1. 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

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 = x
Y = x
20 min
6000 ft

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

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

5. 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

6. 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).

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

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

9. 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

10. 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).

11. 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!

12. 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.

13. 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