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

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

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

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

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 1 y = x + 1 Step 2 4x + y = 6 4x + (x + 1) = 6 Substitute x + 1 for y in the second equation. 5x + 1 = 6 Simplify. Solve for x. Step 3 –1 –1 5x = 5 5 5 x = 1 5x = 5 Subtract 1 from both sides. Divide both sides by 5.

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

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

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

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

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

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

Check It Out! Example 1b Continued Solve the system by substitution. Step 4 x + 8y = 16 Write one of the original equations. x + 8(2) = 16 Substitute 2 for y. x + 16 = 16 Simplify. x = 0 – 16 –16 Subtract 16 from both sides. Write the solution as an ordered pair. Step 5 (0, 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 1 x + y = –7 – y – y x = –y – 7 2(–y – 7) + y = –4 x = –y – 7 Step 2 Substitute –y – 7 for x in the first equation. Distribute 2. 2(–y – 7) + y = –4 –2y – 14 + y = –4

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

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

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. Step 1 y + 6x = 11 – 6x – 6x y = –6x + 11 3x + 2(–6x + 11) = –5 3x + 2y = –5 Step 2 Substitute –6x + 11 for y in the second equation. Distribute 2 to the expression in parentheses. 3x + 2(–6x + 11) = –5

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

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

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. + 2x +2x y = 2x + 8 3x + 2(2x + 8) = 9 3x + 2y = 9 Step 2 Substitute 2x + 8 for y in the second equation. Distribute 2 to the expression in parentheses. 3x + 2(2x + 8) = 9

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 7x = –7 –16 –16 Subtract 16 from both sides. 7x = –7 7 7 Divide both sides by 7. x = –1

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

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.

Example 3 Continued Total paid sign-up fee payment amount for each month. is plus Option 1 t = $50 + $20 m Option 2 t = $30 + $25 m 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.

Example 3 Continued 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. 5 5 m = 4 20 = 5m Step 4 t = 30 + 25m Write one of the original equations. t = 30 + 25(4) Substitute 4 for m. t = 30 + 100 t = 130 Simplify.

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.

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.

Check It Out! Example 3 Continued Total paid payment amount for each month. is fee plus Option 1 t = $60 + $80 m Option 2 t = $160 + $70 m 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.

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

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.

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

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.