Solving Systems Using Substitution

1. Solving Systems Using Substitution ALGEBRA 1 LESSON 9-2 (For help, go to Lessons2-4 and 7-1.) Solve each equation. 1.m– 6 = 4m + 82. 4n= 9 – 2n3.t+ 5 = 10 1 3 For each system, is the ordered pair a solution of both equations? 4. (5, 1) y= –x + 45. (2, 2.4) 4x+ 5y = 20 y = x – 6 2x+ 6y = 10 9-2

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

3. Solving Systems Using Substitution ALGEBRA 1 LESSON 9-2 Solutions(continued) 4. (5, 1) in y= –x + 4 (5, 1) in y= x – 6 1 –5 + 4 1 5 – 6 1 = –1 1 = –1 no no No, (5, 1) is not a solution of both equations. 5. (2, 2.4) in 4x+5y = 20 (2, 2.4) in 2x+6y = 10 4(2) + 5(2.4) 20 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

4. Solving Systems Using Substitution ALGEBRA 1 LESSON 9-2 Solve using substitution. y= 2x + 2 y= –3x + 4 Step 1:Write an equation containing only one variable and solve. y= 2x + 2Start with one equation. –3x + 4 = 2x + 2Substitute –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 + 2Simplify. y = 2.8 9-2

5. 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 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). 9-2

6. Solving Systems Using Substitution ALGEBRA 1 LESSON 9-2 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 –1Add 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. 9-2

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

8. Let v= number of vans and c = number of cars. Drivers v + c = 5 Persons 7 v + 5 c = 31 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? 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

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

10. 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

11. Solving Systems Using Substitution ALGEBRA 1 LESSON 9-2 Solve each system using substitution. 1. 5x+ 4y = 52. 3x+ y = 43. 6m– 2n = 7 y = 5x 2x– y = 6 3m+ n = 4 (0.2, 1) (2, 2) (1.25, 0.25) 9-2