Solving Word Problems Using Two Equations
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The document describes strategies for solving word problems involving two unknowns using a system of two equations. It provides four examples of applying this strategy: 1) Finding two numbers where one is 4 times the other and their sum is 40 2) Finding a two-digit number where the digit sum is 14 and reversing the digits yields a number 18 greater 3) Finding the number of nickels and dimes totaling $1.30 from a collection of 18 coins 4) Finding two people's ages where one is 4 times the other's age currently and will be twice the other's age in 10 years
Solving Word Problems Using Two Equations
- 2. The strategies used to solve problems using two equations are: • Step 1: Represent one of the unknowns as x and the other unknown as y. • Step 2: Translate the information about the variables into two equations using the two unknowns. • Step 3: Solve the system of equations for x and y.
- 3. Solution: Strategy: let x = the smaller number y = the larger number Since one number is 8 more than the other number, the first equation is y = x + 8 Example # 1: One number is 8 more than another number and the sum of the two numbers is 26. Find the numbers.
- 4. Solve the system: y = x + 8 x + y = 26 Substitute the value for y in the second equation and solve for x since y = x + 8 x+ y = 26 x + x + 8 = 26 2x + 8 = 26 2x + 8 – 8 = 26 – 8 2x = 18 ퟐ풙 = ퟐ ퟏퟖ ퟐ x = 9 Example # 1: One number is 8 more than another number and the sum of the two numbers is 26. Find the numbers.
- 5. find the other number: y = x + 8 y = 9 + 8 y = 17 hence, the numbers are 9 and 17. check the second equation. x + y = 26 9 + 17 = 26 26 = 26 Example # 1: One number is 8 more than another number and the sum of the two numbers is 26. Find the numbers.
- 6. Example # 2: The sum of the digits of a two-digit number is 15. If the digits are reversed, the new number is 9 more than the original number. Find the number. Solution: Strategy: Let x = the ten’s digit y = the one’s digit 10x + y = original number 10y + x = new number with digits reversed
- 7. Since the sum of the digits of the number is 15, the first equation is x + y = 15 Since reversing the digits gives a ne number which is 9 more than the original number, the equation is (10x + y) + 9 = (10y + x) Solve the system: x + y = 15 10x + y + 9 = 10y + x Example # 2: The sum of the digits of a two-digit number is 15. If the digits are reversed, the new number is 9 more than the original number. Find the number.
- 8. Solve the first equation for y, substitute in the second equation and find x. x + y = 15 x – x + y = 15 y = 15 –x 10x + y + 9 = 10y + x 10x + (15 - x) + 9 = 10(15 – x) + x 10x +15 – x + 9 = 150 – 10x + x 9x + 24 = 150 – 9x 9x + 9x + 24 = 150 -9x + 9x 18x = 150 – 24 18x = 126 18x = 128 18 18 x = 7 Example # 2: The sum of the digits of a two-digit number is 15. If the digits are reversed, the new number is 9 more than the original number. Find the number.
- 9. Find y: x + y = 15 7 + y = 15 y = 15 – 7 y = 8 Hence the number is 78. Check the information in the second equation. Original number is 78 Reversed number is 87 Since 87 is 9 more than 78, the answer is correct. Example # 2: The sum of the digits of a two-digit number is 15. If the digits are reversed, the new number is 9 more than the original number. Find the number.
- 10. EXAMPLE # 3: A person has 8 coins consisting of quarters and dimes. If the total amount of this change is $1.25, how many of each kind of coin are there? Solution: Strategy: let x = the number of quarters Let y = the number of dimes 25x = the value of the quarters And 10y = the value of the dimes Since there are 8 coins, the first equation is x + y = 8 Since the total values of the quarters plus the dimes is $1.25, the second equation is 25x + 10y = 125 Solve the system: x + y = 8 25x + 10y = 125
- 11. EXAMPLE # 3: A person has 8 coins consisting of quarters and dimes. If the total amount of this change is $1.25, how many of each kind of coin are there? Find the value of y in the first equation. substitute it in the second equation and solve for x. x + y = 8 x – x + y = 8 – x y = 8 – x 25x + 10y = 125 25x + 10(8 – x) = 125 25x + 80 – 10x = 125 15x + 80 = 125 15x + 80 – 80 = 125 – 80 15x = 45 15x = 45 15 15 x = 3
- 12. Example # 3: A person has 8 coins consisting of quarters and dimes. If the total amount of this change is $1.25, how many of each kind of coin are there? Find y: x + y = 8 3 + y = 8 3 – 3 + y = 8 – 3 y = 5 Hence, there are 3 quarters and 5 dimes. Check if their sum is $1.25. 3 quarters = 3 x $0.25 = $0.75 5 dimes = 5 x $0.10 = $0.50 $0.75 + $0.50 = $1.25
- 13. Solution: Strategy: Let x = the amount of $4 coffee used y = the amount of $3 coffee used Since the total amount of the mixture is 20 pounds, the first equation is x + y = 20 Since the cost of the mixture is $3.75, the second equation is 4x + 3y = 20(3.75) Example # 4: A merchant mixes some coffee costing $4 a pound with some coffee costing $3 a pound. How much of each must be used in order to make 20 pounds of mixture costing $3.75 a pound.
- 14. Solve the system: x + y = 20 4x + 3y = 20(3.75) Solve the first equation for x. Substitute in the second equation and solve for y. x + y = 20 x + y – y = 20 - y x = 20 – y Example # 4: A merchant mixes some coffee costing $4 a pound with some coffee costing $3 a pound. How much of each must be used in order to make 20 pounds of mixture costing $3.75 a pound.
- 15. Substitute: 4x + 3y = 20 4(20 – y) + 3y = 20(3.75) 80 – 4y + 3y = 75 80 – y = 75 80 – 80 – y = 75 – 80 -y = -5 -Y = -5 -1 -1 Y = 5 pounds Solve for x: x + y = 20 x + 5 = 20 x + 5 – 5 = 20 – 5 x = 15 pounds Example # 4: A merchant mixes some coffee costing $4 a pound with some coffee costing $3 a pound. How much of each must be used in order to make 20 pounds of mixture costing $3.75 a pound.
- 16. Hence, 15 pounds of the $4 coffee are needed and 5 pounds of of the $3 coffee are needed Check the second equation. 4x + 3y = 20(3.75) 4(15) + 3(5) = 75 60 + 15 = 75 75 = 75 Example # 4: A merchant mixes some coffee costing $4 a pound with some coffee costing $3 a pound. How much of each must be used in order to make 20 pounds of mixture costing $3.75 a pound.
- 17. Use two equations with two unknowns. 1. One number is 4 times another number. If their sum is 40, find the numbers. 2. The sum of the digits of a two-digit number is 14. If the digits are reversed, the new number is 18 more than the original number. Find the number. 3. A person has 18 coins, some of which are nickels and the rest of which are dimes. If the total amount of the coins is $1.30, find the number of nickels and dimes. 4. Matt is 4 times older than mike. In 10 years, he will be twice as old as mike. Find their ages.