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Word problems

Sometimes systems of equations can be used to model word problems.  Let’s jump straight to an example.

Example:  The school that Matt goes to is selling tickets to a choral performance.  On the first day of ticket sales the school sold 12 adult tickets and 3 student tickets for a total of $129.  The school took in $104 on the second day by selling 2 adult tickets and 6 student tickets.  Find the price of an adult ticket and the price of a student ticket.

Solution:  Let a be the price of an adult ticket, and let s represent the price of a student ticket.  On the first day of the performance the 12 adult tickets were sold at the price of a and 3 student tickets were sold at the price of s.  The sum of their sales was $129.  We can model this by

\(12a + 3s = 129\)

Using a similar reasoning, we can model the second day of sales by

\(2a + 6s = 104\)

Combining these two equations gives us a system that we can solve!  We use elimination:

\(12a + 3s = 129\)
\(2a + 6s = 104\)

\( - 24a - 6s =  - 258\)
\(2a + 6s = 104\)

\( - 22a =  - 154\)

\(a = 7\)

That is, an adult ticket cost $7.  Then by substituting \(a = 7\) into the second equation, we have

\(2a + 6s = 104\)
\(2\left( 7 \right) + 6s = 104\)
\(14 + 6s = 104\)
\(6s = 90\)
\(s = 15\)

That is, a student ticket costs $15.

Another Example:  The senior class at High School A and High School B planned separate trips to the water park.  The senior class at High School A rented and filled 8 vans and 4 buses with 256 students.  High School B rented and filled 4 vans and 6 buses with 312 students.  Each van and each bus carried the same number of students.  How many students can a van carry?  How many students can a bus carry?

Solution:  Let v be the number of students a van can carry.  Let b be the number of students a bus can carry.  High School A’s situation can be modeled by

\(8v + 4b = 256\)

Similarly, High School B’s situation can be modeled by

\(4v + 6b = 312\)

We solve the system using elimination

\(8v + 4b = 256\)
\(4v + 6b = 312\)

\(8v + 4b = 256\)
\( - 8v - 12b =  - 624\)

\( - 8b =  - 368\)

\(b = 46\)

That is, a bus can hold 46 students.  Substituting 46 into the first equation gives

\(8v + 4b = 256\)
\(8v + 4\left( {46} \right) = 256\)
\(8v + 184 = 256\)
\(8v = 72\)
\(v = 9\)

That is, each van can hold 9 students.

Below you can download some free math worksheets and practice.


Downloads:
11548 x

This free worksheet contains 10 assignments each with 24 questions with answers.

Example of one question:

Systems-of-Equations-and-Inequalities-Word-problems-easy

Watch below how to solve this example:

 

Downloads:
7598 x

This free worksheet contains 10 assignments each with 24 questions with answers.

Example of one question:

Systems-of-Equations-and-Inequalities-Word-problems-medium

Watch below how to solve this example:

 

Downloads:
5968 x

This free worksheet contains 10 assignments each with 24 questions with answers.

Example of one question:

Systems-of-Equations-and-Inequalities-Word-problems-hard

Watch below how to solve this example:

 
 
 

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