 # 6-2 Solving Systems by Substitution Warm Up Warm Up Lesson Presentation Lesson Presentation California Standards California StandardsPreview.

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6-2 Solving Systems by Substitution Warm Up Warm Up Lesson Presentation Lesson Presentation California Standards California StandardsPreview

6-2 Solving Systems by Substitution Warm Up Solve each equation for x. 1. y = x + 32. y = 3x – 4 Simplify each expression. Evaluate each expression for the given value of x. 5. x + 8 for x = 66. 3(x – 7) for x = 10 x = y – 3 2x – 103. 2(x – 5)4. 12 – 3(x + 1)9 – 3x 12 9 2323

6-2 Solving Systems by Substitution 9.0 Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Student are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. California Standards

6-2 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. Substitution is used 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.

6-2 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 for the other variable. Write the values from steps 3 and 4 as an ordered pair, (x, y), and check.

6-2 Solving Systems by Substitution Solve the system by substitution. Additional 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. Now solve this equation for x. Subtract x from both sides and then divide by 2. Step 3–x 2x = –2 2 x = –1

6-2 Solving Systems by Substitution Solve the system by substitution. Additional 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

6-2 Solving Systems by Substitution You can substitute the value of one variable into either of the original equations to find the value of the other variable. Helpful Hint

6-2 Solving Systems by Substitution Solve the system by substitution. Additional 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. 5x = 5 5 5 x = 1 Step 3–1 5x = 5 5x + 1 = 6 Simplify. Solve for x. Divide both sides by 5. Write the second equation.

6-2 Solving Systems by Substitution Solve the system by substitution. Additional Example 1B Continued Step 4y = x + 1 Write one of the original equations. Substitute 1 for x. y = 1 + 1 y = 2 Step 5 (1, 2) Write the solution as an ordered pair.

6-2 Solving Systems by Substitution Solve the system by substitution. Additional 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

6-2 Solving Systems by Substitution Additional 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.

6-2 Solving Systems by Substitution Check It Out! Example 1a Solve the system by substitution. Check your answer. 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

6-2 Solving Systems by Substitution Check It Out! Example 1a Continued Solve the system by substitution. Check your answer. 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. Check Substitute (–2, 1) into both equations in the system. y = x + 3 1 (–2) + 3 1 y = 2x + 5 1 2(–2) + 5 1 –4 + 5 1

6-2 Solving Systems by Substitution Check It Out! Example 1b Solve the system by substitution. Check your answer. 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. +4 10y = 20 y = 2 10y 20 10 = Divide both sides by 10.

6-2 Solving Systems by Substitution Check It Out! Example 1b Continued Solve the system by substitution. Check your answer. 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.

6-2 Solving Systems by Substitution Solve the system by substitution. Check your answer. Check It Out! Example 1b Continued Check Substitute (0, 2) into both equations in the system. x = 2y – 4 0 2(2) – 4 0 x + 8y = 16 0 + 8(2) 16 16 16

6-2 Solving Systems by Substitution Check It Out! Example 1c Solve the system by substitution. Check your answer. 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

6-2 Solving Systems by Substitution Combine like terms. Step 3 +14 –y = 10 Check It Out! Example 1c Continued –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 Solve the system by substitution. Check your answer.

6-2 Solving Systems by Substitution Check It Out! Example 1c Continued Solve the system by substitution. Check your answer. x – 10 = –7Step 5 +10 x = 3 Add 10 to both sides. Step 6(3, –10) Write the solution as an ordered pair. Check Substitute (3, –10) into both equations in the system. 2x + y = –4 2(3) + (–10) –4 –4 –4 x + y = –7 3 + (–10) –7 –7

6-2 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.

6-2 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

6-2 Solving Systems by Substitution Additional 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 parentheses. 3x + 2(–6x + 11) = –5 3x + 2y = –5Step 2 3x + 2(–6x + 11) = –5

6-2 Solving Systems by Substitution Step 3 Additional Example 2 Continued 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 y + 6x = 11 3x + 2y = –5 Solve by substitution.

6-2 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. Additional Example 2 Continued y + 6x = 11 3x + 2y = –5 Solve by substitution. Write one of the original equations.

6-2 Solving Systems by Substitution Check It Out! Example 2 –2x + y = 8 3x + 2y = 9 Solve by substitution. Check your answer. 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 parentheses. 3x + 2(2x + 8) = 9

6-2 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. Check your answer. 3x + 4x + 16 = 9

6-2 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. Check your answer. Subtract 2 from each side. Simplify. Write one of the original equations.

6-2 Solving Systems by Substitution Check It Out! Example 2 Continued –2x + y = 8 3x + 2y = 9 Solve by substitution. Check your answer. Check Substitute (–1, 6) into both equations in the system. 3x + 2y = 9 3(–1) + 2(6) 9 –3 + 12 9 9 –2x + y = 8 –2(–1) + (6) 8 8 8 2 + (6) 8

6-2 Solving Systems by Substitution Additional Example 3: 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.

6-2 Solving Systems by Substitution Additional Example 3 Continued Step 1 t = 50 + 20m t = 30 + 25m Both equations are solved for t. Step 2 50 + 20m = 30 + 25m Substitute 50 + 20m for t in the second equation. Total paid is sign-up fee plus monthly fee months.times Option 1 t=\$50 + \$20 m  Option 2 t = \$30+\$25m 

6-2 Solving Systems by Substitution Step 3 50 + 20m = 30 + 25m Solve for m. Subtract 20m from both sides. –20m – 20m 50 = 30 + 5m Subtract 30 from both sides. –30 –30 20 = 5m Divide both sides by 5. Write one of the original equations. Step 4t = 30 + 25m t = 30 + 25(4) t = 30 + 100 t = 130 Substitute 4 for m. Simplify. Additional Example 3 Continued 5 m = 4 20 = 5m

6-2 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. Additional Example 3 Continued Option 1: t = 50 + 20(12) = 290 Option 2: t = 30 + 25(12) = 330 If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.

6-2 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.

6-2 Solving Systems by Substitution Check It Out! Example 3 Continued Step 1 t = 60 + 80m t = 160 + 70m Both equations are solved for t. Step 2 60 + 80m = 160 + 70m Substitute 60 + 80m for t in the second equation. Total paid isfeeplus monthly fee months. times Option 1t=\$60+\$80m  Option 2t=\$160+\$70m 

6-2 Solving Systems by Substitution Step 3 60 + 80m = 160 + 70m Solve for m. Subtract 70m from both sides. –70m –70m 60 + 10m = 160 Subtract 60 from both sides. Divide both sides by 10. –60 10m = 100 10 m = 10 Write one of the original equations. Step 4t = 160 + 70m t = 160 + 70(10) t = 160 + 700 t = 860 Substitute 10 for m. Simplify. Check It Out! Example 3 Continued

6-2 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 = 60 + 80(6) = 540 Option 2: t = 160 + 270(6) = 580 b. If you plan to move in 6 months, which is the cheaper option? Explain.

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

6-2 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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