Applications of Linear Equations in Two Variables

1. 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.

2. 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.

3. 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

4. x + 2y = 5.75 2x + 5y = 13.00 Isolate x in the first equation. x + 2y = 5.75 x = -2y + 5.75 2 x + 5 y = 13.00 Substitute x = -2y + 5.75 into the second equation. 2(-2y + 5.75) + 5y = 13.00 -4y + 11.50 + 5y = 13.00 Solve for y. y + 11.50 = 13.00 y = 1.50

5. x = -2y + 5.75 First equation after solving for x Substitute y = 1.50 into this equation. x = -2(1.50) + 5.75 x = -3.00 + 5.75 x = 2.75 The 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 TRUE 2 popcorn + 5 drinks = 2($2.75) + 5($1.50) = $13.00 TRUE

6. Application Involving Principal and Interest I = Prt where P is the principal. r is the annual interest rate, and t is the time in years

7. 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%

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

9. Multiply by -25 x + y = 6000 -25x - 25y = -150,000 25x + 35y = 195,000 x + y = 6000 0.025x + 0.035y = 195 25x + 35y = 195,000 10y = 45,000 Multiply by 1000 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

10. 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

11. 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.

12. 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. 20% 10% 40% x liters of solution y liters of solution 30 liters of solution + = 0.10x L of pure alcohol 0.40y L of pure alcohol .20(30)L of pure alcohol

13. 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 in 10% solution 40% solution 20% solution 0.10x + 0.40y = 6

14. We will solve the system with the addition method by first clearing decimals Multiply by -1 x + y = 30 0.10x + 0.40y = 6 -x –y = -30 x +4 y = 60 x + 4y = 60 3y = 30 Multiply by 10 3y = 30 After eliminating the x – variable, solve for y y = 10 10 L of 40% solution is needed Substitute y = 10 into either of the original equations x + (10) = 30 x + y =30 x = 20 20 L of 10% solution is needed 10 L of 40% solution must be mixed with 20 L of 10% solution

15. The following formula relates the distance traveled to the rate and time of travel. d = rt distance = rate x time

16. 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.

17. To set up two equation in p and w, recall that d = rt From 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

18. Multiply by 3 2p + 2w = 600 3p - 3w = 600 6p + 6w = 1800 6p -6w = 1200 Multiply by 2 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.