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.

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.

Now, we find the value of variable.

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

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3