i need help with algebra

Question

The Langham Creek Choir is selling tickets for their mid-year concert. Rachel paid $52 for two adult, two student, and one child tickets. RJ paid $56 for one adult, two student, and three child tickets. Hong paid $44 for one adult and fou child tickets. What is the price for each type of tickets?

Answer 1

Because we're given 3 descriptions for 3 different ticket situations, we know this is a system of equations. I find it easiest to write the equations if I first organize the information in a table format. Once you have your equations, it's a matter of choosing either substitution or elimination. I chose elimination for this problem; I explain why, and show the rest of the steps, in the video. Have a great day!

Answer 2

2a+ 2s+ c= 52 (i)

a+ 2s+ 3c= 56 (ii)

a+ 4c= 44 (iii)

Subtracting (ii) from (i)

2a+ 2s + c= 52

-a- 2s - 3c=-56

a - 2c = -4 (iv)

subtracting (iii) from (iv)

a- 2c= -4

-a-4c=-44

-6c=-48

c=8 (v)

Plugging in c=8 in (iv)

a - 2c = -4

a - 2(8) = -4

a - 16 = -4

a = -4 +16

a= 12

Plugging in a = 12 and c = 8 in (i)

2(12) + 2s + 8 = 52

32 + 2s= 52

2s = 52 - 32

2s = 20

s = 10

Th price of the ticket for a child is $8 , student $10 and adult $12 .

Answer 3

Suppose a, s, c are the price of the 3 tickets

Then

2a + 2s + c = 52 (1)

a + 2s + 3c = 56 (2)

a + 4c = 44 (3)

From (3), a = 44 - 4c

Sub into (1) : 88 -8c + 2s + c = 52

2s - 7c = -36 (4)

Sub into (2) : 44 -4c + 2s + 3c = 56

2s -c = 12 (5)

Subtract (5) from (4) : 2s - 7c - (2s -c) = -36 - 12

-6c = -48

c = 8

2s - c = 12

2s - 8 = 12

2s = 20

s = 10

a = 44 - 4c = 44 - 32 = 12

a = 12, s = 10, c = 8

Answer 4

2A+2S +C = 52

A+2S+3C = 56

A+4C =44

subtract 2nd equation from 1st to get

A -2C =-4

subtract from 3rd equation

6C=48

C =48/6= $8 per Child

A=44-32= $12 per Adult

S=$10 per Student