UNIT 3.2 SOLVING UNIT 3.2 SOLVING SYSTEMS ALGEBRAICALLYSYSTEMS ALGEBRAICALLY

Lesson Quiz: Part I

Solve the system using your graphing calculator.

1.x + y = 1

3x –2y = 8(2, –1)

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.A.Write and graph a system of equations that could be used to model the situation.B.In how many minutes will the planes be at the same altitude?C.What will the altitude be?

Y = 10000-200x

Y = 5000 + 50x

20 min6000 ft

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

Objectives

Use substitution to solve the system of equations.

Example 1A: Solving Linear Systems by Substitution

y = x – 1

x + y = 7

Use substitution to solve the system of equations.

y = 2x – 1

3x + 2y = 26Step 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

3x + 2(2x–1) = 26

3x + 4x – 2 = 26

7x = 28x = 4

Substitute (2x –1) for y in the other equation.Combine like terms.

Check It Out! Example 1a

Step 3 Substitute the x-value into one of the original equations to solve for y.

y = 2x – 1

y = 2(4) – 1

y = 7

Substitute x = 4.

The solution is the ordered pair (4, 7).

Check It Out! Example 1a Continued

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

Use elimination to solve the system of equations.

Example 2A: Solving Linear Systems by Elimination

3x + 4y = 3

4x – 2y = –18

Use elimination to solve the system of equations. 2x + 6y = –8

5x –3y = 88

Check It Out! Example 2a

Step 1 Find the value of one variable.

The y-terms have opposite coefficients.

First part of the solution

12x = 168

2x + 6y = –8

10x – 6y = 176

Add the equations to eliminate y.

x = 14

Check It Out! Example 2a Continued

28 + 6y = –8

6y = –36

Second part of the solution

Step 2 Substitute the x-value into one of the original equations to solve for y.

2(14) + 6y = –8

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!

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 = –182 = –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

Lesson Quiz

Use substitution or elimination to solve each system of equations.

3x + y = 1

y = x + 91.

(–2, 7)

5x – 4y = 10

3x – 4y = –22.

(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

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