7.5 Using an Elimination Strategy to Solve a System of Linear Equations

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1 7.5 Using an Elimination Strategy to Solve a System of Linear Equations FOCUS Use elimination to solve a linear system. The solution of this linear system is: x 2 and y 1 We can add the equations: 2x 3y 7 x y 3 3x 4y 10 2x + 3y = y 4 x + y = 3 2 (2, 1) x 0 2 When we graph the new equation on the same grid, we see it also passes through (2, 1). We can use this property to solve a linear system. 3x + 4y = y (2, 1) x 2 Example 1 Solving a Linear System by Adding to Eliminate a Variable Solve this linear system. 4x 2y x 2y 26 2 Solution 4x 2y x 2y 26 2 The coefficients of y in both equations are opposite integers: 2 and 2 So, use elimination. Add the equations to eliminate y. 4x 2y 10 1 (5x 2y 26) 2 9x 36 Solve for x. Divide each side by 9. x 4 To find the value of y when x 4, substitute in one of the original equations. Choose equation 1. When we add equations to eliminate one variable, we solve by elimination. 4x 2y (4) 2y y 10 Isolate 2y. Subtract 16 from each side. 2y y 6 Solve for y. Divide each side by 2. y 3 The solution is: x 4 and y 3 2y 2y Copyright 2011 Pearson Canada Inc. 407

2 In each equation, substitute: x 4 and y 3 4x 2y 10 L.S. 4x 2y 4(4) 2(3) R.S. 10 5x 2y 26 L.S. 5x 2y 5(4) 2(3) For each equation, L.S. R.S. So, the solution of the linear system is x 4 and y 3. R.S. 26 Check 1. Solve this linear system. 2x 8y x 3y 8 2 The coefficients of x in both equations are opposite integers. So, use elimination. Add the equations to eliminate x. 2x 8y 18 1 ( 2x 3y 8) 2 Divide each side by. y To find the value of x when y, substitute in equation 2. 2x 3y 8 2 2x 3( ) 8 2x 8 x The solution is: x and y In each equation, substitute: x and y 2x 8y 18 L.S. R.S. 2x 3y 8 L.S. R.S. For each equation, L.S. R.S. So, the solution of the linear system is x and y Copyright 2011 Pearson Canada Inc.

3 Sometimes we have to multiply one or both equations by a number before we can eliminate a variable by adding. Example 2 Solving a Linear System by Multiplying then Eliminating Solve this linear system. 4x 2y 2 1 3x y 6 2 Solution 4x 2y 2 1 3x y 6 2 Use elimination. No coefficients of like terms are opposite integers. Look at the y-terms: 2y and y To make the coefficients opposite integers, multiply equation 2 by 2. 2 equation 2: Like terms have the same variable. 2(3x y 6) 6x 2y 12 3 Add equations 1 and 3 to eliminate y. 4x 2y 2 1 (6x 2y 12) 3 10x 10 Solve for x. Divide each side by 10. x 1 To find the value of y when x 1, substitute in equation 1. 4x 2y 2 1 4( 1) 2y 2 4 2y 2 Isolate 2y. Add 4 to each side. 2y 2 4 2y 6 Solve for y. Divide each side by 2. y 3 The solution is: x 1 and y 3 When we multiply an equation by a number, we produce an equivalent equation. We could substitute x 1 in either equation to find y. In the original equations, substitute: x 1 and y 3 4x 2y 2 L.S. 4x 2y 4( 1) 2( 3) R.S. 2 3x y 6 L.S. 3x y 3( 1) ( 3) For each equation, L.S. R.S. So, the solution of the linear system is x 1 and y 3. R.S Copyright 2011 Pearson Canada Inc. 409

4 Check 1. Solve this linear system. 3x 10y x y 3 2 Use elimination. No coefficients of like terms are opposite integers. Look at the y-terms: 10y and y To make the coefficients opposite integers, multiply equation 2 by. equation 2: (2x y 3) 3 Add equations 1 and 3 to eliminate y. 3x 10y Divide each side by. x To find the value of y when x, substitute in equation 2. 2x y 3 2 2( ) y 3 Scan the equations, and look for numbers that are easiest to work with. y The solution is: x and y In the original equations, substitute: x and y 3x 10y 16 2x y 3 L.S. R.S. L.S. R.S. L.S. L.S. For each equation, L.S. R.S. So, the solution of the linear system is x and y Copyright 2011 Pearson Canada Inc.

5 Example 3 Using a Linear System to Solve a Problem a) Create a linear system to model this situation: Ava and Ethan sold flower bulbs for a school fundraiser. Ava sold 5 bags of tulip bulbs and 2 bags of daffodil bulbs. She raised $80. Ethan sold 1 bag of tulip bulbs and 4 bags of daffodil bulbs. He raised $70. b) Solve this problem: What is the cost of a bag of each type of bulb? Solution a) Cost of 5 bags of tulip bulbs cost of 2 bags of daffodil bulbs $80 Cost of 1 bag of tulip bulbs cost of 4 bags of daffodil bulbs $70 Let the cost of a bag of tulip bulbs be t dollars. Let the cost of a bag of daffodil bulbs be d dollars. A linear system that models the situation is: 5t 2d 80 1 t 4d 70 2 b) Use elimination. No coefficients of like terms have opposite integers. Look at the t-terms: 5t and t To make the coefficients opposite integers, multiply equation 2 by 5. ( 5) equation 2: 5(t 4d 70) 5t 20d Add equations 1 and 3 to eliminate t. 5t 2d 80 1 ( 5t 20d 350) 3 18d 270 d 15 Solve for d. Divide each side by 18. To find the value of t when d 15, substitute in equation 2. t 4d 70 2 t 4(15) 70 t t t 10 The solution is: t 10 and d 15 Use the data in the problem to verify the solution. Solve for t. Subtract 60 from each side. Cost of 5 bags of tulip bulbs at $10 each and 2 bags of daffodil bulbs at $15 each is: $50 $30 $80 Cost of 1 bag of tulip bulbs at $10 and 4 bags of daffodil bulbs at $15 each is: $10 $60 $70 The total costs match the data in the problem. So, the solution is correct. A bag of tulip bulbs costs $10 and a bag of daffodil bulbs costs $15. We could substitute d 15 in either equation to find t Copyright 2011 Pearson Canada Inc. 411

6 Check 1. a) Create a linear system to model this situation: A store sells running shoes and hiking boots. The store makes $25 profit from the sale of a pair of running shoes and $50 profit from the sale of a pair of hiking boots. One afternoon, the company sold 14 pairs of footwear for a total profit of $450. Let the number of pairs of running shoes sold be r. Let the number of pairs of hiking boots sold be h. Running shoes Hiking boots Total Profit per pair ($) Number sold Total profit ($) The number of pairs sold is represented by equation 1: The total profit in dollars is represented by equation 2: A linear system that models the situation is: 1 2 b) Solve this problem: How many pairs of running shoes and how many pairs of hiking boots were sold? Use elimination. No coefficients of like terms have opposite integers. We could have multiplied equation 1 by 50, then Look at the r-terms: and added to eliminate h. To make the coefficients opposite integers, multiply equation 1 by. equation 1: ( ) 3 Add equations 2 and 3 to eliminate r. h To find the value of r when h, substitute in equation 1. r The solution is: r and h Copyright 2011 Pearson Canada Inc.

7 Use the data in the problem to verify the solution. pairs of running shoes at a profit of each pairs of hiking boots at a profit of each pairs for a total profit of The totals match the data in the problem. So, the solution is correct. pairs of running shoes and pairs of hiking boots were sold. Practice 1. To solve each linear system by elimination, what would you multiply equation 1 by? a) x 2y 2 1 b) 2x y 6 1 4x 6y 8 2 3x 5y 9 2 In equation 2, the coefficient of x is: Its opposite is: So, multiply equation 1 by. In equation 2, the coefficient of y is: Its opposite is: So, multiply equation 1 by. 2. Solve this linear system using elimination. 5x 7y x 7y 8 2 The coefficients of y in both equations are opposite integers. So, add the equations to eliminate y. 5x 7y 24 1 (3x 7y 8) 2 x To find the value of y when x, substitute in equation 2. 3x 7y 8 2 3( ) 7y 8 7y 8 y The solution is: x and y Copyright 2011 Pearson Canada Inc. 413

8 In each equation, substitute: x and y 5x 7y 24 3x 7y 8 L.S. R.S. L.S. R.S. For each equation, L.S. R.S. So, the solution of the linear system is x and y. 3. Solve this linear system using elimination. 3x 2y 2 1 2x 5y 16 2 No coefficients of like terms are opposite integers. Look at the y-terms: 2y and 5y To make the coefficients opposite integers, multiply equation 1 by and equation 2 by. Sometimes, we have to multiply both equations by numbers to get opposite coefficients. equation 1: ( 3x 2y 2) 3 equation 2: (2x 5y 16) 4 Add equations 3 and 4 to eliminate y. x To find the value of y when x, substitute in equation 1. 3x 2y 2 1 3( ) 2y 2 y The solution is: x and y In the original equations, substitute: x and y 3x 2y 2 L.S. R.S. 2x 5y 16 L.S. R.S. L.S. L.S. For each equation, L.S. R.S. So, the solution of the linear system is x and y Copyright 2011 Pearson Canada Inc.

9 4. a) Create a linear system to model this situation: A dog groomer charges $50 to groom a small dog and $75 to groom a large dog. One Saturday, the groomer groomed 9 dogs for a total income of $500. Let the number of small dogs groomed be s. Let the number of large dogs groomed be L. Small dog Large dog Total Cost per grooming ($) Number groomed Total income ($) Let the number of dogs groomed be represented by equation 1: Let the total income in dollars be represented by equation 2: A linear system that models the situation is: 1 2 b) Solve this problem: How many small dogs and how many large dogs were groomed? The solution is: s and L Use the data in the problem to verify these numbers. small dogs and large dogs were groomed Copyright 2011 Pearson Canada Inc. 415