ALGEBRA 1 LESSON 7 2 Solving Systems Using

ALGEBRA 1 LESSON 7 -2 Solving Systems Using Substitution Solve using substitution. y = 2 x + 2 y = – 3 x + 4 Step 1: Write an equation containing only one variable and solve. y = 2 x + 2 Start with one equation. – 3 x + 4 = 2 x + 2 Substitute – 3 x + 4 for y in that equation. 4 = 5 x + 2 Add 3 x to each side. 2 = 5 x 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 y = 0. 8 + 2 y = 2. 8 Substitute 0. 4 for x in either equation. Simplify. 7 -2

ALGEBRA 1 LESSON 7 -2 Solving Systems Using Substitution (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 = – 3 x + 4 since y = 2 x + 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

ALGEBRA 1 LESSON 7 -2 Solving Systems Using Substitution Solve using substitution. – 2 x + y = – 1 4 x + 2 y = 12 Step 1: Solve the first equation for y because it has a coefficient of 1. – 2 x + y = – 1 y = 2 x – 1 Add 2 x to each side. Step 2: Write an equation containing only one variable and solve. 4 x + 2 y = 12 4 x + 2(2 x – 1) = 12 4 x + 4 x – 2 = 12 8 x = 14 Start with the other equation. Substitute 2 x – 1 for y in that equation. Use the Distributive Property. Combine like terms and add 2 to each side. x = 1. 75 Divide each side by 8. 7 -2

ALGEBRA 1 LESSON 7 -2 Solving Systems Using Substitution (continued) Step 3: Solve for y in the other equation. – 2(1. 75) + y = 1 – 3. 5 + y = – 1 y = 2. 5 Substitute 1. 75 for x. Simplify. Add 3. 5 to each side. Since x = 1. 75 and y = 2. 5, the solution is (1. 75, 2. 5). 7 -2

ALGEBRA 1 LESSON 7 -2 Solving Systems Using Substitution 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 Persons v 7 v + + c 5 c = 5 = 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

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

ALGEBRA 1 LESSON 7 -2 Solving Systems Using Substitution (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

ALGEBRA 1 LESSON 7 -2 Solve each system using substitution. 1. 5 x + 4 y = 5 2. 3 x + y = 4 3. 6 m – 2 n = 7 (0. 2, 1)y = 5 x 2 x – (2, y = 2) 6 7 -2 3 m + n 0. 25) =4 (1. 25,

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

Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -3 Solutions 1. y = 4 x – 3 y = 2 x + 13 Substitute 4 x – 3 for y in the second equation. y = 2 x + 13 4 x – 3 = 2 x + 13 4 x – 2 x – 3 = 2 x – 2 x + 13 2 x – 3 = 13 2 x = 16 x=8 y = 4 x – 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 + 5 x = 4 y = 7 x – 20 Substitute 7 x – 20 for y in the first equation. y + 5 x = 4 7 x – 20 + 5 x = 4 12 x – 20 = 4 12 x = 24 x=2 y = 7 x – 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 = – 2 x + 2 3 x – 17 = 2 y Substitute – 2 x + 2 for y in the second equation. 3 x – 17 = 2 y 3 x – 17 = 2(– 2 x + 2) 3 x – 17 = – 4 x + 4 7 x – 17 = 4 7 x = 21 x = 3 y = – 2 x + 2 = – 2(3) + 2 = – 6 + 2 – 4 Since x = 3 and y = – 4, the solution is (3, – 4). 7 -3