Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 Solve using substitution. y = 2x + 2 y = –3x + 4 Step 1: Write an equation containing only one variable and solve. y = 2x + 2 Start with one equation. –3x + 4 = 2x + 2 Substitute –3x + 4 for y in that equation. 4 = 5x + 2 Add 3x to each side. 2 = 5x Subtract 2 from each side. 0.4 = x Divide each side by 5. Step 2: Solve for the other variable. y = 2(0.4) + 2 Substitute 0.4 for x in either equation. y = 0.8 + 2 Simplify. y = 2.8 7-2

Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 (continued) Since x = 0.4 and y = 2.8, the solution is (0.4, 2.8). Check: See if (0.4, 2.8) satisfies y = –3x + 4 since y = 2x + 2 was used in Step 2. 2.8 –3(0.4) + 4 Substitute (0.4, 2.8) for (x, y) in the equation. 2.8 –1.2 + 4 2.8 = 2.8 7-2

Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 Solve using substitution. –2x + y = –1 4x + 2y = 12 Step 1: Solve the first equation for y because it has a coefficient of 1. –2x + y = –1 y = 2x –1 Add 2x to each side. Step 2: Write an equation containing only one variable and solve. 4x + 2y = 12 Start with the other equation. 4x + 2(2x –1) = 12 Substitute 2x –1 for y in that equation. 4x + 4x –2 = 12 Use the Distributive Property. 8x = 14 Combine like terms and add 2 to each side. x = 1.75 Divide each side by 8. 7-2

Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 (continued) Step 3: Solve for y in the other equation. –2(1.75) + y = 1 Substitute 1.75 for x. –3.5 + y = –1 Simplify. y = 2.5 Add 3.5 to each side. Since x = 1.75 and y = 2.5, the solution is (1.75, 2.5). 7-2

Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 A youth group with 26 members is going to the beach. There will also be five chaperones that will each drive a van or a car. Each van seats 7 persons, including the driver. Each car seats 5 persons, including the driver. How many vans and cars will be needed? Let v = number of vans and c = number of cars. Drivers v + c = 5 Persons 7 v + 5 c = 31 Solve using substitution. Step 1: Write an equation containing only one variable. v + c = 5 Solve the first equation for c. c = –v + 5 7-2

Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 (continued) Step 2: Write and solve an equation containing the variable v. 7v + 5c = 31 7v + 5(–v + 5) = 31 Substitute –v + 5 for c in the second equation. 7v – 5v + 25 = 31 Solve for v. 2v + 25 = 31 2v = 6 v = 3 Step 3: Solve for c in either equation. 3 + c = 5 Substitute 3 for v in the first equation. c = 2 7-2

Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 (continued) Three vans and two cars are needed to transport 31 persons. Check: Is the answer reasonable? Three vans each transporting 7 persons is 3(7), of 21 persons. Two cars each transporting 5 persons is 2(5), or 10 persons. The total number of persons transported by vans and cars is 21 + 10, or 31. The answer is correct. 7-2

Solve each system using substitution. ALGEBRA 1 LESSON 7-2 Solve each system using substitution. 1. 5x + 4y = 5 2. 3x + y = 4 3. 6m – 2n = 7 y = 5x 2x – y = 6 3m + n = 4 (0.2, 1) (2, 2) (1.25, 0.25) 7-2

Solving Systems Using Substitution ALGEBRA 1 LESSON 7-3 Solve each system using substitution. 1. y = 4x – 3 2. y + 5x = 4 3. y = –2x + 2 y = 2x + 13 y = 7x – 20 3x – 17 = 2y 7-3

Solving Systems Using Substitution ALGEBRA 1 LESSON 7-3 Solutions 1. y = 4x – 3 y = 2x + 13 Substitute 4x – 3 for y in the second equation. 4x – 3 = 2x + 13 4x – 2x – 3 = 2x – 2x + 13 2x – 3 = 13 2x = 16 x = 8 y = 4x – 3 = 4(8) – 3 = 32 – 3 = 29 Since x = 8 and y = 29, the solution is (8, 29). 7-3

Solving Systems Using Substitution ALGEBRA 1 LESSON 7-3 Solutions (continued) 2. y + 5x = 4 y = 7x – 20 Substitute 7x – 20 for y in the first equation. y + 5x = 4 7x – 20 + 5x = 4 12x – 20 = 4 12x = 24 x = 2 y = 7x – 20 = 7(2) – 20 = 14 – 20 = –6 Since x = 2 and y = –6, the solution is (2, –6). 7-3

Solving Systems Using Substitution ALGEBRA 1 LESSON 7-3 Solutions (continued) 3. y = –2x + 2 3x – 17 = 2y Substitute –2x + 2 for y in the second equation. 3x – 17 = 2(–2x + 2) 3x – 17 = –4x + 4 7x – 17 = 4 7x = 21 x = 3 y = –2x + 2 = –2(3) + 2 = –6 + 2  –4 Since x = 3 and y = –4, the solution is (3, –4). 7-3