Lesson 6 2 WarmUp ALGEBRA 1 Solving Systems

Lesson 6 -2 Warm-Up ALGEBRA 1

“Solving Systems Using Substitution” (6 -2) How do you use the substitution method to find a solution for a system of equations? Method 2: Substitution – Since the right side of both equations equal y, then the right sides of both equations equal each other. This will create a one variable equation which can be solved in terms of that variable (x = ___) Example: Solve y = 2 x – 3 and y = x – 1 Step 1: Use substitution to write an equation with one variable, x, and solve for that variable. (2 x – 3) = x - 1 Substitute 2 x – 3 for y in the second equation. -x -x Subtract x from both sides. x-3=0 -1 +3 +3 Add 3 to both sides. x = 2 Step 2: Solve for the other variable, y, using either equation. y = 2 x - 3 Given y = 2(2) - 3 Substitute 2 for x y= 1 Simplify Since x = 2 and y = 1, the solution is (2, 1) Step 3: Check if the solution makes both equations true statements. y = 2 x - 3 y=x-1 Given 1 = 2(2) – 3 1 = (2) – 1 Substitute 1 for y and 2 for x 1 = 1 ALGEBRA 1

Solving Systems Using Substitution LESSON 6 -2 Additional Examples 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 Start with one equation. (– 3 x + 4) = 2 x + 2 Substitute – 3 x + 4 for y in that equation. +3 x Addition Property of Equality 4 = 5 x + 2 Simplify. -2 -2 Subtraction Property of Equality 2 = 5 x Simplify. 0. 4 = x Division Property of Equality Step 2: Solve for the other variable. y = 2(0. 4) + 2 Substitute 0. 4 for x in either equation. y = 0. 8 + 2 y = 2. 8 Simplify. ALGEBRA 1

Solving Systems Using Substitution LESSON 6 -2 Additional Examples (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. y – 3 x + 4 2. 8 – 3(0. 4) + 4 2. 8 – 1. 2 + 4 Given. Substitute (0. 4, 2. 8) for (x, y) in the equation. 2. 8 = 2. 8 ALGEBRA 1

“Solving Systems Using Substitution” (6 -2) How do you use the To find the solution for a system of equations that are not in y equals form (where (6 -1) substitution method y is isolated), solve one equation for y (if possible, choose the one in which the to find a solution for a system of equations if the equations aren’t in y = ___ form? coefficient and constant won’t be fractions) and substitute for y in the other equation. Example: Solve 3 y + 2 x = 4 and -6 x + y = – 7 Step 1: Solve the second equation for y because it has a coefficient of 1. -6 x + y = – 7 Given +6 x Add 6 x to both sides. y = 6 x - 7 Step 2: Use substitution to write an equation with a single variable, x. 3 y + 2 x = 4 Given 3(6 x - 7) + 2 x = 4 18 x – 21 + 2 x = 4 20 x - 21 = 4 +21 20 x + 0 = 25 20 20 x = 1. 25 y = 6 x - 7 (Substitute) Distributive Property Combine x terms Addition Property of Equality Simplify Division Property of Equality ALGEBRA 1

“Solving Systems Using Substitution” (6 -2) Step 3: Solve for the other variable, y, using either equation. (6 -1)Given y = 6 x - 7 y = 6(1. 25) - 7 y = 0. 5 x = 1. 25 (Substitute) Simplify Since x = 1. 25 and y = 0. 5, the solution is (0. 5, 1. 25) Step 4: Check if the solution makes both equations true statements 3 y + 2 x = 4 -6 x + y = – 7 3(0. 5) + 2(1. 25) = 4 -6(1. 25) + (0. 5) = – 7 Substitute (y = 1. 25 and x = 0. 5) 4 = 4 -7 = -7 ALGEBRA 1

Solving Systems Using Substitution LESSON 6 -2 Additional Examples 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 (1 y). – 2 x + y = – 1 +2 x Addition Property of Equality y = 2 x – 1 Simplify. Step 2: Write an equation containing only one variable and solve. 4 x + 2 y = 12 Start with the other equation. 4 x + 2(2 x – 1) = 12 Substitute 2 x – 1 for y in that equation. 4 x + 4 x – 2 = 12 Use the Distributive Property. 8 x =__ 14 Combine like terms and add 2 to each side. __ 8 8 Division Property of Equality x = 1. 75 Simplify Step 3: Solve for y in the other equation. – 2(1. 75) + y = – 1 Substitute 1. 75 for x. – 3. 5 + y = – 1 Simplify. +3. 5 Add 3. 5 to each side. y = 2. 5 Simplify Since x = 1. 75 and y = 2. 5, the solution is (1. 75, 2. 5). ALGEBRA 1

Solving Systems Using Substitution LESSON 6 -2 Additional Examples 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 ALGEBRA 1

Solving Systems Using Substitution LESSON 6 -2 Additional Examples (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 + 25 = 31 - 25 -25 2 v = 6 v =3 Substitute –v + 5 for c in the second equation. Solve for v. Combine like terms Subtraction Property of Equality Divide both sides by 3 Step 3: Solve for c in either equation. (3) + c = 5 Substitute 3 for v in the first equation. c=2 ALGEBRA 1

Solving Systems Using Substitution LESSON 6 -2 Additional Examples (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), or 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. ALGEBRA 1

Solving Systems Using Substitution LESSON 6 -2 Lesson Quiz Solve each system using substitution. 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) ALGEBRA 1