Systems of Linear Equations Copyright Cengage Learning All

Systems of Linear Equations Copyright © Cengage Learning. All rights reserved. 4

SECTION 4. 4 Applications Copyright © Cengage Learning. All rights reserved.

A Blueprint for Problem Solving 3

Blueprint for Problem Solving The word problems in this section have two unknown quantities. We will write two equations in two variables (each of which represents one of the unknown quantities), which of course is a system of equations. We then solve the system by one of the methods: graphical method, elimination method, or substitution method. Here are the steps to follow in solving these word problems. 4

Blueprint for Problem Solving 5

Example 1 Number Problem One number is 2 more than 5 times another number. Their sum is 20. Find the two numbers. Solution: We apply the steps in our blueprint. Step 1: We know that the two numbers have a sum of 20 and that one of them is 2 more than 5 times the other. We don’t know what the numbers themselves are. 6

Example 1 – Solution cont’d Step 2: Let x represent one of the numbers and y represent the other. “One number is 2 more than 5 times another” translates to y = 5 x + 2 “Their sum is 20” translates to x + y = 20 7

Example 1 – Solution cont’d Step 3: The system that describes the situation must be x + y = 20 y = 5 x + 2 Step 4: Since the 2 nd equation is already in slope intercept form, we can solve this system by substituting the expression 5 x + 2 in the second equation for y in the first equation: x + (5 x + 2) = 20 6 x + 2 = 20 8

Example 1 – Solution cont’d 6 x = 18 x=3 Using x = 3 in either of the first two equations and then solving for y, we get y = 17. Step 5: So 17 and 3 are the numbers we are looking for. Step 6: The number 17 is 2 more than 5 times 3, and the sum of 17 and 3 is 20. 9

Example 2 Interest Problem Mr. Hicks had $15, 000 to invest. He invested part at 6% and the rest at 7%. If he earns $980 in interest, how much did he invest at each rate? Solution: Remember, step 1 is done mentally. Step 1: We do not know the specific amounts invested in the two accounts. We do know that their sum is $15, 000 and that the interest rates on the two accounts are 6% and 7%. 10

Example 2 – Solution cont’d Step 2: Let x = the amount invested at 6% and y = the amount invested at 7%. Because Mr. Hicks invested a total of $15, 000, we have x + y = 15, 000 The interest he earns comes from 6% of the amount invested at 6% and 7% of the amount invested at 7%. To find 6% of x, we multiply x by 0. 06, which gives us 0. 06 x. 11

Example 2 – Solution cont’d To find 7% of y, we multiply 0. 07 times y and get 0. 07 y. Step 3: The system is x + y = 15, 000 0. 06 x + 0. 07 y = 980 12

Example 2 – Solution cont’d Step 4: Since both equations are given in standard form, we know that we should use elimination method. Substituting y = 8, 000 into the first equation and solving for x, we get x = 7, 000. Step 5: He invested $7, 000 at 6% and $8, 000 at 7%. 13

Example 2 – Solution cont’d Step 6: Checking our solutions in the original problem, we have the following: The sum of $7, 000 and $8, 000 is $15, 000, the total amount he invested. To complete our check, we find the total interest earned from the two accounts: 14

Example 3 Coin Problem John has $1. 70 all in dimes and nickels. He has a total of 22 coins. How many of each kind does he have? Solution: Step 1: We know that John has 22 coins that are dimes and nickels. We know that a dime is worth 10 cents and a nickel is worth 5 cents. We do not know the specific number of dimes and nickels he has. 15

Example 3 – Solution cont’d Step 2: Let x = the number of nickels and y = the number of dimes. The total number of coins is 22, so x + y = 22 The total amount of money he has is $1. 70, which comes from nickels and dimes. 16

Example 3 – Solution cont’d Step 3: The system that represents the situation is Step 4: We multiply the first equation by – 5 and the second by 100 to eliminate the variable x: y = 12 17

Example 3 – Solution cont’d Substituting y = 12 into our first equation, we get x = 10. Step 5: John has 12 dimes and 10 nickels. Step 6: Twelve dimes and 10 nickels total 22 coins. 18

Example 4 Mixture Problem How much 20% alcohol solution and 50% alcohol solution must be mixed to get 12 gallons of 30% alcohol solution? Solution: To solve this problem we must first understand that a 20% alcohol solution is 20% alcohol and 80% water. Step 1: We know there are two solutions that together must total 12 gallons. 20% of one of the solutions is alcohol and the rest is water, whereas the other solution is 50% alcohol and 50% water. We do not know how many gallons of each individual solution we need. 19

Example 4 – Solution cont’d Step 2: Let x = the number of gallons of 20% alcohol solution needed and y = the number of gallons of 50% alcohol solution needed. Because the total number of gallons we will end up with is 12, and this 12 gallons must come from the two solutions we are mixing, our first equation is x + y = 12 To obtain our second equation, we look at the amount of alcohol in our two original solutions and our final solution. 20

Example 4 – Solution cont’d The amount of alcohol in the x gallons of 20% solution is 0. 20 x, and the amount of alcohol in y gallons of 50% solution is 0. 50 y. The amount of alcohol in the 12 gallons of 30% solution is 0. 30(12). Because the amount of alcohol we start with must equal the amount of alcohol we end up with, our second equation is 0. 20 x + 0. 50 y = 0. 30(12) 21

Example 4 – Solution cont’d The information we have so far can also be summarized with a table. Sometimes by looking at a table like the one that follows, it is easier to see where the equations come from. Step 3: Our system of equations is x + y = 12 0. 20 x + 0. 50 y = 0. 30(12) 22

Example 4 – Solution cont’d Step 4: Multiplying each side of the 2 nd equation by 10 gives us an equivalent equation that is a little easier to work with. 0. 20 x + 0. 50 y = 0. 30(12) 2 x + 5 y = 36 x + y = 12 × (-2) × 10 2 x + 5 y = 3(12) 2 x + 5 y = 36 -2 x – 2 y = -24 3 y = 12 Therefore y = 4, and because x + 4 = 12, x must be 8. 23

Example 4 – Solution cont’d Step 5: It takes 8 gallons of 20% alcohol solution and 4 gallons of 50% alcohol solution. Step 6: Eight gallons of 20% alcohol solution plus four gallons of 50% alcohol solution will equal twelve gallons of 30% alcohol solution. 24