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Using Algebraic Methods to Solve Linear Systems. 3-2. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 2. Holt Algebra 2. Warm Up Determine if the given ordered pair is an element of the solution set of. 2 x – y = 5. 3 y + x = 6. no. 2. ( –1, 1). 1. (3, 1).

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

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

  3. Objectives Solve systems of equations by substitution. Solve systems of equations by elimination.

  4. Vocabulary substitution elimination

  5. 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. y= x – 1 x + y = 7 Example 1A: Solving Linear Systems by Substitution Use substitution to solve the system of equations. 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 Substitute (x –1) for y in the other equation. 2x– 1 = 7 Combine like terms. 2x = 8 x = 4

  7. 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 Substitute x = 4. y = 3 The solution is the ordered pair (4, 3).

  8. Example 1A Continued Check A graph or table supports your answer.

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

  10. By either method, the solution is . Example 1B Continued Substitute the value into one of the original equations to solve for the other variable. Method 1 Method 2 Substitute the value to solve for the other variable. 2y + (3) = 4 • + x = 4 1 + x = 4 2y = 1 x = 3 Second part of the solution

  11. y= 2x – 1 3x + 2y = 26 Check It Out! Example 1a Use substitution to solve the system of equations. 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 Substitute (2x –1) for y in the other equation. 3x + 2(2x–1) = 26 3x + 4x – 2 = 26 Combine like terms. 7x = 28 x = 4

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

  13. Check It Out! Example 1a Continued Check A graph or table supports your answer.

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

  15. Check It Out! Example 1b Continued Method 1 Isolate y. Method 2 Isolate x. 5x + 6(–2x + 2) = –9 Combine like terms. 5x – 12x + 12 = –9 10 – 5y + 12y = –18 –7x = –21 10 + 7y = –18 7y = –28 x = 3 y = –4 First part of the solution.

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

  17. Reading Math The elimination method is sometimes called the addition method or linear combination. 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.

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

  19. 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 Second part of the solution y = 5 The solution to the system is (–2, 5).

  20. 3x + 5y = –16 2x+ 3y = –9 6x + 10y = –32 2(3x + 5y) = 2(–16) –3(2x + 3y) = –3(–9) –6x – 9y = 27 Example 2B: Solving Linear Systems by Elimination Use elimination to solve the system of equations. 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. y = –5 First part of the solution

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

  22. 2x + 3y = –9 3x + 5y = –16 –16 2(3) + 3(–5) –9 3(3) + 5(–5)   –16 –16 –9 –9 Example 2B: Solving Linear Systems by Elimination Check Substitute 3 for x and –5 for y in each equation.

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

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

  25. 5x – 3y = 42 8x+ 5y = 28 –8(5x– 3y) = –8(42) –40x+ 24y = –336 40x + 25y = 140 5(8x + 5y) = 5(28) Check It Out! Example 2b Use elimination to solve the system of equations. 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. 49y = –196 First part of the solution y = –4

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

  27. 8x + 5y = 28 5x– 3y = 42 42 8(6) + 5(–4) 28 5(6) – 3(–4)   42 42 28 28 Check It Out! Example 2b Check Substitute 6 for x and –4 for y in each equation.

  28. Remember! 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. 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.

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

  30. –32= –32  Check It Out! Example 3a Classify the system and determine the number of solutions. 56x + 8y = –32 7x + y = –4 Because isolating y is straightforward, use substitution. 7x + y = –4 y = –4 – 7x Solve the second equation for y. Substitute (–4 –7x) for y in the first equation. 56x+ 8(–4 – 7x) = –32 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.

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

  32. Example 4: Zoology Application 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? Let x present the amount of beef mix in the mixture. Let y present the amount of bacon mix in the mixture.

  33. Amount of beef mix amount of bacon mix plus equals 60. = x + y 60 Protein of beef mix protein of bacon mix protein in mixture. plus equals = + 0.18x 0.09y 0.15(60) Example 4 Continued Write one equation based on the amount of dog food: Write another equation based on the amount of protein:

  34. x + y = 60 0.18x+0.09y = 9 Example 4 Continued Solve the system. First equation x + y = 60 y = 60 – x Solve the first equation for y. 0.18x + 0.09(60 – x) = 9 Substitute (60 – x) for y. 0.18x + 5.4 – 0.09x = 9 Distribute. Simplify. 0.09x = 3.6 x = 40

  35. Example 4 Continued Substitute x into one of the original equations to solve for y. Substitute the value of x into one equation. 40 + y = 60 y = 20 Solve for y. The mixture will contain 40 lb of the beef mix and 20 lb of the bacon mix.

  36. Check It Out! Example 4 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. Let y represent the amount of the Kona beans in the blend.

  37. Amount of Sumatra beans amount of Kona beans equals 50. plus = + x y 50 Cost of Sumatra beans cost of Kona beans cost of beans. equals plus + = 13y 10(50) 5x Check It Out! Example 4 Continued Write one equation based on the amount of each bean: Write another equation based on cost of the beans:

  38. x + y = 50 5x + 13y = 500 Check It Out! Example 4 Continued Solve the system. First equation x + y = 50 y = 50 – x Solve the first equation for y. 5x + 13(50 – x) = 500 Substitute (50 – x) for y. 5x + 650 – 13x = 500 Distribute. –8x = –150 Simplify. x = 18.75

  39. Check It Out! Example 4 Continued Substitute x into one of the original equations to solve for y. Substitute the value of x into one equation. 18.75 + y = 50 y = 31.25 Solve for y. The mixture will contain 18.75 lb of the Sumatra beans and 31.25 lb of the Kona beans.

  40. Lesson Quiz Use substitution or elimination to solve each system of equations. 3x + y = 1 5x – 4y = 10 1. 2. y = x + 9 3x – 4y = –2 (–2, 7) (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; children’s: $17

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