ALGEBRA 1 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 (an x term on both sides) which can be solved in terms of that variable (x = ___) Example: Solve y = 2x – 3 and y = x – 1 Step 1: Use substitution to write an equation with one variable, x, and solve for that variable. (2x – 3) = x - 1Substitute 2x – 3 for y in the second equation. -x - x Subtract x from both sides. x - 3 = Add 3 to both sides. x = 2 Step 2: Solve for the other variable, y, using either equation. y = 2x - 3 Given y = 2(2) - 3 x = 2 (substitute) 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 = 2x - 3 y = x - 1 Given 1 = 2(2) – 31 = 2 – 1 y = 1 and x = 2 (substitute) 1 = 1 

ALGEBRA 1 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 Simplify. 0.4 = xDivide each side by 5. Step 2:Solve for the other variable. y = 2(0.4) + 2Substitute 0.4 for x in either equation. 2 = 5x Simplify. y = Simplify. y = 2.8 Solving Systems Using Substitution LESSON 6-2 Additional Examples +3x +3xAdd 3x to each side Subtract 2 from each side.

ALGEBRA 1 (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 –3(0.4) + 4 Substitute (0.4, 2.8) for (x, y) in the equation. 2.8 – = 2.8 Solving Systems Using Substitution LESSON 6-2 Additional Examples y –3x + 4 Given.

ALGEBRA 1 “Solving Systems Using Substitution” (6-2) (6-1) How do you use the substitution method to find a solution for a system of equations if the equations aren’t in y = ___ form? To find the solution for a system of equations that are not in y equals form (where y is isolated), solve one equation for y (if possible, choose the one in which the coefficient and constant won’t be fractions) and substitute for y in the other equation. Example: Solve 3y + 2x = 4 and -6x + y = –7 Step 1: Solve the second equation for y because it has a coefficient of 1. -6x + y = –7 Given +6x +6x Add 6x to both sides. y = 6x - 7 Step 2: Use substitution to write an equation with a single variable, x. 3y + 2x = 4 Given 3(6x - 7) + 2x = 4 y = 6x - 7 (Substitute) 18x – x = 4Distributive Property 20x - 21 = 4Combine x terms Add 21 to each side. 20x = 25Simplify Divide both sides by 20. x = 1.25 

ALGEBRA 1 “Solving Systems Using Substitution” (6-2) (6-1) Step 3: Solve for the other variable, y, using either equation. y = 6x - 7Given y = 6(1.25) - 7x = 1.25 (Substitute) y = 0.5 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 3y + 2x = 4 -6x + 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 Solve using substitution. –2x + y = –1 4x + 2y = 12 Step 1:Solve the first equation for y because it has a coefficient of 1 (1y). Step 2:Write an equation containing only one variable and solve. 4x + 2y = 12 Start with the other equation. 4x + 4x – 2 = 12 Use the Distributive Property. –2x + y = –1 y = 2x –1 Simplify. 4x + 2(2x –1) = 12 Substitute 2x –1 for y in that equation. 8x = 14 Combine like terms and add 2 to each side. x = 1.75 Simplify Solving Systems Using Substitution LESSON 6-2 Additional Examples 8x = 14 Divide each side by x +2x Add 2x to each side.

ALGEBRA 1 (continued) Step 3:Solve for y in the other equation. Since x = 1.75 and y = 2.5, the solution is (1.75, 2.5). –2(1.75) + y = –1Substitute 1.75 for x. –3.5 + y = –1 Simplify. y = 2.5 Simplify Solving Systems Using Substitution LESSON 6-2 Additional Examples Add 3.5 to each side.

ALGEBRA 1 Let v = number of vans and c = number of cars. Driversv + c = 5 Persons 7 v + 5 c = 31 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 Solving Systems Using Substitution LESSON 6-2 Additional Examples

ALGEBRA 1 (continued) Step 2:Write and solve an equation containing the variable v. 7v + 5c = 31 7v – 5v + 25 = 31Solve for v. Step 3:Solve for c in either equation. 3 + c = 5Substitute 3 for v in the first equation. 7v + 5(–v + 5) = 31 Substitute –v + 5 for c in the second equation. 2v + 25 = 31 Combine like terms Subtract 25 from both sides v = 3 Divide both sides by 3 c = 2 Solving Systems Using Substitution LESSON 6-2 Additional Examples 2v = 6

ALGEBRA 1 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 , or 31. The answer is correct. (continued) Solving Systems Using Substitution LESSON 6-2 Additional Examples

ALGEBRA 1 Solve each system using substitution. 1.5x + 4y = 52.3x + y = 4 3.6m – 2n = 7 y = 5x2x – y = 6 3m + n = 4 (0.2, 1) (2,  2) (1.25, 0.25) Solving Systems Using Substitution LESSON 6-2 Lesson Quiz