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Holt Algebra Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz

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Holt Algebra Solving Systems by Substitution Warm Up Solve each equation for x. 1. y = x y = 3x – 4 Simplify each expression. x = y – 3 2x – (x – 5) – 3(x + 1)9 – 3x

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Holt Algebra Solving Systems by Substitution Warm Up Continued Evaluate each expression for the given value of x. 5. x + 8 for x = (x – 7) for x =

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Holt Algebra Solving Systems by Substitution Solve linear equations in two variables by substitution. Objective

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Holt Algebra Solving Systems by Substitution Sometimes it is difficult to identify the exact solution to a system by graphing. In this case, you can use a method called substitution. The goal when using substitution is to reduce the system to one equation that has only one variable. Then you can solve this equation by the methods taught in Chapter 2.

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Holt Algebra Solving Systems by Substitution Solving Systems of Equations by Substitution Step 2 Step 3 Step 4 Step 5 Step 1 Solve for one variable in at least one equation, if necessary. Substitute the resulting expression into the other equation. Solve that equation to get the value of the first variable. Substitute that value into one of the original equations and solve. Write the values from steps 3 and 4 as an ordered pair, (x, y), and check.

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Holt Algebra Solving Systems by Substitution Solve the system by substitution. Example 1A: Solving a System of Linear Equations by Substitution y = 3x y = x – 2 Step 1 y = 3x y = x – 2 Both equations are solved for y. Step 2 y = x – 2 3x = x – 2 Substitute 3x for y in the second equation. Solve for x. Subtract x from both sides and then divide by 2. Step 3–x 2x = –2 2 x = –1

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Holt Algebra Solving Systems by Substitution Solve the system by substitution. Example 1A Continued Step 4y = 3x Write one of the original equations. Substitute –1 for x. y = 3( – 1) y = –3 Step 5 ( – 1, –3) Check Substitute (–1, –3) into both equations in the system. Write the solution as an ordered pair. y = 3x –3 3(–1) –3 y = x – 2 –3 –1 – 2 –3

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Holt Algebra Solving Systems by Substitution Solve the system by substitution. Example 1B: Solving a System of Linear Equations by Substitution y = x + 1 4x + y = 6 Step 1 y = x + 1 The first equation is solved for y. Step 2 4x + y = 6 4x + (x + 1) = 6 Substitute x + 1 for y in the second equation. Subtract 1 from both sides. Step 3–1 5x = x = 1 5x = 5 5x + 1 = 6 Simplify. Solve for x. Divide both sides by 5.

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Holt Algebra Solving Systems by Substitution Solve the system by substitution. Example1B Continued Step 4y = x + 1 Write one of the original equations. Substitute 1 for x. y = y = 2 Step 5 (1, 2) Check Substitute (1, 2) into both equations in the system. Write the solution as an ordered pair. y = x x + y = 6 4(1)

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Holt Algebra Solving Systems by Substitution Solve the system by substitution. Example 1C: Solving a System of Linear Equations by Substitution x + 2y = –1 x – y = 5 Step 1 x + 2y = –1 Solve the first equation for x by subtracting 2y from both sides. Step 2 x – y = 5 (–2y – 1) – y = 5 Substitute – 2y – 1 for x in the second equation. –3y – 1 = 5 Simplify. −2y x = –2y – 1

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Holt Algebra Solving Systems by Substitution Example 1C Continued Step 3 –3y – 1 = 5 Add 1 to both sides. +1 –3y = 6 –3 y = –2 Solve for y. Divide both sides by –3. Step 4x – y = 5 x – (–2) = 5 x + 2 = 5 –2 x = 3 Step 5 (3, –2) Write one of the original equations. Substitute –2 for y. Subtract 2 from both sides. Write the solution as an ordered pair.

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Holt Algebra Solving Systems by Substitution Check It Out! Example 1a Solve the system by substitution. y = x + 3 y = 2x + 5 Both equations are solved for y. Step 1y = x + 3 y = 2x + 5 Substitute 2x + 5 for y in the first equation. Solve for x. Subtract x and 5 from both sides. –x – 5 –x – 5 x = –2 Step 32x + 5 = x + 3 Step 2 2x + 5 = x + 3 y = x + 3

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Holt Algebra Solving Systems by Substitution Check It Out! Example 1a Continued Solve the system by substitution. Step 4y = x + 3 Write one of the original equations. Substitute –2 for x. y = –2 + 3 y = 1 Step 5 (–2, 1) Write the solution as an ordered pair.

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Holt Algebra Solving Systems by Substitution Check It Out! Example 1b Solve the system by substitution. x = 2y – 4 x + 8y = 16 The first equation is solved for x. Step 1 x = 2y – 4 Substitute 2y – 4 for x in the second equation. Simplify. Then solve for y. (2y – 4) + 8y = 16 x + 8y = 16Step 2 Step 310y – 4 = 16 Add 4 to both sides y = 20 y = 2 10y = Divide both sides by 10.

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Holt Algebra Solving Systems by Substitution Check It Out! Example 1b Continued Solve the system by substitution. Step 4 x + 8y = 16 Write one of the original equations. Substitute 2 for y. x + 8(2) = 16 x + 16 = 16 x = 0 – 16 –16 Simplify. Subtract 16 from both sides. Step 5 (0, 2) Write the solution as an ordered pair.

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Holt Algebra Solving Systems by Substitution Check It Out! Example 1c Solve the system by substitution. 2x + y = –4 x + y = –7 Solve the second equation for x by subtracting y from each side. Substitute –y – 7 for x in the first equation. Distribute 2. 2(–y – 7) + y = –4 x = –y – 7Step 2 Step 1 x + y = –7 – y x = –y – 7 2(–y – 7) + y = –4 –2y – 14 + y = –4

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Holt Algebra Solving Systems by Substitution Combine like terms. Step –y = 10 Check It Out! Example 1c Continued Solve the system by substitution. –2y – 14 + y = –4 Add 14 to each side. –y – 14 = –4 y = –10 Step 4 x + y = –7 Write one of the original equations. Substitute –10 for y. x + (–10) = –7 x – 10 = – 7

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Holt Algebra Solving Systems by Substitution Check It Out! Example 1c Continued Solve the system by substitution. x – 10 = –7Step x = 3 Add 10 to both sides. Step 6(3, –10) Write the solution as an ordered pair.

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Holt Algebra Solving Systems by Substitution Sometimes you substitute an expression for a variable that has a coefficient. When solving for the second variable in this situation, you can use the Distributive Property.

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Holt Algebra Solving Systems by Substitution When you solve one equation for a variable, you must substitute the value or expression into the other original equation, not the one that had just been solved. Caution

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Holt Algebra Solving Systems by Substitution Example 2: Using the Distributive Property y + 6x = 11 3x + 2y = –5 Solve by substitution. Solve the first equation for y by subtracting 6x from each side. Step 1 y + 6x = 11 – 6x y = –6x + 11 Substitute –6x + 11 for y in the second equation. Distribute 2 to the expression in parenthesis. 3x + 2(–6x + 11) = –5 3x + 2y = –5Step 2 3x + 2(–6x + 11) = –5

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Holt Algebra Solving Systems by Substitution Step 3 Example 2 Continued y + 6x = 11 3x + 2y = –5 Solve by substitution. 3x + 2(–6x) + 2(11) = –5 –9x + 22 = –5 Simplify. Solve for x. Subtract 22 from both sides. –9x = –27 – 22 –22 Divide both sides by –9. –9x = –27 –9 x = 3 3x – 12x + 22 = –5

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Holt Algebra Solving Systems by Substitution Step 4 y + 6x = 11 Substitute 3 for x. y + 6(3) = 11 Subtract 18 from each side. y + 18 = 11 –18 –18 y = –7 Step 5(3, –7) Write the solution as an ordered pair. Simplify. Example 2 Continued y + 6x = 11 3x + 2y = –5 Solve by substitution. Write one of the original equations.

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Holt Algebra Solving Systems by Substitution Check It Out! Example 2 –2x + y = 8 3x + 2y = 9 Solve by substitution. Solve the first equation for y by adding 2x to each side. Step 1 –2x + y = 8 + 2x +2x y = 2x + 8 Substitute 2x + 8 for y in the second equation. 3x + 2(2x + 8) = 9 3x + 2y = 9Step 2 Distribute 2 to the expression in parenthesis. 3x + 2(2x + 8) = 9

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Holt Algebra Solving Systems by Substitution Step 3 3x + 2(2x) + 2(8) = 9 7x + 16 = 9 Simplify. Solve for x. Subtract 16 from both sides. 7x = –7 –16 Divide both sides by 7. 7x = –7 7 x = –1 Check It Out! Example 2 Continued –2x + y = 8 3x + 2y = 9 Solve by substitution. 3x + 4x + 16 = 9

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Holt Algebra Solving Systems by Substitution Step 4 –2x + y = 8 Substitute –1 for x. –2(–1) + y = 8 y + 2 = 8 –2 y = 6 Step 5(–1, 6) Write the solution as an ordered pair. Check It Out! Example 2 Continued –2x + y = 8 3x + 2y = 9 Solve by substitution. Subtract 2 from each side. Simplify. Write one of the original equations.

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Holt Algebra Solving Systems by Substitution Example 2: Consumer Economics Application Jenna is deciding between two cell-phone plans. The first plan has a $50 sign-up fee and costs $20 per month. The second plan has a $30 sign-up fee and costs $25 per month. After how many months will the total costs be the same? What will the costs be? If Jenna has to sign a one-year contract, which plan will be cheaper? Explain. Write an equation for each option. Let t represent the total amount paid and m represent the number of months.

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Holt Algebra Solving Systems by Substitution Example 2 Continued Total paid is sign- up fee plus payment amount for each month. Option 1t=$50 + $20 m Option 2 t = $30 + $25 m Step 1 t = m t = m Both equations are solved for t. Step m = m Substitute m for t in the second equation.

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Holt Algebra Solving Systems by Substitution Step m = m Solve for m. Subtract 20m from both sides. –20m – 20m 50 = m Subtract 30 from both sides. –30 –30 20 = 5m Divide both sides by 5. Write one of the original equations. Step 4t = m t = (4) t = t = 130 Substitute 4 for m. Simplify. Example 2 Continued 5 m = 4 20 = 5m

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Holt Algebra Solving Systems by Substitution Step 5 (4, 130) Write the solution as an ordered pair. In 4 months, the total cost for each option would be the same $130. Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330. Example 2 Continued Option 1: t = (12) = 290 Option 2: t = (12) = 330 If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.

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Holt Algebra Solving Systems by Substitution Check It Out! Example 3 One cable television provider has a $60 setup fee and $80 per month, and the second has a $160 equipment fee and $70 per month. a. In how many months will the cost be the same? What will that cost be. Write an equation for each option. Let t represent the total amount paid and m represent the number of months.

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Holt Algebra Solving Systems by Substitution Total paid is fee plus payment amount for each month. Option 1t=$60 + $80 m Option 2 t = $160 + $70 m Check It Out! Example 3 Continued Step 1 t = m t = m Both equations are solved for t. Step m = m Substitute m for t in the second equation.

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Holt Algebra Solving Systems by Substitution Step m = m Solve for m. Subtract 70m from both sides. –70m –70m m = 160 Subtract 60 from both sides. Divide both sides by 10. –60 10m = m = 10 Write one of the original equations. Step 4t = m t = (10) t = t = 860 Substitute 10 for m. Simplify. Check It Out! Example 3 Continued

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Holt Algebra Solving Systems by Substitution Step 5(10, 860) Write the solution as an ordered pair. In 10 months, the total cost for each option would be the same, $860. The first option is cheaper for the first six months. Check It Out! Example 3 Continued Option 1: t = (6) = 540 Option 2: t = (6) = 580 b. If you plan to move in 6 months, which is the cheaper option? Explain.

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Holt Algebra Solving Systems by Substitution Lesson Quiz: Part I Solve each system by substitution (1, 2) (–2, –4) y = 2x x = 6y – 11 3x – 2y = –1 –3x + y = –1 x – y = 4

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Holt Algebra Solving Systems by Substitution Lesson Quiz: Part II 4. Plumber A charges $60 an hour. Plumber B charges $40 to visit your home plus $55 for each hour. For how many hours will the total cost for each plumber be the same? How much will that cost be? If a customer thinks they will need a plumber for 5 hours, which plumber should the customer hire? Explain. 8 hours; $480; plumber A: plumber A is cheaper for less than 8 hours.

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