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Solving Systems by Substitution

This lesson covers solving systems of linear equations in two variables by substitution. Learn how to solve for one variable and substitute the result into the other equation. Includes examples and practice problems.

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Solving Systems by Substitution

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  1. 5-2 Solving Systems by Substitution Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 1 Holt Algebra 1

  2. Warm Up Solve for x. 1.y = x + 3 2.y = 3x – 4 Simplify each expression. x = y – 3 2x – 10 3. 2(x– 5) 4. 12 – 3(x + 1) 9 – 3x

  3. Warm Up Continued Evaluate each expression for the given value of x. 5.x + 8 for x = 6 6. 3(x –7) for x =10 12 9

  4. Objective Solve systems of linear equations in two variables by substitution.

  5. 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.

  6. Solve for one variable in at least one equation, if necessary. Step 1 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.

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

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

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

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

  11. −2y −2y x = –2y– 1 Step 2 x– y = 5 (–2y – 1) –y= 5 Example 1C: Solving a System of Linear Equations by Substitution Solve the system by substitution. x + 2y= –1 x – y = 5 Step 1x + 2y = –1 Solve the first equation for x by subtracting 2y from both sides. Substitute –2y –1 for x in the second equation. –3y –1 = 5 Simplify.

  12. +1 +1 –3y = 6 –3y = 6 –3 –3 y = –2 Step 4 x –y = 5 x – (–2) = 5 x + 2 = 5 –2 –2 x =3 Example 1C Continued Step 3 Solve for y. –3y –1= 5 Add 1 to both sides. Divide both sides by –3. Write one of the original equations. Substitute –2 for y. Subtract 2 from both sides. Write the solution as an ordered pair. Step 5 (3, –2)

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

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

  15. Step 2 x + 8y = 16 (2y –4) + 8y = 16 Step 3 10y – 4 = 16 +4 +4 10y 20 10y = 20 = 10 10 Check It Out! Example 1b Solve the system by substitution. x= 2y – 4 x + 8y = 16 Step 1x = 2y –4 The first equation is solved for x. Substitute 2y – 4 for x in the second equation. Simplify. Then solve for y. Add 4 to both sides. Divide both sides by 10. y = 2

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

  17. – y – y x = –y –7 Step 2 x = –y –7 2(–y – 7) + y = –4 Distribute 2. 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. Step 1x + y = –7 Substitute –y – 7 for x in the first equation. 2(–y – 7) + y = –4 –2y – 14 + y = –4

  18. x + y = –7 Step 4 +14 +14 –y = 10 Check It Out! Example 1c Continued Solve the system by substitution. Step 3 –2y –14 + y = –4 Combine like terms. –y –14 = –4 Add 14 to each side. y = –10 Write one of the original equations. x +(–10) = –7 Substitute –10 for y. x –10 = – 7

  19. +10 +10 Check It Out! Example 1c Continued Solve the system by substitution. Step 5 x –10 = –7 Add 10 to both sides. x = 3 Step 6 (3, –10) Write the solution as an ordered pair.

  20. 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.

  21. Caution When you solve one equation for a variable, you must substitute the value or expression into the other originalequation, not the one that had just been solved.

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

  23. – 22 –22 –9x = –27 –9x = –27 –9–9 Example 2 Continued y + 6x = 11 Solve by substitution. 3x + 2y = –5 Step 3 Simplify. Solve for x. 3x + 2(–6x) +2(11) = –5 3x –12x + 22 = –5 –9x + 22 = –5 Subtract 22 from both sides. Divide both sides by –9. x =3

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

  25. + 2x +2x y = 2x + 8 Step 2 3x + 2y = 9 3x+ 2(2x + 8) = 9 Distribute 2 to the expression in parentheses. Check It Out! Example 2 –2x + y = 8 Solve by substitution. 3x + 2y = 9 Step 1–2x + y = 8 Solve the first equation for y by adding 2x to each side. Substitute 2x + 8 for y in the second equation. 3x + 2(2x + 8) = 9

  26. –16 –16 7x = –7 7x = –7 77 Check It Out! Example 2 Continued –2x + y = 8 Solve by substitution. 3x + 2y = 9 Step 3 3x + 2(2x) +2(8) = 9 Simplify. Solve for x. 3x + 4x + 16 = 9 7x + 16 = 9 Subtract 16 from both sides. Divide both sides by 7. x =–1

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

  28. 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.

  29. + Option 1 t = $50 m $20 $30 $25 t + Option 2 = m Step 1 t = 50 + 20m t = 30 + 25m Step 2 50 + 20m = 30 + 25m Example 3 Continued Total paid sign-up fee for each month. payment amount is plus Both equations are solved for t. Substitute 50 + 20m for t in the second equation.

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

  31. Example 3 Continued Write the solution as an ordered pair. Step 5 (4, 130) In 4 months, the total cost for each option would be the same $130. If Jenna has to sign a one-year contract, which plan will be cheaper? Explain. Option 1: t = 50 + 20(12) = 290 Option 2: t = 30 + 25(12) = 330 Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330.

  32. Check It Out! Example 3 One cable television provider has a $60 setup fee and charges $80 per month, and the second has a $160 equipment fee and charges $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.

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

  34. –70m –70m 60 + 10m = 160 –60 –60 10 10 m =10 Step 4 t = 160 + 70m t = 160 + 700 t = 860 Check It Out! Example 3 Continued Step 3 60 + 80m = Solve for m. Subtract 70m from both sides. 160 + 70m Subtract 60 from both sides. 10m = 100 Divide both sides by 10. Write one of the original equations. t = 160 + 70(10) Substitute 10 for m. Simplify.

  35. Check It Out! Example 3 Continued 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. b. If you plan to move in 6 months, which is the cheaper option? Explain. Option 1: t = 60 + 80(6) = 540 Option 2: t = 160 + 70(6) = 580 The first option is cheaper for the first six months.

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

  37. 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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