Applications of Systems of Equations

Solving application problems using systems of equations involves setting up two equations with two variables and solving the newly formed system by either substitution or elimination.  You will have two unknown quantities.  Each unknown quantity will be identified using a unique variable.  In other words, assign one unknown using the variable “x” and the second unknown using the variable “y”.  Then you must form two equations using those variables.  Using the two equations and either the method of substitution or elimination, you will solve finding the value of each of the unknown variables.
Here are a few examples.

1)  During the 1996-1997 National Basketball Association season, the Boston Celtics played 82 games.  They lost 52 more games than they won.  What was their win-loss record that year?

First identify the unknowns using the variables, x and y … let x = wins and let y = losses

Next, set up two equations relating the two variables.  The problem states that the Boston Celtics played a total of 82 games, obviously they “won” some and they “loss” the rest.  This information gives us the first equation, x + y = 82.  Reading the second sentence, losses (y) were (=) (+) 52 more than wins (x).  This information gives us the second equation, y = x + 52.  Now, using these two equations solve using either substitution or addition.  I choose substitution since the second equation is already solved for y.

     x = wins
     y = losses

     x + y = 82
     y = x + 52

     x + x + 52 = 82
     2x + 52 = 82
     2x = 30
     x = 15 games won
     y = x + 52 = 15 + 52 = 67 games lost

     The win-loss record was 15 – 67.

2)  Josh and Langston found that the width of their basketball court was 44 feet less than the length.  If the perimeter was 288 feet, what were the length and the width of their court?

     x = width
     y = length

     x = y – 44                                     …   since the width (x) was (=) length (y) less (–) 44 feet (44)
     2x + 2y = 288                               …   use the formula 2 width + 2 length = Perimeter
     2(y – 44) + 2y = 288          … I would use substitution since the first equation is already solved for x
     2y – 88 + 2y = 288
     4y – 88 = 288
     4y = 376
     y = 94 feet long
     x = y – 44 = 94 – 44 = 50 feet wide

3)  Julie wanted to frame several family photos, including some of her recent wedding.  She went to a discount store and purchased two 11 x 14 frames and three 8 x 10 frames costing $22 (before taxes.)  Later she returned to the store and purchased one 11 x 14 frame and two 8 x 10 frames costing $13 (before taxes.)  How much did Julie pay for each of the different sized frames?

     x = cost per each 11 x 14 frame
     y = cost per each 8 x 10 frame

 2x + 3y = 22 ... purchase of two(2)11 x 14 frames(x) and(+) three(3) 8 x 10 frames(y) at(=)$22 (22)

 1x + 2y = 13     … purchase of one(1)11 x 14 frame(x) and(+) two(2) 8 x 10 frames(y) at(=)$13(13)

I choose to use the method of elimination.  I choose to multiply the second line by –2 and then add the two equations.

     –2[1x + 2y = 13] = – 2x – 4y = – 26

         2x + 3y = 22
      – 2x – 4y = – 26
              – y   = – 4
                 y   =  $4 for each 8 x 10 frame
                 x + 2y = 13 … x + 2(4) = 13 … x = $5 for each 11 x 14 frame

4)  How many gallons each of 25% alcohol and 35% alcohol should be mixed to get 20 gallons of 32% alcohol?

     x = amount of 25% alcohol
     y = amount of 35% alcohol

     x + y = 20   … since the mixture by volume will be a total of 20 gallons
    25x + 35y = 32(20)   …%(amount) + %(amount) = %(total amount)

Choose your method and solve … x = 6 gallons of 25% alcohol
                                                     y = 14 gallons of 35% alcohol

5)  A freight train and an express train leave towns 390 kilometers apart, traveling toward one another.  The freight train travels 30 kilometers per hour slower than the express train.  They pass one another 3 hours later.  What are their speeds?

     Let x = speed of the freight train
     Let y = speed of the express train

     x = y – 30   …   freight train (x) travels (=) the express train’s speed (y) but 30 kilometers per hour  
                               slower (–30)

     3x + 3y = 390   … since rate times time equals distance (r Ÿ t = d) and the total distance between the
                                   trains is 390 kilometers

     x = 50 kilometers per hour for the speed of the freight train
     y = 80 kilometers per hour for the speed of the express train

General Algebra Tips

The views and opinions expressed in this page are strictly those of Mary Lou Baker.
The contents of this page have not been reviewed or approved by Columbia State Community College.

This page was edited on 19-Sep-2007