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Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems 3-2 Using Algebraic Methods to Solve Linear Systems Holt Algebra 2 Warm Up.

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Presentation on theme: "Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems 3-2 Using Algebraic Methods to Solve Linear Systems Holt Algebra 2 Warm Up."— Presentation transcript:

1 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems 3-2 Using Algebraic Methods to Solve Linear Systems Holt Algebra 2 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz Holt McDougal Algebra 2

2 3-2 Using Algebraic Methods to Solve Linear Systems Warm Up Determine if the given ordered pair is an element of the solution set of 2x – y = 5 3y + x = 6 1. (3, 1) yes 2. (–1, 1) no Solve each equation for y. 3. x + 3y = 2x + 4y – x y = 3y + 2x – 1 y = –x + 4 y = 2x + 3.

3 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Solve systems of equations by substitution. Solve systems of equations by elimination. Objectives

4 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems substitution elimination Vocabulary

5 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems The graph shows a system of linear equations. As you can see, without the use of technology, determining the solution from the graph is not easy. You can use the substitution method to find an exact solution. In substitution, you solve one equation for one variable and then substitute this expression into the other equation.

6 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Use substitution to solve the system of equations. Example 1A: Solving Linear Systems by Substitution y = x – 1 x + y = 7 Step 1 Solve one equation for one variable. The first equation is already solved for y: y = x – 1. Step 2 Substitute the expression into the other equation. x + y = 7 x + (x – 1) = 7 2x – 1 = 7 2x = 8 x = 4 Substitute (x –1) for y in the other equation. Combine like terms.

7 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Step 3 Substitute the x-value into one of the original equations to solve for y. Example 1A Continued y = x – 1 y = (4) – 1 y = 3 Substitute x = 4. The solution is the ordered pair (4, 3).

8 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Example 1A Continued Check A graph or table supports your answer.

9 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Use substitution to solve the system of equations. Example 1B: Solving Linear Systems by Substitution 2y + x = 4 3x – 4y = 7 Method 1 Isolate y. 2y + x = 4 5x = 15 3x + 2x – 8 = 7 First equation. Method 2 Isolate x. Isolate one variable. Second equation. Substitute the expression into the second equation. Combine like terms. 2y + x = 4 x = 3 x = 4 – 2y 3x – 4y= 7 3(4 – 2y)– 4y = 7 12 – 6y – 4y = 7 12 – 10y = 7 –10y = –5 5x – 8 = 7 First part of the solution 3x –4 +2 = 7 y = + 2 3x – 4y = 7

10 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Substitute the value into one of the original equations to solve for the other variable. 2y + (3) = 4 2y = 1 Example 1B Continued Substitute the value to solve for the other variable. Second part of the solution 2 + x = x = 4 x = 3 By either method, the solution is. Method 1Method 2

11 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Use substitution to solve the system of equations. y = 2x – 1 3x + 2y = 26 Step 1 Solve one equation for one variable. The first equation is already solved for y: y = 2x – 1. Step 2 Substitute the expression into the other equation. 3x + 2y = 26 3x + 2(2x–1) = 26 3x + 4x – 2 = 26 7x = 28 x = 4 Substitute (2x –1) for y in the other equation. Combine like terms. Check It Out! Example 1a

12 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Step 3 Substitute the x-value into one of the original equations to solve for y. y = 2x – 1 y = 2(4) – 1 y = 7 Substitute x = 4. The solution is the ordered pair (4, 7). Check It Out! Example 1a Continued

13 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Check A graph or table supports your answer. Check It Out! Example 1a Continued

14 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Use substitution to solve the system of equations. 5x + 6y = –9 2x – 2 = –y Method 1 Isolate y. 2x – 2 = –y First equation. Isolate one variable. Substitute the expression into the second equation. 5x + 6y = –9 Check It Out! Example 1b Method 2 Isolate x. y = –2x + 2 Second equation. 2x – 2 = –y x = 1 – y 5(1 – y)+ 6y = –9 5x + 6y = –9 5x + 6(–2x + 2) = –9

15 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Combine like terms. Method 1 Isolate y.Method 2 Isolate x. –7x = –21 5x + 6(–2x + 2) = –9 x = 3 5x – 12x + 12 = –9 First part of the solution. Check It Out! Example 1b Continued 10 – 5y + 12y = – y = –18 7y = –28 y = –4

16 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Substitute the value into one of the original equations to solve for the other variable. 5(3) + 6y = – y = –9 Example 1b Continued Substitute the value to solve for the other variable. Second part of the solution 5x + (–24) = –9 x = 3 By either method, the solution is (3, –4). 6y = –24 y = –4 5x + 6(–4) = –9 5x = 15

17 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems You can also solve systems of equations with the elimination method. With elimination, you get rid of one of the variables by adding or subtracting equations. You may have to multiply one or both equations by a number to create variable terms that can be eliminated. The elimination method is sometimes called the addition method or linear combination. Reading Math

18 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Use elimination to solve the system of equations. Example 2A: Solving Linear Systems by Elimination 3x + 2y = 4 4x – 2y = –18 Step 1 Find the value of one variable. 3x + 2y = 4 + 4x – 2y = –18 The y-terms have opposite coefficients. First part of the solution 7x = –14 x = –2 Add the equations to eliminate y.

19 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Example 2A Continued Step 2 Substitute the x-value into one of the original equations to solve for y. 3(–2) + 2y = 4 2y = 10 y = 5 Second part of the solution The solution to the system is (–2, 5).

20 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Use elimination to solve the system of equations. Example 2B: Solving Linear Systems by Elimination 3x + 5y = –16 2x + 3y = –9 Step 1 To eliminate x, multiply both sides of the first equation by 2 and both sides of the second equation by –3. Add the equations. First part of the solution y = –5 2(3x + 5y) = 2(–16) –3(2x + 3y) = –3(–9) 6x + 10y = –32 –6x – 9y = 27

21 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Example 2B Continued Second part of the solution 3x + 5(–5) = –16 3x = 9 3x – 25 = –16 x = 3 Step 2 Substitute the y-value into one of the original equations to solve for x. The solution for the system is (3, –5).

22 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Example 2B: Solving Linear Systems by Elimination Check Substitute 3 for x and –5 for y in each equation. 3x + 5y = –16 2x + 3y = –9 –16 3(3) + 5(–5) –16 2(3) + 3(–5)–9

23 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Use elimination to solve the system of equations. 4x + 7y = –25 –12x –7y = 19 Check It Out! Example 2a Step 1 Find the value of one variable. The y-terms have opposite coefficients. First part of the solution –8x = –6 4x + 7y = –25 – 12x – 7y = 19 Add the equations to eliminate y. x =

24 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Check It Out! Example 2a Continued 3 + 7y = –25 7y = –28 Second part of the solution Step 2 Substitute the x-value into one of the original equations to solve for y. 4 ( ) + 7y = –25 y = –4 The solution to the system is (, –4).

25 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Use elimination to solve the system of equations. 5x – 3y = 42 8x + 5y = 28 Step 1 To eliminate x, multiply both sides of the first equation by –8 and both sides of the second equation by 5. Add the equations. First part of the solution y = –4 Check It Out! Example 2b 49y = –196 –8(5x – 3y) = –8(42) 5(8x + 5y) = 5(28) –40x + 24y = –336 40x + 25y = 140

26 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Second part of the solution 5x – 3(–4) = 42 5x = 30 5x + 12 = 42 x = 6 Step 2 Substitute the y-value into one of the original equations to solve for x. The solution for the system is (6,–4). Check It Out! Example 2b

27 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Check Substitute 6 for x and –4 for y in each equation. 5x – 3y = 42 8x + 5y = (6) – 3(–4) 42 8(6) + 5(–4)28 Check It Out! Example 2b

28 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems In Lesson 3–1, you learned that systems may have infinitely many or no solutions. When you try to solve these systems algebraically, the result will be an identity or a contradiction. An identity, such as 0 = 0, is always true and indicates infinitely many solutions. A contradiction, such as 1 = 3, is never true and indicates no solution. Remember!

29 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Classify the system and determine the number of solutions. Example 3: Solving Systems with Infinitely Many or No Solutions 3x + y = 1 2y + 6x = –18 Because isolating y is straightforward, use substitution. Substitute (1–3x) for y in the second equation. Solve the first equation for y. 3x + y = 1 2(1 – 3x) + 6x = –18 y = 1 –3x 2 – 6x + 6x = –18 2 = –18 Distribute. Simplify. Because 2 is never equal to –18, the equation is a contradiction. Therefore, the system is inconsistent and has no solution. x

30 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Classify the system and determine the number of solutions. 56x + 8y = –32 7x + y = –4 Because isolating y is straightforward, use substitution. Substitute (–4 –7x) for y in the first equation. Solve the second equation for y. 7x + y = –4 56x + 8(–4 – 7x) = –32 y = –4 – 7x 56x – 32 – 56x = –32 Distribute. Simplify. Because –32 is equal to –32, the equation is an identity. The system is consistent, dependent and has infinite number of solutions. Check It Out! Example 3a –32 = –32

31 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Classify the system and determine the number of solutions. 6x + 3y = –12 2x + y = –6 Because isolating y is straightforward, use substitution. Substitute (–6 – 2x) for y in the first equation. Solve the second equation. 2x + y = –6 6x + 3(–6 – 2x)= –12 y = –6 – 2x 6x –18 – 6x = –12 Distribute. Simplify. Because –18 is never equal to –12, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions. Check It Out! Example 3b –18 = –12 x

32 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems A veterinarian needs 60 pounds of dog food that is 15% protein. He will combine a beef mix that is 18% protein with a bacon mix that is 9% protein. How many pounds of each does he need to make the 15% protein mixture? Example 4: Zoology Application Let x present the amount of beef mix in the mixture. Let y present the amount of bacon mix in the mixture.

33 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Example 4 Continued Write one equation based on the amount of dog food: Amount of beef mix plus amount of bacon mix equals xy = Write another equation based on the amount of protein: Protein of beef mix plus protein of bacon mix equals 0.18x0.09y protein in mixture. 0.15(60) + =

34 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Solve the system. x + y = x +0.09y = 9 x + y = 60 y = 60 – x First equation 0.18x (60 – x) = x – 0.09x = x = 3.6 x = 40 Solve the first equation for y. Substitute (60 – x) for y. Distribute. Simplify. Example 4 Continued

35 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Substitute the value of x into one equation. Substitute x into one of the original equations to solve for y y = 60 y = 20 Solve for y. The mixture will contain 40 lb of the beef mix and 20 lb of the bacon mix. Example 4 Continued

36 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems A coffee blend contains Sumatra beans which cost $5/lb, and Kona beans, which cost $13/lb. If the blend costs $10/lb, how much of each type of coffee is in 50 lb of the blend? Let x represent the amount of the Sumatra beans in the blend. Check It Out! Example 4 Let y represent the amount of the Kona beans in the blend.

37 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Write one equation based on the amount of each bean: Amount of Sumatra beans plus amount of Kona beans equals xy = Write another equation based on cost of the beans: Cost of Sumatra beans plus cost of Kona beans equals 5x5x 13y cost of beans. 10(50) += Check It Out! Example 4 Continued

38 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Solve the system. x + y = 50 5x + 13y = 500 x + y = 50 y = 50 – x First equation 5x + 13(50 – x) = 500 5x – 13x = 500 –8x = –150 x = Solve the first equation for y. Substitute (50 – x) for y. Distribute. Simplify. Check It Out! Example 4 Continued

39 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Substitute the value of x into one equation. Substitute x into one of the original equations to solve for y y = 50 y = Solve for y. The mixture will contain lb of the Sumatra beans and lb of the Kona beans. Check It Out! Example 4 Continued

40 Holt McDougal Algebra Using Algebraic Methods to Solve Linear Systems Lesson Quiz Use substitution or elimination to solve each system of equations. 3x + y = 1 y = x (–2, 7) 5x – 4y = 10 3x – 4y = –2 2. (6, 5) 3. The Miller and Benson families went to a theme park. The Millers bought 6 adult and 15 children tickets for $423. The Bensons bought 5 adult and 9 children tickets for $293. Find the cost of each ticket. adult: $28; childrens: $17


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