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High School Math Algebra High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

To state: The solutions for the system of equation. x + 2 y + z = 4 2 x − y + 4 z = − 8 − 3 x + y − 2 z = − 1 The resultant solution is 3 , 2 , − 3 . Given information: The given system equations is: x + 2 y + z = 4 2 x − y + 4 z = − 8 − 3 x + y − 2 z = − 1 Explanation: Write the original system equations: x + 2 y + z = 4 2 x − y + 4 z = − 8 − 3 x + y − 2 z = − 1 Let’s number the equation, x + 2 y + z = 4 ⋯ 1 2 x − y + 4 z = − 8 ⋯ 2 − 3 x + y − 2 z = − 1 ⋯ 3 Multiply equation (2) by 2, 2 x − y + 4 z = − 8 × 2 4 x − 2 y + 8 z = − 16 ⋯ 4 Add equation (1) and (4) to eliminate variable y , x + 2 y + z = 4 4 x − 2 y + 8 z = − 16 ¯ 5 x + 9 z = − 12 Number the resultant equation, 5 x + 9 z = − 12 ⋯ 5 Multiply equation (3) by -2, − 3 x + y − 2 z = − 1 × − 2 6 x − 2 y + 4 z = 2 ⋯ 6 Add equation (1) and (6) to eliminate variable y , x + 2 y + z = 4 6 x − 2 y + 4 z = 2 ¯ 7 x + 5 z = 6 Number the resultant equation, 7 x + 5 z = 6 ⋯ 7 Now multiply equation (5) by 7 and equation (7) by -5, 5 x + 9 z = − 12 × 7 35 x + 63 z = − 84 ⋯ 8 7 x + 5 z = 6 × − 5 − 35 x − 25 z = − 30 ⋯ 9 Add equation (8) and (9) to eliminate variable x , 35 x + 63 z = − 84 − 35 x − 25 z = − 30 ¯ 38 z = − 114 z = − 3 Now find the value of variable y by substituting the value of x in equation (7) and simplify, 7 x + 5 z = 6 7 x + 5 − 3 = 6 7 x − 15 = 6 7 x = 6 + 15 7 x = 21 x = 3 Now find the value of variable y by substituting the values of y and x in equation (1) and simplify, x + 2 y + z = 4 3 + 2 y + − 3 = 4 2 y = 4 y = 2 Therefore, the solution for the given system of equation is 3 , 2 , − 3 .

To state: The solutions for the system of equation. x + 2 y + z = 4 2 x − y + 4 z = − 8 − 3 x + y − 2 z = − 1 The resultant solution is 3 , 2 , − 3 . Given information: The given system equations is: x + 2 y + z = 4 2 x − y + 4 z = − 8 − 3 x + y − 2 z = − 1 Explanation: Write the original system equations: x + 2 y + z = 4 2 x − y + 4 z = − 8 − 3 x + y − 2 z = − 1 Let’s number the equation, x + 2 y + z = 4 ⋯ 1 2 x − y + 4 z = − 8 ⋯ 2 − 3 x + y − 2 z = − 1 ⋯ 3 Multiply equation (2) by 2, 2 x − y + 4 z = − 8 × 2 4 x − 2 y + 8 z = − 16 ⋯ 4 Add equation (1) and (4) to eliminate variable y , x + 2 y + z = 4 4 x − 2 y + 8 z = − 16 ¯ 5 x + 9 z = − 12 Number the resultant equation, 5 x + 9 z = − 12 ⋯ 5 Multiply equation (3) by -2, − 3 x + y − 2 z = − 1 × − 2 6 x − 2 y + 4 z = 2 ⋯ 6 Add equation (1) and (6) to eliminate variable y , x + 2 y + z = 4 6 x − 2 y + 4 z = 2 ¯ 7 x + 5 z = 6 Number the resultant equation, 7 x + 5 z = 6 ⋯ 7 Now multiply equation (5) by 7 and equation (7) by -5, 5 x + 9 z = − 12 × 7 35 x + 63 z = − 84 ⋯ 8 7 x + 5 z = 6 × − 5 − 35 x − 25 z = − 30 ⋯ 9 Add equation (8) and (9) to eliminate variable x , 35 x + 63 z = − 84 − 35 x − 25 z = − 30 ¯ 38 z = − 114 z = − 3 Now find the value of variable y by substituting the value of x in equation (7) and simplify, 7 x + 5 z = 6 7 x + 5 − 3 = 6 7 x − 15 = 6 7 x = 6 + 15 7 x = 21 x = 3 Now find the value of variable y by substituting the values of y and x in equation (1) and simplify, x + 2 y + z = 4 3 + 2 y + − 3 = 4 2 y = 4 y = 2 Therefore, the solution for the given system of equation is 3 , 2 , − 3 .

Solution Summary: The author states the solution for the system of equations: (3,2,-3).

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

15th Edition

ISBN: 9780133281163

Author: Prentice Hall

Publisher: Prentice Hall

See similar textbooks

Expert Solution & Answer

Chapter 3.5, Problem 34PPSE

Solution

To state: The solutions for the system of equation.

x+2y+ z=42x−y+4z=−8−3x+y−2z=−1

The resultant solution is 3,2,−3 .

Given information:

The given system equations is:

x+2y+ z=42x−y+4z=−8−3x+y−2z=−1

Explanation:

Write the original system equations:

x+2y+ z=42x−y+4z=−8−3x+y−2z=−1

Let’s number the equation,

x+2y+ z=4⋯12x−y+4z=−8⋯2−3x+y−2z=−1⋯3

Multiply equation (2) by 2,

2x−y+4z=−8×24x−2y+8z=−16⋯4

Add equation (1) and (4) to eliminate variable y ,

x+2y+ z=44x−2y+8z=−16¯ 5x+9z=−12

Number the resultant equation,

5x+9z=−12⋯5

Multiply equation (3) by -2,

−3x+y−2z=−1×−26x−2y+4z=2⋯6

Add equation (1) and (6) to eliminate variable y ,

x+2y+ z=46x−2y+4z=2¯ 7x+5z=6

Number the resultant equation,

7x+5z=6⋯7

Now multiply equation (5) by 7 and equation (7) by -5,

5x+9z=−12×735x+63z=−84⋯87x+5z=6×−5−35x−25z=−30⋯9

Add equation (8) and (9) to eliminate variable x ,

35x+63z=−84−35x−25z=−30¯ 38z=−114z=−3

Now find the value of variable y by substituting the value of x in equation (7) and simplify,

7x+5z=67x+5−3=67x−15=67x=6+157x=21x=3

Now find the value of variable y by substituting the values of y and x in equation (1) and simplify,

x+2y+z=43+2y+−3=42y=4y=2

Therefore, the solution for the given system of equation is 3,2,−3 .

Chapter 3.5, Problem 33PPSE

Chapter 3.5, Problem 35PPSE

Chapter 3 Solutions

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

Ch. 3.1 - Prob. 1LC Ch. 3.1 - Prob. 2LC Ch. 3.1 - Prob. 3LC Ch. 3.1 - Prob. 4LC Ch. 3.1 - Prob. 5LC Ch. 3.1 - Prob. 6LC Ch. 3.1 - Prob. 7PPSE Ch. 3.1 - Prob. 8PPSE Ch. 3.1 - Prob. 9PPSE Ch. 3.1 - Prob. 10PPSE

Ch. 3.1 - Prob. 11PPSE Ch. 3.1 - Prob. 12PPSE Ch. 3.1 - Prob. 13PPSE Ch. 3.1 - Prob. 14PPSE Ch. 3.1 - Prob. 15PPSE Ch. 3.1 - Prob. 16PPSE Ch. 3.1 - Prob. 17PPSE Ch. 3.1 - Prob. 18PPSE Ch. 3.1 - Prob. 19PPSE Ch. 3.1 - Prob. 20PPSE Ch. 3.1 - Prob. 21PPSE Ch. 3.1 - Prob. 22PPSE Ch. 3.1 - Prob. 23PPSE Ch. 3.1 - Prob. 24PPSE Ch. 3.1 - Prob. 25PPSE Ch. 3.1 - Prob. 26PPSE Ch. 3.1 - Prob. 27PPSE Ch. 3.1 - Prob. 28PPSE Ch. 3.1 - Prob. 29PPSE Ch. 3.1 - Prob. 30PPSE Ch. 3.1 - Prob. 31PPSE Ch. 3.1 - Prob. 32PPSE Ch. 3.1 - Prob. 33PPSE Ch. 3.1 - Prob. 34PPSE Ch. 3.1 - Prob. 35PPSE Ch. 3.1 - Prob. 36PPSE Ch. 3.1 - Prob. 37PPSE Ch. 3.1 - Prob. 38PPSE Ch. 3.1 - Prob. 39PPSE Ch. 3.1 - Prob. 40PPSE Ch. 3.1 - Prob. 41PPSE Ch. 3.1 - Prob. 42PPSE Ch. 3.1 - Prob. 43PPSE Ch. 3.1 - Prob. 44PPSE Ch. 3.1 - Prob. 45PPSE Ch. 3.1 - Prob. 46PPSE Ch. 3.1 - Prob. 47PPSE Ch. 3.1 - Prob. 48PPSE Ch. 3.1 - Prob. 49PPSE Ch. 3.1 - Prob. 50PPSE Ch. 3.1 - Prob. 51PPSE Ch. 3.1 - Prob. 52PPSE Ch. 3.1 - Prob. 53PPSE Ch. 3.1 - Prob. 54PPSE Ch. 3.1 - 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