Solving Systems Using Solve each equation Substitution ALGEBRA

Solving Systems Using Solve each equation. Substitution ALGEBRA 1 LESSON 9 -2 (For help, go to Lessons 2 -4 and 7 -1. ) 1. m – 6 = 4 m + 8 2. 4 n = 9 – 2 n 1 3. 3 t + 5 = 10 For each system, is the ordered pair a solution of both equations? 4. (5, 1) y = –x + 4 5. (2, 2. 4) 4 x + 5 y = 20 y=x– 6 2 x + 6 y = 10 9 -2

Solving Systems Using Substitution ALGEBRA 1 LESSON 9 -2 Solutions 1. m – 6 = 4 m + 8 2. m – 6 = 4 m – m + 8 – 6 = 3 m + 8 4 n + 2 n = 9 – 2 n + 2 n 6 n = 9 n = 11 – 14 = 3 m – 4 2 = m 2 3 3. 4 n = 9 – 2 n 1 t + 5 = 10 3 1 t =5 3 t = 15 9 -2

Solving Systems Using Substitution Solutions (continued) ALGEBRA 1 LESSON 9 -2 4. (5, 1) in y = –x + 4 1 (5, 1) in y = x – 6 – 5 + 4 1 =/ – 1 1 5– 6 1 =/ – 1 no no No, (5, 1) is not a solution of both equations. 5. (2, 2. 4) in 4 x + 5 y = 20 4(2) + 5(2. 4) 20 (2, 2. 4) in 2 x + 6 y = 10 2(2) + 6(2. 4) 10 8 + 12 20 4 + 14. 4 10 20 = 20 18. 4 =/ 10 yes no No, (2, 2. 4) is not a solution of both equations. 9 -2

Solving Systems Using Substitution ALGEBRA 1 LESSON 9 -2 Solve using substitution. y = – 3 x + 4 y = 2 x + 2 Step 1: Write an equation containing only one variable and solve. y = 2 x + 2 – 3 x + 4 = 2 x + 2 4 = 5 x + 2 2 = 5 x 0. 4 = x Start with one equation. Substitute – 3 x + 4 for y in that equation. Add 3 x to each side. Subtract 2 from each side. 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. 9 -2

Solving Systems Using Substitution ALGEBRA 1 LESSON 9 -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 = – 3 x + 4 since y = 2 x + 2 was used in Step 2. 2. 8 – 3(0. 4) + 4 2. 8 – 1. 2 + 4 2. 8 = 2. 8 Substitute (0. 4, 2. 8) for (x, y) in the equation. 9 -2

Solving Systems Using Substitution ALGEBRA 1 LESSON 9 -2 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 x = 1. 75 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. Divide each side by 8. 9 -2

Solving Systems Using Substitution ALGEBRA 1 LESSON 9 -2 (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). 9 -2

Solving Systems Using Substitution ALGEBRA 1 LESSON 9 -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 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 9 -2

Solving Systems Using Substitution ALGEBRA 1 LESSON 9 -2 (continued) Step 2: Write and solve an equation containing the variable v. 7 v + 5 c = 31 7 v + 5(–v + 5) = 31 7 v – 5 v + 25 = 31 2 v = 6 v =3 Substitute –v + 5 for c in the second equation. Solve for v. Step 3: Solve for c in either equation. 3+c=5 c=2 Substitute 3 for v in the first equation. 9 -2

Solving Systems Using Substitution ALGEBRA 1 LESSON 9 -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. 9 -2

Solving Systems Using Substitution Solve each system using substitution. ALGEBRA 1 LESSON 9 -2 1. 5 x + 4 y = 5 2. 3 x + y = 4 3. 6 m – 2 n = 7 y = 5 x 2 x – y = 6 3 m + n = 4 (0. 2, 1) (2, 2) (1. 25, 0. 25) 9 -2