3-2 Warm Up Lesson Presentation Lesson Quiz Using Algebraic Methods to Solve Linear Systems 3-2 Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

Lesson Quiz: Part I (2, –1) Y = 10000-200x Y = 5000 + 50x 20 min Solve the system using your graphing calculator. x + y = 1 1. (2, –1) 3x –2y = 8 2. Lynn is piloting a plane at an altitude of 10,000 ft. She begins to descend at a rate of 200ft per minute. Miguel is flying a different plane at an altitude of 5000 ft. At the same time that Lynn begins to descend, Miguel begins to climb at a rate of 50 ft. per minute. Write and graph a system of equations that could be used to model the situation. In how many minutes will the planes be at the same altitude? What will the altitude be? Y = 10000-200x Y = 5000 + 50x 20 min 6000 ft

Objectives Solve systems of equations by substitution. Solve systems of equations by elimination.

Example 1A: Solving Linear Systems by Substitution Use substitution to solve the system of equations. y = x – 1 x + y = 7

Check It Out! Example 1a Use substitution to solve the system of equations. y = 2x – 1 3x + 2y = 26 Step 1 Solve one equation for one variable. The first equation is already solved for y: y = 2x – 1. Step 2 Substitute the expression into the other equation. 3x + 2y = 26 Substitute (2x –1) for y in the other equation. 3x + 2(2x–1) = 26 3x + 4x – 2 = 26 Combine like terms. 7x = 28 x = 4

Check It Out! Example 1a Continued Step 3 Substitute the x-value into one of the original equations to solve for y. y = 2x – 1 y = 2(4) – 1 Substitute x = 4. y = 7 The solution is the ordered pair (4, 7).

You can also solve systems of equations with the elimination method You can also solve systems of equations with the elimination method. With elimination, you get rid of one of the variables by adding or subtracting equations. You may have to multiply one or both equations by a number to create variable terms that can be eliminated. The elimination method is sometimes called the addition method or linear combination. Reading Math

Example 2A: Solving Linear Systems by Elimination Use elimination to solve the system of equations. 3x + 4y = 3 4x – 2y = –18

Check It Out! Example 2a Use elimination to solve the system of equations. 2x + 6y = –8 5x –3y = 88 Step 1 Find the value of one variable. 2x + 6y = –8 10x – 6y = 176 The y-terms have opposite coefficients. 12x = 168 Add the equations to eliminate y. x = 14 First part of the solution

Check It Out! Example 2a Continued Step 2 Substitute the x-value into one of the original equations to solve for y. 2(14) + 6y = –8 28 + 6y = –8 6y = –36 Second part of the solution y = –6 The solution to the system is (14 , –6).

In Lesson 3–1, you learned that systems may have infinitely many or no solutions. When you try to solve these systems algebraically, the result will be an identity or a contradiction. An identity, such as 0 = 0, is always true and indicates infinitely many solutions. A contradiction, such as 1 = 3, is never true and indicates no solution. Remember!

Example 3: Solving Systems with Infinitely Many or No Solutions Classify the system and determine the number of solutions. 3x + y = 1 2y + 6x = –18 Because isolating y is straightforward, use substitution. 3x + y = 1 y = 1 –3x Solve the first equation for y. 2(1 – 3x) + 6x = –18 Substitute (1–3x) for y in the second equation. 2 – 6x + 6x = –18 Distribute. 2 = –18 x Simplify. Because 2 is never equal to –18, the equation is a contradiction. Therefore, the system is inconsistent and has no solution.

Lesson Quiz Use substitution or elimination to solve each system of equations. 3x + y = 1 5x – 4y = 10 1. 2. y = x + 9 3x – 4y = –2 (–2, 7) (6, 5) 3. The Miller and Benson families went to a theme park. The Millers bought 6 adult and 15 children tickets for $423. The Bensons bought 5 adult and 9 children tickets for $293. Find the cost of each ticket. adult: $28; children’s: $17