What is the price of a child's ticket and a senior citizen ticket?

Question

The circus is coming to town. Sue is selling tickets. The first day she sold 4 senior citizen tickets and 2 children's tickets for a total of $80. The second day she received $145 for 5 senior citizen tickets and 7 children's tickets. What is the price of a child's ticket and a senior citizen ticket?

Answer 1

Hello, Jeffrey

 

Lets set the problem up.

 

Let s represent the senior tickets

Let c represent the children tickets

 

Translating into algebra: She sold 4 senior tickets is 4s, 5 senior tickets 5s               

                                    She sold 2 children's ticket 2c, 7 children's tickets 7c

 

We are almost there, Jeff.  But next, we need to set up the system of linear equations from the information provided.

 

4s+2c= 80

5s+7c=145

                         

Now that we have the equations, we need to decide which method is the easiest to solve for x and y.  In this case, in my opinion, the elimination method is the easiest. And probably getting rid of the s is the best way to approach this problem.  Again, this is my opinion.  I am going to eliminate the s's by multiplying the first equation by -5 and the second equation by 4.  If I do that, I will end up with -20s and 20s.  These two will cancel each other out.  Ok, lets do it.

 

-5(4s+2c=80)= -20s-10c= -400

4(5s+7c=145)= 20s+28c= 580

                              18c=180

                                 c=10

So, children's tickets price is 10. Next, I am going to plug in 10 for c into the first original equation.  Why?? The numbers are smaller.. :).

 

4s+2c= 80

4s+2(10)=80

4s+20=80

    -20   -20

     4s=60

      s=15

The price for the senior citizen ticket is 15.

 

So, it is 10 for the children's tickets and 15 for the senior's tickets..

 

Ok, Jeffrey.. Glad to help.. and don't fall behind.. And please rate my answer.

 

D. Y. Taylor

 

 

          

Answer 2

Hello, Jeffrey

the first day she sold 4 senior citizen tickets and 2 children's tickets for a total of $80. 

The second day she received $145 for 5 senior citizen tickets and 7 children's tickets.

 

these two sentences are the most important.

you can suppose the price or senior citizen ticket =x dollars, children's ticket= y dollars.

 

you can list two equations:

 

4x+2y=80

5x+7y=145

so I times the first one by 3.5 to cancel out the y

4x+2y=80 time 3.5

equals 14x+7y=280

then I subtract 5x+7y=145 from this new equation 14x+7y=280

I can get 9x=135

x=15

then I plug in the solution for x to get the solution for y

4*15+2y=80 i just chose the first equation

y=10

so the price for a child's ticket is 10 dollars, the price for a senior citizen ticket is 15 dollars.

 

Best wishes!

Answer 3

What the top two answers are doing is combining several steps.

when you have two equations to solve with two variables you first solve for one variable in one equation.

the 2 equations

4x+2y=80
5x+7y=145

(4x+2y) - 4x = (80) - 4x           solve for y in terms of x in first equation by first subtracting 4x from each side

1/2(2y) = (80 - 4x)1/2              simplify and divide by 2

y = 40 - 2x                              you now have x in terms of y

 

5x + 7(40 - 2x) = 145                        put the above value of y in the second equation and solve for x

5x + 280 -14x = 145

(5x + 280 -14x) - 280  = (145) - 280

-9x = -135

x = 15

so that y = 40 - 2x = 40 - 2(15) = 10

y = 10 children's price

x = 15 adult's price 

double  check both equations

 

4x+2y=80          4(15) + 2(10) = 80            80 = 80
5x+7y=145        5(15) + 7(10) = 145         145 = 145