ in terms of $a$ from the first equation.
$c = 21 - a$

STEP 4

Now, we will substitute $c = 21 - a$ into the second equation to find $a$ .
$6a + 4(21 - a) = 104$

STEP 5

Distribute the $4$ in the second equation.
$6a + 84 - 4a = 104$

STEP 6

Combine like terms in the second equation.
$2a + 84 = 104$

STEP 7

Subtract $84$ from both sides of the equation to isolate $2a$ .
$2a = 104 - 84$

STEP 8

Calculate the difference to find the value of $2a$ .
$2a = 20$

STEP 9

Divide both sides of the equation by $2$ to find the value of $a$ .
$a = \frac{20}{2}$

STEP 10

Calculate the value of $a$ .
$a = 10$

STEP 11

Now that we have the value of $a$ , we can find the value of $c$ using the equation $c = 21 - a$ .
$c = 21 - 10$

STEP 12

Calculate the value of $c$ .
$c = 11$

STEP 13

We have found that the number of adult tickets sold is $a = 10$ and the number of child tickets sold is $c = 11$ . However, we need to check the other options provided to ensure there are no other solutions that also satisfy the system of equations.

STEP 14

Check the option "# of adult tickets $=11$ , # of child tickets $=10$ ".
$\begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array}$
$\begin{array}{c} 11+10=21 \\ 6(11)+4(10)=104 \end{array}$

STEP 15

Verify the first equation with the given option.
$11 + 10 = 21$
$21 = 21$
The first equation is satisfied.

STEP 16

Verify the second equation with the given option.
$6(11) + 4(10) = 104$
$66 + 40 = 104$
$106 = 104$
The second equation is not satisfied, so this option is incorrect.

STEP 17

Check the option "# of adult tickets $=12$ , # of child tickets $=9$ ".
$\begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array}$
$\begin{array}{c} 12+9=21 \\ 6(12)+4(9)=104 \end{array}$

STEP 18

Verify the first equation with the given option.
$12 + 9 = 21$
$21 = 21$
The first equation is satisfied.

STEP 19

Verify the second equation with the given option.
$6(12) + 4(9) = 104$
$72 + 36 = 104$
$108 = 104$
The second equation is not satisfied, so this option is incorrect.

STEP 20

Check the option "# of adult tickets $=10$ , # of child tickets $=11$ ".
$\begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array}$
$\begin{array}{c} 10+11=21 \\ 6(10)+4(11)=104 \end{array}$

STEP 21

Verify the first equation with the given option.
$10 + 11 = 21$
$21 = 21$
The first equation is satisfied.

STEP 22

Verify the second equation with the given option.
$6(10) + 4(11) = 104$
$60 + 44 = 104$
$104 = 104$
The second equation is satisfied, so this option is correct.

STEP 23

Since we have found a solution that satisfies both equations, we can conclude that the number of adult tickets sold is 10 and the number of child tickets sold is 11.
The solution is: \# of adult tickets $= 10$ \# of child tickets $= 11$

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Sell 21 tickets for a school play, with adult tickets at 6andchildticketsat6 and child tickets at 6andchildticketsat4. Solve for the number of adult and child tickets sold.aaa = number of adult tickets, ccc = number of child tickets a+c=21a + c = 21a+c=21 6a+4c=1046a + 4c = 1046a+4c=104

STEP 1

STEP 2

We are given the system of equations:a+c=216a+4c=104 \begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array} a+c=216a+4c=104​We will solve this system to find the values of aaa and ccc.

STEP 3

First, we will express ccc in terms of aaa from the first equation.c=21−a c = 21 - a c=21−a

STEP 4

Now, we will substitute c=21−ac = 21 - ac=21−a into the second equation to find aaa.6a+4(21−a)=104 6a + 4(21 - a) = 104 6a+4(21−a)=104

STEP 5

Distribute the 444 in the second equation.6a+84−4a=104 6a + 84 - 4a = 104 6a+84−4a=104

STEP 6

Combine like terms in the second equation.2a+84=104 2a + 84 = 104 2a+84=104

STEP 7

Subtract 848484 from both sides of the equation to isolate 2a2a2a.2a=104−84 2a = 104 - 84 2a=104−84

STEP 8

Calculate the difference to find the value of 2a2a2a.2a=20 2a = 20 2a=20

STEP 9

Divide both sides of the equation by 222 to find the value of aaa.a=202 a = \frac{20}{2} a=220​

STEP 10

Calculate the value of aaa.a=10 a = 10 a=10

STEP 11

Now that we have the value of aaa, we can find the value of ccc using the equation c=21−ac = 21 - ac=21−a.c=21−10 c = 21 - 10 c=21−10

STEP 12

Calculate the value of ccc.c=11 c = 11 c=11

STEP 13

We have found that the number of adult tickets sold is a=10a = 10a=10 and the number of child tickets sold is c=11c = 11c=11. However, we need to check the other options provided to ensure there are no other solutions that also satisfy the system of equations.

STEP 14

Check the option "# of adult tickets =11=11=11, # of child tickets =10=10=10".a+c=216a+4c=104 \begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array} a+c=216a+4c=104​11+10=216(11)+4(10)=104 \begin{array}{c} 11+10=21 \\ 6(11)+4(10)=104 \end{array} 11+10=216(11)+4(10)=104​

STEP 15

Verify the first equation with the given option.11+10=21 11 + 10 = 21 11+10=2121=21 21 = 21 21=21The first equation is satisfied.

STEP 16

Verify the second equation with the given option.6(11)+4(10)=104 6(11) + 4(10) = 104 6(11)+4(10)=10466+40=104 66 + 40 = 104 66+40=104106=104 106 = 104 106=104The second equation is not satisfied, so this option is incorrect.

STEP 17

Check the option "# of adult tickets =12=12=12, # of child tickets =9=9=9".a+c=216a+4c=104 \begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array} a+c=216a+4c=104​12+9=216(12)+4(9)=104 \begin{array}{c} 12+9=21 \\ 6(12)+4(9)=104 \end{array} 12+9=216(12)+4(9)=104​

STEP 18

Verify the first equation with the given option.12+9=21 12 + 9 = 21 12+9=2121=21 21 = 21 21=21The first equation is satisfied.

STEP 19

Verify the second equation with the given option.6(12)+4(9)=104 6(12) + 4(9) = 104 6(12)+4(9)=10472+36=104 72 + 36 = 104 72+36=104108=104 108 = 104 108=104The second equation is not satisfied, so this option is incorrect.

STEP 20

Check the option "# of adult tickets =10=10=10, # of child tickets =11=11=11".a+c=216a+4c=104 \begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array} a+c=216a+4c=104​10+11=216(10)+4(11)=104 \begin{array}{c} 10+11=21 \\ 6(10)+4(11)=104 \end{array} 10+11=216(10)+4(11)=104​

STEP 21

Verify the first equation with the given option.10+11=21 10 + 11 = 21 10+11=2121=21 21 = 21 21=21The first equation is satisfied.

STEP 22

Verify the second equation with the given option.6(10)+4(11)=104 6(10) + 4(11) = 104 6(10)+4(11)=10460+44=104 60 + 44 = 104 60+44=104104=104 104 = 104 104=104The second equation is satisfied, so this option is correct.

STEP 23

Since we have found a solution that satisfies both equations, we can conclude that the number of adult tickets sold is 10 and the number of child tickets sold is 11.The solution is: \# of adult tickets =10= 10=10 \# of child tickets =11= 11=11

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Sell 21 tickets for a school play, with adult tickets at $6 and child tickets at$ 4. Solve for the number of adult and child tickets sold.
$a$ = number of adult tickets, $c$ = number of child tickets $a + c = 21$ $6a + 4c = 104$

We are given the system of equations:
$\begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array}$
We will solve this system to find the values of $a$ and $c$ .

First, we will express $c$ in terms of $a$ from the first equation.
$c = 21 - a$

Now, we will substitute $c = 21 - a$ into the second equation to find $a$ .
$6a + 4(21 - a) = 104$

Distribute the $4$ in the second equation.
$6a + 84 - 4a = 104$

Combine like terms in the second equation.
$2a + 84 = 104$

Subtract $84$ from both sides of the equation to isolate $2a$ .
$2a = 104 - 84$

Calculate the difference to find the value of $2a$ .
$2a = 20$

Divide both sides of the equation by $2$ to find the value of $a$ .
$a = \frac{20}{2}$

Calculate the value of $a$ .
$a = 10$

Now that we have the value of $a$ , we can find the value of $c$ using the equation $c = 21 - a$ .
$c = 21 - 10$

Calculate the value of $c$ .
$c = 11$

We have found that the number of adult tickets sold is $a = 10$ and the number of child tickets sold is $c = 11$ . However, we need to check the other options provided to ensure there are no other solutions that also satisfy the system of equations.

Check the option "# of adult tickets $=11$ , # of child tickets $=10$ ".
$\begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array}$
$\begin{array}{c} 11+10=21 \\ 6(11)+4(10)=104 \end{array}$

Verify the first equation with the given option.
$11 + 10 = 21$
$21 = 21$
The first equation is satisfied.

Verify the second equation with the given option.
$6(11) + 4(10) = 104$
$66 + 40 = 104$
$106 = 104$
The second equation is not satisfied, so this option is incorrect.

Check the option "# of adult tickets $=12$ , # of child tickets $=9$ ".
$\begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array}$
$\begin{array}{c} 12+9=21 \\ 6(12)+4(9)=104 \end{array}$

Verify the first equation with the given option.
$12 + 9 = 21$
$21 = 21$
The first equation is satisfied.

Verify the second equation with the given option.
$6(12) + 4(9) = 104$
$72 + 36 = 104$
$108 = 104$
The second equation is not satisfied, so this option is incorrect.

Check the option "# of adult tickets $=10$ , # of child tickets $=11$ ".
$\begin{array}{c} a+c=21 \\ 6a+4c=104 \end{array}$
$\begin{array}{c} 10+11=21 \\ 6(10)+4(11)=104 \end{array}$

Verify the first equation with the given option.
$10 + 11 = 21$
$21 = 21$
The first equation is satisfied.

Verify the second equation with the given option.
$6(10) + 4(11) = 104$
$60 + 44 = 104$
$104 = 104$
The second equation is satisfied, so this option is correct.

Since we have found a solution that satisfies both equations, we can conclude that the number of adult tickets sold is 10 and the number of child tickets sold is 11.
The solution is: \# of adult tickets $= 10$ \# of child tickets $= 11$