1. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Warm Up Determine if the given ordered pair is an element of the solution set of 2x – y = 5 3y + x = 6 1. (3, 1) yes 2. (–1, 1) no Solve each equation for y. 3. x + 3y = 2x + 4y – 4 4. 6x + 5 + y = 3y + 2x – 1 y = –x + 4 y = 2x + 3.

2. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Solve systems of equations by substitution. Solve systems of equations by elimination. Objectives

3. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems substitution elimination Vocabulary

4. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems The graph shows a system of linear equations. As you can see, without the use of technology, determining the solution from the graph is not easy. You can use the substitution method to find an exact solution. In substitution, you solve one equation for one variable and then substitute this expression into the other equation.

5. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Use substitution to solve the system of equations. Example 1A: Solving Linear Systems by Substitution y = x – 1 x + y = 7 Step 1 Solve one equation for one variable. The first equation is already solved for y: y = x – 1. Step 2 Substitute the expression into the other equation. x + y = 7 x + (x – 1) = 7 2x – 1 = 7 2x = 8 x = 4 Substitute (x –1) for y in the other equation. Combine like terms.

6. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Step 3 Substitute the x-value into one of the original equations to solve for y. Example 1A Continued y = x – 1 y = (4) – 1 y = 3 Substitute x = 4. The solution is the ordered pair (4, 3).

7. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Use substitution to solve the system of equations. Example 1B: Solving Linear Systems by Substitution 2y + x = 4 3x – 4y = 7 Method 1 Isolate y. 2y + x = 4 5x = 15 3x + 2x – 8 = 7 First equation. Method 2 Isolate x. Isolate one variable. Second equation. Substitute the expression into the second equation. Combine like terms. 2y + x = 4 x = 3 x = 4 – 2y 3x – 4y= 7 3(4 – 2y)– 4y = 7 12 – 6y – 4y = 7 12 – 10y = 7 –10y = –5 5x – 8 = 7 First part of the solution 3x –4 +2 = 7 y = + 2 3x – 4y = 7

8. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Substitute the value into one of the original equations to solve for the other variable. 2y + (3) = 4 2y = 1 Example 1B Continued Substitute the value to solve for the other variable. Second part of the solution 2 + x = 4 1 + x = 4 x = 3 By either method, the solution is. Method 1Method 2

9. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems 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

10. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Use elimination to solve the system of equations. Example 2A: Solving Linear Systems by Elimination 3x + 2y = 4 4x – 2y = –18 Step 1 Find the value of one variable. 3x + 2y = 4 + 4x – 2y = –18 The y-terms have opposite coefficients. First part of the solution 7x = –14 x = –2 Add the equations to eliminate y.

11. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Example 2A Continued Step 2 Substitute the x-value into one of the original equations to solve for y. 3(–2) + 2y = 4 2y = 10 y = 5 Second part of the solution The solution to the system is (–2, 5).

12. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Use elimination to solve the system of equations. Example 2B: Solving Linear Systems by Elimination 3x + 5y = –16 2x + 3y = –9 Step 1 To eliminate x, multiply both sides of the first equation by 2 and both sides of the second equation by –3. Add the equations. First part of the solution y = –5 2(3x + 5y) = 2(–16) –3(2x + 3y) = –3(–9) 6x + 10y = –32 –6x – 9y = 27

13. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Example 2B Continued Second part of the solution 3x + 5(–5) = –16 3x = 9 3x – 25 = –16 x = 3 Step 2 Substitute the y-value into one of the original equations to solve for x. The solution for the system is (3, –5).

14. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems 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!

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

16. Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems Lesson Quiz Use substitution or elimination to solve each system of equations. 3x + y = 1 y = x + 9 1. (–2, 7) 5x – 4y = 10 3x – 4y = –2 2. (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