Need help solving systems of equations with 3 variables.

Question

Tickets for a concert cost $2 for children, $3 for students, and $4 for adults. Ticket sales totaled $522 and 177 people attended the concert. Twice as many students as adults attended. Find how many of each type of ticket were sold. Assume that everyone who bought a ticket attended the concert. 

Answer 1

Hi Cheryl

51 children tickets were sold

84 students tickets were sold

42 adult tickets were sold

Please see below

Your problem describes the following:

Ticket prices

$2 for children

$3 for students

$4 for adults

Total Attendance 177 people

Total Ticket sales 522 dollars

It also states that "Twice as many students as adults attended."

Let adults = x

Let students = y

Let children = z

So there are two totals to be applied to the data and one important substitution to keep in mind

"Twice as many students as adults attended."

So y = 2x we will use this later

Equation 1 for total attendance

x + y + z = 177

Equation 2 for total ticket sales

4x + 3y + 2z = 522

Eliminate z by first multiplying Equation 1 by negative 2

-2x -2y - 2z = -354

Combine this with Equation 2 to eliminate z

-2x - 2y - 2z = -354

4x + 3y + 2z = 522

This leaves

2x + y = 168

Remember "Twice as many students as adults attended."

That says y = 2x so

2x + y = 168

Can be expressed in terms of x only as

2x + 2x = 168

4x = 168

x = 168/4

x = 42

Since we've made use of our substitution

we can use it again in Equation 1

x + y + z = 177

Can be written in terms of x and z only since y = 2x

x + 2x + z = 177

3x + z = 177

We now know that x = 42 and we can use that to solve for z in

3x + z = 177

3(42) + z = 177

126 + z = 177

z = 177 -126

z =51

Since y = 2x then

2(42) = 84 = y

Lets check everything

x + y + z = 177

42 + 84 + 51 = 177

Next

4x + 3y + 2z = 522

4(42) + 3(84) + 51(2) = 522

168 + 252 + 102 = 522

I hope you find this useful if you have any questions send me an email.

Answer 2

Use elimination.

Let x = # of children, let y = # of students, let z = # of adults

The 3 equations are:

1) 2x + 3y + 4z = 522

2) x + y + z = 177

3) 0x + 1y - 2z = 0

Since the last equation does not have an "x", let's eliminate "x" from the first two equations by multiplying equation 2 by -2:

2A) -2x - 2y - 2z = -354

Then adding equations 1) and 2A:

2x + 3y + 4z = 522

-2x - 2y - 2z = -354

-------------------------

4) y + 2z = 168

Multiply equation 4 by -1:

4A) -y - 2z = -168

Add equation 3 and 4A:

y - 2z = 0

-y - 2z = -168

-4z = -168

z = 42

Using equation 3, plug in z = 42:

y - 2(42) = 0

y = 84

Using equation 2) plug in y and z:

x + 84 + 42 = 177

x = 51

So, the number of children is 51, the number of students is 84, the number of adults is 42