# solving systems with three variables Algebra II help?.?

i don't need it worked out or anything. just set up the system.

A Stadium has 49,000 seats. Seats sell for \$25 in section A, \$20 in section B, and \$15 in section C. The number of seat in section A equals the total number of seats in Sections B and Section C. Suppose the stadium takes in \$1,052,000 from each sold-out event. How many seats does each section hold?

Update:

thanx you so much! i really wish i understood this Alg stuff. i'm better with geomerty.

Relevance

Describing revenue:

25A+20B+15C=1,052,000

Describing available seats:

A+B+C=49,000

Describing the relationship between number of seats:

A= B+C

or...

B+C-A=0

So those are your three equations.

• Let:

a = seats in section A

b = seats in section B

c = seats in section C

a = b + c (1)

25a + 20b + 15c = 1052000 (2)

a + b + c = 49000 (3)

Substitute (1) into (3)

a + a = 49000

2a = 49000

Therefore a = 24500

Substitute a into (1) and (2)

b + c = 24500 (4)

25(24500) + 20b + 15c = 1,052,000

612500 + 20b + 15c = 1,052,000

20b + 15c = 439,500 (5)

I have now created a system of two equations which will allow me to find b and c

Make b the subject in (4)

b = 24500 - c

Substitute b into 5

20(24500 - c) + 15c = 439,500

490,000 - 20c + 15c = 439,500

-5c = -50,500

c = 10,100

Therefore

b + c = 24500

b + 10,100 = 24500

b = 14400

• permit: 2x - 3y + z = 6 (First Equation) x - y + z = 2 (2nd Equation) x - y - 2z = 8 (0.33 Equation) Then; Subtract equation 2 from equation 3 subsequently, (x - y - 2z = 8) - (x - y + z = 2) ______________ -3z = 6 z = -2 (answer) Subtract equation 2 from equation a million (2x - 3y + z = 6) - (x - y + z = 2) ______________ x - 2y = 4 x = 4 + 2y (equation 4) evaluate equation 2...then replace the values of z and x. x - y + z = 2 4 + 2y - y + (-2) = 2 y = 0 (answer) evaluate equation 4..then replace the cost of y x = 4 + 2(0) x = 4 (answer) good luck!

• Here it is...

A + B + C = 49,000

25A + 20B +15C = 1,052,000

A = B + C