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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 by substitution. { 3 x + y − 2 z = 22 x + 5 y + z = 4 x = − 3 z The solution is ( 6 , 0 , − 2 ) . Given information: The given system equations is: { 3 x + y − 2 z = 22 x + 5 y + z = 4 x = − 3 z Explanation: Write the original system equations: { 3 x + y − 2 z = 22 x + 5 y + z = 4 x = − 3 z Let’s number the equation, { 3 x + y − 2 z = 22 ⋯ ( 1 ) x + 5 y + z = 4 ⋯ ( 2 ) x = − 3 z ⋯ ( 3 ) Now substitute x = − 3 z in equation (1), 3 x + y − 2 z = 22 3 ( − 3 z ) + y − 2 z = 22 − 9 z + y − 2 z = 22 y − 11 z = 22 ⋯ ( 4 ) Now substitute x = − 3 z in equation (2), x + 5 y + z = 4 − 3 z + 5 y + z = 4 5 y − 2 z = 4 ⋯ ( 5 ) Multiply the equation (4) by 5, ( y − 11 z = 22 ) × 5 5 y − 55 z = 110 ⋯ ( 6 ) Add equations (5) and (6), { 5 y − 2 z = 4 5 y − 55 z = − 110 _ 53 z = − 106 z = − 2 Now calculate the value of variable y by substituting the value of z in equation (5), 5 y − 2 z = 4 5 y − 2 ( − 2 ) = 4 5 y + 4 = 4 5 y = 0 y = 0 Now calculate the value of variable x by substituting the value of z in equation (3), x = − 3 z x = − 3 ( − 2 ) x = 6 Therefore, the solution of the system of equation is ( 6 , 0 , − 2 ) .

To state: The solutions for the system of equation by substitution. { 3 x + y − 2 z = 22 x + 5 y + z = 4 x = − 3 z The solution is ( 6 , 0 , − 2 ) . Given information: The given system equations is: { 3 x + y − 2 z = 22 x + 5 y + z = 4 x = − 3 z Explanation: Write the original system equations: { 3 x + y − 2 z = 22 x + 5 y + z = 4 x = − 3 z Let’s number the equation, { 3 x + y − 2 z = 22 ⋯ ( 1 ) x + 5 y + z = 4 ⋯ ( 2 ) x = − 3 z ⋯ ( 3 ) Now substitute x = − 3 z in equation (1), 3 x + y − 2 z = 22 3 ( − 3 z ) + y − 2 z = 22 − 9 z + y − 2 z = 22 y − 11 z = 22 ⋯ ( 4 ) Now substitute x = − 3 z in equation (2), x + 5 y + z = 4 − 3 z + 5 y + z = 4 5 y − 2 z = 4 ⋯ ( 5 ) Multiply the equation (4) by 5, ( y − 11 z = 22 ) × 5 5 y − 55 z = 110 ⋯ ( 6 ) Add equations (5) and (6), { 5 y − 2 z = 4 5 y − 55 z = − 110 _ 53 z = − 106 z = − 2 Now calculate the value of variable y by substituting the value of z in equation (5), 5 y − 2 z = 4 5 y − 2 ( − 2 ) = 4 5 y + 4 = 4 5 y = 0 y = 0 Now calculate the value of variable x by substituting the value of z in equation (3), x = − 3 z x = − 3 ( − 2 ) x = 6 Therefore, the solution of the system of equation is ( 6 , 0 , − 2 ) .

Solution Summary: The author states the solutions for the system of equation by substitution.

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, Problem 25E

Solution

To state: The solutions for the system of equation by substitution.

{3x+y−2z=22x+5y+z=4x=−3z

The solution is (6,0,−2) .

Given information:

The given system equations is:

{3x+y−2z=22x+5y+z=4x=−3z

Explanation:

Write the original system equations:

{3x+y−2z=22x+5y+z=4x=−3z

Let’s number the equation,

{3x+y−2z=22⋯(1)x+5y+z=4⋯(2)x=−3z⋯(3)

Now substitute x=−3z in equation (1),

3x+y−2z=223(−3z)+y−2z=22−9z+y−2z=22y−11z=22⋯(4)

Now substitute x=−3z in equation (2),

x+5y+z=4−3z+5y+z=45y−2z=4⋯(5)

Multiply the equation (4) by 5,

(y−11z=22)×55y−55z=110⋯(6)

Add equations (5) and (6),

{5y−2z=45y−55z=−110_53z=−106z=−2

Now calculate the value of variable y by substituting the value of z in equation (5),

5y−2z=45y−2(−2)=45y+4=45y=0y=0

Now calculate the value of variable x by substituting the value of z in equation (3),

x=−3zx=−3(−2)x=6

Therefore, the solution of the system of equation is (6,0,−2) .

Chapter 3, Problem 24E

Chapter 3, Problem 26E

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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Prob. 7GR Ch. 3 - Prob. 8GR Ch. 3 - Prob. 9GR Ch. 3 - Prob. 10GR Ch. 3 - Prob. 11GR Ch. 3 - Prob. 12GR Ch. 3 - Prob. 13GR Ch. 3 - Prob. 14GR Ch. 3 - Prob. 15GR Ch. 3 - Prob. 16GR Ch. 3 - Prob. 17GR Ch. 3 - Prob. 18GR Ch. 3 - Prob. 19GR Ch. 3 - Prob. 20GR Ch. 3 - Prob. 21GR Ch. 3 - Prob. 22GR Ch. 3 - Prob. 23GR Ch. 3 - Prob. 1MCQ Ch. 3 - Prob. 2MCQ Ch. 3 - Prob. 3MCQ Ch. 3 - Prob. 4MCQ Ch. 3 - Prob. 5MCQ Ch. 3 - Prob. 6MCQ Ch. 3 - Prob. 7MCQ Ch. 3 - Prob. 8MCQ Ch. 3 - Prob. 9MCQ Ch. 3 - Prob. 10MCQ Ch. 3 - Prob. 11MCQ Ch. 3 - Prob. 12MCQ Ch. 3 - Prob. 13MCQ Ch. 3 - Prob. 14MCQ Ch. 3 - Prob. 15MCQ Ch. 3 - Prob. 16MCQ Ch. 3 - Prob. 17MCQ Ch. 3 - Prob. 18MCQ Ch. 3 - Prob. 1CV Ch. 3 - Prob. 2CV Ch. 3 - Prob. 3E Ch. 3 - Prob. 4E Ch. 3 - Prob. 5E Ch. 3 - Prob. 6E Ch. 3 - Prob. 7E Ch. 3 - Prob. 8E Ch. 3 - Prob. 9E Ch. 3 - Prob. 10E Ch. 3 - Prob. 11E Ch. 3 - Prob. 12E Ch. 3 - Prob. 13E Ch. 3 - Prob. 14E Ch. 3 - Prob. 15E Ch. 3 - Prob. 16E Ch. 3 - Prob. 17E Ch. 3 - Prob. 18E Ch. 3 - Prob. 19E Ch. 3 - 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