Tickets for a play cost 2 pounds for a child, and 4 pounds for an adult. one adult brought 4 children with him and the remaining adults each brought 2 children with them. The total ticket sales were 60 pounds. how many adults and children were present in that play?

Solve using augmented matrix.

Respuesta :

Answer:

Number of adults = 7

Number of children = 16

Step-by-step explanation:

Tickets for a play cost 2 pounds for a child, and 4 pounds for an adult.

Let x number of adults and y number of children.

1 child ticket cost = 2 pound

y children ticket cost = 2y pound

1 adult ticket cost = 4 pound

x adults ticket cost = 4x pound

Total number of ticket sales were 60 pounds

Therefore, 4x + 2y = 60  ------------- (1)

One adult brought 4 children with him and the remaining adults each brought 2 children with them.

Remaining number of adult whose brought 2 children = x-1

Number children = 2(x-1)

Total number of children = 2(x-1)+4

Therefore, y=2x+2 ---------------------(2)

System of equation,

 2x + y = 30

-2x + y = 2

Using augmented matrix to solve system of equation.

[tex]\begin{bmatrix}2&1&\ |30\\-2&1&|2\end{bmatrix}\\\\R_2\rightarrow R_2+R_1\\\\\begin{bmatrix}2&1& |30\\0&2&|32\end{bmatrix}\\\\R_2\rightarrow\dfrac{1}{2}R_2\\[/tex]

[tex]\begin{bmatrix}2&1&\ |30\\0&1&|16\end{bmatrix}\\\\R_1\rightarrow R_1-R_2\\\\\begin{bmatrix}2&0&\ |14\\0&1&|16\end{bmatrix}\\\\\\[/tex]

[tex]R_1\rightarrow \dfrac{1}{2}R_1\\\\\begin{bmatrix}1&0&|7\\0&1&|16\end{bmatrix}\\\\[/tex]

Now, we find the value of variable.

[tex]x=7\text{ and }y=16[/tex]

Hence, Number of adults are 7 and Number of children are 16.