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

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Presentation on theme: "When solving an application that involves two unknowns, sometimes it is convenient to use a system of linear equations in two variables."— Presentation transcript:

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

2 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 2 x + 5 y = 13.00Substitute 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 = 5.75 2x + 5y = 13.00

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

6 I = Prtwhere P is the principal. r is the annual interest rate, and t is the time in years

7 Example 2 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% 2.5%Account3.5% accountTotal Principalxy6000 Interest (I=Pr)0.025x0.03y195 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 Multiply by 1000 x + y = 6000 0.025x + 0.035y = 19525x + 35y = 195,000 -25x - 25y = -150,000 25x + 35y = 195,000 -25x - 25y = -150,000 25x + 35y = 195,000 10y = 45,000 After eliminating the x-variable, solve for y. 10y = 45,000 10 10 y = 4500The amount invested in the 3.5% account is $4500

10 x + y = 6000Substitute 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. 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

13 10% Alcohol40% Alcohol20% Alcohol Number of liters of solution xy30 Number of liters of pure alcohol 0.10x0.40y0.20(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 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 x + y = 30 0.10x + 0.40y = 6 Multiply by 10 x + 4y = 60 Multiply by -1 -x –y = -30 x +4 y = 60 3y = 30 After eliminating the x – variable, solve for y y = 1010 L of 40% solution is needed x + y =30 Substitute y = 10 into either of the original equations x + (10) = 30 x = 2020 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. DistanceRateTime With a tail wind 600p + w2 Against a head wind 600p - w3

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


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