Lesson Using systems of equations to solve problems on tickets

Using systems of equations to solve problems on tickets

Problem 1

Let "x" be the number of adult tickets sold and let "y" be the number of student tickets sold.

Then from the condition you have this system of two equations in two unknowns 

 x +  y =  55,    (1)  
8x + 5y = 365.    (2)


Express  x = 55 - y  from (1) and substitute it into equation (2). You will get

8*(55 - y) + 5y = 365.


Simplify and solve for y:

440 - 8y + 5y = 365,

-3y = 365 - 440  ====>  -3y = -75  ====>  y =  = 25.


Then x = 55 - 25 = 30.

Problem 2

Let "a" be the number of adult tickets sold and let "c" be the number of adult tickets sold.

Then from the condition you have this system of two equations in two unknowns 

 a +  c =  700,    (1)  
5a + 3c = 2900.    (2)


Express  a = 700 - c  from (1) and substitute it into equation (2). You will get

5*(700 - c) + 3c = 2900.


Simplify and solve for c:

3500 - 5c + 3c = 2900,

-2c = 2900 - 3500  ====>  -2c = - 600  ====>  c =  = 300.


Then  a = 700 - 300 = 400.

Problem 3

Let "x" be the number of the adult    ticket bought, and
let "y" be the number of the children ticket bought.

Then you have a system of two linear equations in two unknowns

   x +    y = 104,           (1)
225x + 175y = 20250.         (2)  (this is the "value" equation in cents)

From the equation (1), express y = 104-x, and substitute this expression into the equation (2). You will get

225*x + 175(104-x) = 20250.  (3)


The equation (3) is a single equation for the single unknown "x".

Reduce all terms in both sides by the factor of 25 :

9x + 7*(104-x) = 810.

9x + 728 - 7x = 810,

2x = 810 - 728,

2x = 82   --->  x =  = 41.

Answer.  41 adult tickets and 104-41 = 63 children tickets.