Astadium has 50,000 seats. seats sell for $42 in section a, $36 in section b, and $30 in section c. the number of seats in section a equals the total number of seats in sections b and c. suppose the stadium takes in $1,888,800 from each sold out event. how many seats does each section hold?

Section A has 25,000 seats.

Section B has 14,800 seats.

Section C has 10,200 seats.

Step-by-step explanation:

Let the seats in section A be = x

Let the seats in section B be = y

Let the seats in section C be = z

The equations forms as follows:

          .....(1)

       .......(2)

......(3)

Substituting the value of x in (1) to get equation in two terms.

=>

taking out 2 common, we get;

.........(4)

And substituting the value of x in (2), we get

=>

=>

Taking out 2 common, we get;

     ........(5)

Multiplying (4) by 39 and subtracting (5) from (4), we get

We get z = 10200

And

We get y = 14800

Also

We get x = 25000

Therefore,

Section A has 25,000 seats.

Section B has 14,800 seats.

Section C has 10,200 seats.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3