Applications of Linear Equations in Two Variables

Applications Involving Cost

When solving an application that involves two unknowns, sometimes it is convenient to use a system of linear equations in

two variables.

At a movie theater a couple buys one large popcorn and two drinks for $5.75. A group of teenagers buys two large popcorns and five drinks for $13.00. Find the cost of one

large popcorn and the cost of one drink.

Solution:In this application we have two unknowns, which we can represent by x and y. Let x represent the cost of one large popcorn. Let y represent the cost of one drink.

We must now write two equations. Each of the first two sentences in the problem gives a relationship

between x and y

( Cost of 1 ) + (cost of 2) = (total) x + 2y = 5.75 large popcorn drinks cost

( Cost of 2 ) + (cost of 2) = (total) 2x + 5y = 13.00 large popcorns drinks cost

2 x + 5 y = 13.00 Substitute x = -2y + 5.75 into the second equation.

x + 2y = 5.75 x = -2y + 5.75 Isolate x in the first equation.

2(-2y + 5.75) + 5y = 13.00-4y + 11.50 + 5y = 13.00

y + 11.50 = 13.00 y = 1.50

Solve for y.

x + 2y = 5.752x + 5y = 13.00

x = -2y + 5.75 First equation after solving for xSubstitute y = 1.50 into this equation.

x = -2(1.50) + 5.75

x = -3.00 + 5.75

x = 2.75The cost of one large popcorn is $2.75 and the cost of one drink is

$1.50

Check by verifying that the solutions meet the specified conditions

1 popcorn + 2 drinks = 1($2.75) + 2 ($1.50) = $5.75 TRUE2 popcorn + 5 drinks = 2($2.75) + 5($1.50) = $13.00 TRUE

Application Involving Principal and Interest

I = Prt where P is the principal.r is the annual interest rate, andt is the time in years

Example 2

Using a System of Linear equations

Involving Investments Joanne has a total of $6000 to deposit in two accounts. One account earns 3.5% simple interest and the other earns 2.5% simple interest. If the total amount of interest at the end of 1 year is $195, find the amount she deposited in each account.

Solution:Let x represent the principal deposited in the 2.5%Let y represent the principal deposited in the 3.5%

(Principal) + (principal) = (total ) x + y = 6000 invested invested principal at 2.5% at 3.5%

( Interest) + (interest) = (total ) 0.025x + 0.035y = 195 earned earned interest at 2.5% at 3.5%

2.5%Account 3.5% account Total

Principal x y 6000

Interest (I=Pr) 0.025x 0.03y 195

We will choose the addition method to solve the system of equations. First multiply the second equation by 1000 to clear decimals.

Multiply by -25

x + y = 6000

Multiply by 1000

x + y = 60000.025x + 0.035y = 195 25x + 35y = 195,000

-25x - 25y = -150,000 25x + 35y = 195,000

10y = 45,000

10y = 45,000 After eliminating the x-variable, solve for y.

10y = 45,000 10 10

y = 4500 The amount invested in the 3.5% account is $4500

x + y = 6000 Substitute y = 4500 into the equation x + y =6000

x + 4500 = 6000

x = 1500 The amount invested in the 2.5% account is $1500.

To check the solution, verify that the conditions of the problem have been met.

1. The sum of $1500 and $4500 is $6000 as desired. TRUE

2. The interest earned on $1500 at 2.5% is: 0.025($1500) = $37.50 The interest earned on $4500 at 3.5% is: 0.035($4500) = $157.50 $195.00

TRUE

Example 3

Using a System of linear Equations in a Mixture Application

A 10% Alcohol solution is mixed with a 40% alcohol solution to produce 30L of a 20% alcohol solution. Find the number of liters of 10% solution and the number of liters of 40% solution required for this mixture.

Solution:

Each solution contains a percentage of alcohol plus some other mixing agent such as water. Before we set up a system of equations to model this situation, it is helpful to have background understanding of the problem.

10%

x liters of solution

0.10x L of pure alcohol

+

40%

y liters of solution

0.40y L of pure alcohol

=

20%

.20(30)L of pure alcohol

30 liters of solution

10% Alcohol 40% Alcohol 20% Alcohol

Number of liters of solution

x y 30

Number of liters of pure alcohol

0.10x 0.40y 0.20(30) = 6

From the first row, we have

( Amount of ) + ( amount of ) = ( total amount ) 10% solution 40% solution of 20% solution x + y = 30

From the second row, we have

( Amount of ) + ( amount of ) = (total amount of) alcohol in alcohol in alcohol in10% solution 40% solution 20% solution

0.10x + 0.40y = 6

We will solve the system with the addition method by first clearing decimals

x + y = 300.10x + 0.40y = 6

Multiply by 10

x + 4y = 60Multiply by -1

-x –y = -30 x +4 y = 60 3y = 30

3y = 30 After eliminating the x – variable, solve for y

y = 10 10 L of 40% solution is needed

x + y =30 Substitute y = 10 into either of the original equationsx + (10) = 30

x = 20 20 L of 10% solution is needed

10 L of 40% solution must be mixed with 20 L of 10% solution

The following formula relates the distance traveled to the rate and time of travel.

d = rt distance = rate x time

Example 4

A plane travels with a tail wind from Kansas City, Missouri, to Denver, Colorado, a distance of 600 miles in 2 hours. The return trip against a head wind takes 3 hours. Find the speed of the plane in still air, and find the speed of the wind.

Solution

Let p represent the speed of the plane in still air.Let w represent the speed of the wind.

Distance Rate Time

With a tail wind

600 p + w 2

Against a head wind

600 p - w 3

To set up two equation in p and w, recall that d = rtFrom the first row, we have

(Distance ) = (rate with )(time traveled ) with the wind the wind with the wind 600 = (p + w)2

From the second row, we have

(Distance ) = (rate against)(time traveled ) against the wind the wind against the wind 600 = (p – w)3

Using the distributive property to clear parentheses, produces the following system

2p + 2w = 6003p - 3w = 600

Multiply by 36p + 6w = 1800

Multiply by 26p -6w = 1200

12p = 3000 12p = 3000 12p = 3000 12 12 p = 250

The speed of the plane in still air is 250 mph.

Substitute p = 250 into the first equation 2(250) + 2w = 600

500 + 2w = 600 2w = 100 w = 50

The speed of the plane in still air is 250 mph. The speed of the wind is 50 mph.