1. 3-2 Solving Systems Algebraically Hubarth Algebra II

2. Solve using substitution. y = 2x + 2 y = -x + 5 Step 1:Write an equation containing only one variable and solve. y = 2x + 2 –x + 5= 2x + 2Substitute –x + 5 for y in that equation. 5 = 3x + 2Add x to each side. 1 = xDivide each side by 3. Step 2:Solve for the other variable. y = 2(1) + 2Substitute 1 for x in either equation. 3 = 3x Subtract 2 from each side. y = 2 + 2 Simplify. y = 4 Ex 1 Solve Systems by Substitution y = -x + 5 The solution is ( 1, 4)

3. Ex 2 Using Substitution and the Distributive Property Step 1:Solve the first equation for x because it has a coefficient of -1. Step 2:Write an equation containing only one variable and solve. y = 3 Divide each side by 6. Step 3:Solve for y in the other equation. -x + 2(3) = 8 Substitute 3 for y. -x + 6 =8 Simplify and divide each side by -1. x = -2 Subtract 2 from each side. (-2, 3)

4. Ex 3 Solve System by Using Substitution y = 4x – 1 5 = 6x – y 5 = 6x – (4x – 1) 5 = 6x – 4x + 1 5 = 2x +1 () ( ) 4 = 2x x = 2 y = 4 (2) – 1 y = 8 – 1 y = 7 (2, 7) is the solution of the system

5. Solve by elimination.2x + 3y = 11 –2x + 9y = 1 Step 1:Eliminate x because the sum of the coefficients is 0. 2x + 3y = 11 –2x + 9y =1 0 + 12y = 12Addition Property of Equality y = 1Solve for y. Step 2:Solve for the eliminated variable x using either original equation. 2x + 3y = 11Choose the first equation. 2x + 3(1) = 11Substitute 1 for y. 2x + 3 = 11 Solve for x. 2x = 8 x = 4 Ex 4 Adding Equations Since x = 4 and y = 1, the solution is (4, 1). Check:See if (4, 1) makes the equation not used in Step 2 true. –2(4) + 9(1) 1 Substitute 4 for x and 1 for y into the second equation. –8 + 9 1 1 = 1

6. Define:Let a = number of adults Let s = number of students Relate:total number at the gametotal amount collected Write: a + s = 1139 5 a +1 s = 3067 On a special day, tickets for a minor league baseball game cost $5 for adults and $1 for students. The attendance that day was 1139, and $3067 was collected. Write and solve a system of equations to find the number of adults and the number of students that attended the game. Ex 5 Application Step 2:Solve for the eliminated variable using either of the original equations. a + s = 1139Choose the first equation. 482 + s = 1139Substitute 482 for a. s = 657Solve for s. There were 482 adults and 657 students at the game. a + s = 1139 5a + s = 3067 –4a + 0 = –1928Subtraction Property of Equality a = 482Solve for a. Step 1:Eliminate one variable.

7. Solve by elimination.3x + 6y = –6 –5x – 2y = –14 Step 1:Eliminate one variable. Start with the given system. 3x + 6y = –6 –5x – 2y = –14 Add the equations to eliminate y. 3x + 6y = –6 –15x – 6y = –42 –12x + 0 = –48 To prepare to eliminate y, multiply the second equation by 3. 3x + 6y = –6 3(–5x – 2y = –14) Step 2:Solve for x. –12x = –48 x = 4 Ex 6 Multiplying One Equation Step 3:Solve for the eliminated variable using either of the original equations. 3x + 6y = –6Choose the first equation. 3(4) + 6y = –6Substitute 4 for x. 12 + 6y = –6Solve for y. 6y = –18 y = –3 The solution is (4, –3).

8. Solve by elimination.3x + 5y = 10 5x + 7y = 10 Step 1:Eliminate one variable. Subtract the equations to eliminate x. 15x + 25y = 50 -15x - 21y = -30 0 + 4y = 20 Start with the given system. 3x + 5y = 10 5x + 7y = 10 To prepare to eliminate x, multiply one equation by 5 and the other equation by -3. 5(3x + 5y = 10) -3(5x + 7y = 10) Step 2:Solve for y. 4y = 20 y = 5 Ex 7 Multiplying Both Equations 3x + 5y = 10Use the first equation. 3x + 5(5) = 10Substitute 5 for y. 3x + 25 = 10 3x = –15 x = –5 The solution is (–5, 5). Step 3:Solve for the eliminated variable x using either of the original equations.

9. Practice 1. Solve by Elimination a.x + y = 10 x – y = 8 b. -2x + 15y = -32 7x – 5y = 17 c. 15x + 3y = 9 10x + 7y = -4 2. At Renaldi’s Pizza, a soda and two slices of pizza-of-the-day costs $10.25. A soda and four slices of the pizza-of-the-day costs $18.75. Find the cost of each item? (9, 1)(1, -2) x = 1.75, y = 4.25 3. x – 5 = y 2x – 3y = 7 (8, 3) 4. x = 4y – 1 3x + 5y = 31 (7, 2) (-1, 3)

10. Practice 1.y = x – 3 x + y = 5 2. 2x + y = 1 x = -2y + 5 3. x – 5 = y 2x – 3y = 7 4. x = 4y – 1 3x + 5y = 31 (4, 1)(-1, 3) (8, 3) (7, 2)