1. Questions

Algebra I DONT KNOW HOW TO DO 3 VARIABLES

Solving Linear Equations and Inequalities

Solve by eliminating x (this is solving 3 linear equations) then substitute to the other two equation

x + y + 5z =2 (1)
4x - 3y + 5z =3 (2)
3x - 2y + 5z=1 (3)

my teacher gave me the answer she said the solution set for this should be {(7,5-2)}

pleeeeeeeeeeeeeaaaseee :(

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Clyde

  1. x = 2 - y - 5z

    4(2 - y - 5z) -3y + 5z = 3
    3(2 - y - 5z) -2y + 5z = 1

    8 - 4y - 20z - 3y + 5z = 3
    6 - 3y - 15z - 2y + 5z = 1

    -7y - 15z = -5
    -5y - 20z = -5 or y = 1 - 4z

    -7(1-4z) - 15z = -5

    7 + 28z -15 z = -5

    13 z = 2

    SO I get:
    z = 2/13
    y = 13/13 - 8/13 = 5/13
    x = 26/13 - 5/13 -10/13 = 11/13

    check
    11/13 + 5/13 +10/13 = 26/13 = 2 yes
    4(11/13)-3(5/13)+5(2/13)
    =(44-15+10)/13 = 39/13 = 3
    yes
    33/13 -10/13 + 10/13 =33/13 = 1
    yes
    so I disagree

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    Damon

  2. x + y + 5 z = 2
    4 x - 3 y + 5 z = 3
    3 x - 2 y + 5 z = 1

    All three equations contain member 5 z

    Equation ( 1 )

    x + y + 5 z = 2 Subtract ( x + y ) to both sides

    x + y + 5 z - ( x + y ) = 2 -( x + y )

    5 z = 2 - x - y

    Equation ( 2 )

    4 x - 3 y + 5 z = 3 Subtract ( 4 x - 3 y ) to both sides

    4 x - 3 y + 5 z - ( 4 x - 3 y ) = 3 - ( 4 x - 3 y )

    5 z = 3 - 4 x - ( - 3 y )

    5 z = 3 - 4 x + 3 y

    Equation ( 3 )

    3 x - 2 y + 5 z = 1 Subtract ( 3 x - 2 y ) to both sides

    3 x - 2 y + 5 z - ( 3 x - 2 y ) = 1 - ( 3 x - 2 y )

    5 z = 1 - 3 x - ( - 2 y )

    5 z = 1 - 3 x + 2 y

    5 z = 5 z = 5 z

    2 - x - y = 3 - 4 x + 3 y = 1 - 3 x + 2 y Add x to all three sides

    2 - x - y + x = 3 - 4 x + 3 y + x = 1 - 3 x + 2 y + x

    2 - x + x - y = 3 - 4 x + x + 3 y = 1 - 3 x + x + 2 y

    2 + 0 - y = 3 - 4 x + x + 3 y = 1 - 3 x + x + 2 y

    2 - y = 3 - 3 x + 3 y = 1 - 2 x + 2 y Add y to both sides

    2 - y + y = 3 - 3 x + 3 y + y = 1 - 2 x + 2 y + y

    2 - 0 = 3 - 3 x + 4 y = 1 - 2 x + 3 y

    2 = 3 - 3 x + 4 y = 1 - 2 x + 3 y Subtract 2 to all three sides

    2 - 2 = 3 - 3 x + 4 y - 2 = 1 - 2 x + 3 y - 2

    0 = 3 - 2 - 3 x + 4 y = 1 - 2 - 2 x + 3 y

    0 = 1 - 3 x + 4 y = - 1 - 2 x + 3 y

    Now you have system of 2 equations :

    1 - 3 x + 4 y = 0

    and

    - 1 - 2 x + 3 y = 0

    1 - 3 x + 4 y = 0 Add 3 x to both sides

    1 - 3 x + 4 y + 3 x = 0 + 3 x

    1 - 3 x + 3 x + 4 y + 3 = 0 + 3 x

    1 + 0 + 4 y = 3 x

    1 + 4 y = 3 x Subtract 1 to both sides

    1 + 4 y - 1 = 3 x

    1 - 1 + 4 y = 3 x - 1

    0 + 4 y = 3 x - 1

    4 y = 3 x - 1

    - 1 - 2 x + 3 y = 0 Add 2 x to both sides

    - 1 - 2 x + 3 y + 2 x = 0 + 2 x

    - 1 - 2 x + 2 x + 3 y = 2 x

    - 1 + 0 + 3 y = 2 x

    - 1 + 3 y = 2 x Add 1 to both sides

    - 1 + 3 y + 1 = 2 x + 1

    - 1 + 1 + 3 y = 2 x + 1

    0 + 3 y = 2 x + 1

    3 y = 2 x + 1

    You again have system of 2 equations :

    4 y = 3 x - 1

    and

    3 y = 2 x + 1

    4 y = 3 x - 1 Multiply both sides by 3

    4 y * 3 = ( 3 x - 1 ) * 3

    12 y = 3 x * 3 - 1 * 3

    12 y = 9 x - 3

    3 y = 2 x + 1 Multiply both sides by 4

    3 y * 4 = 2 x * 4 + 1 * 4

    12 y = 8 x + 4

    12 y = 12 y

    9 x - 3 = 8 x + 4 Subtract 8 x to both sides

    9 x - 3 - 8 x = 8 x + 4 - 8 x

    9 x - 8 x - 3 = 8 x - 8 x + 4

    x - 3 = 0 + 4

    x - 3 = 4 Add 3 to both sides

    x - 3 + 3 = 4 + 3

    x + 0 = 7

    x = 7

    Replace value of x in equation :

    4 y = 3 x - 1

    4 y = 3 * 7 - 1

    4 y = 21 - 1

    4 y = 20 Divide both sides by 4

    4 y / 4 = 20 / 4

    y = 5

    Replace values x and y in equation :

    5 z = 2 - x - y

    5 z = 2 - 7 - 5

    5 z = - 10 Divide both sides by 5

    5 z / 5 = - 10 / 5

    z = - 2

    The soluitions are :

    x = 7 , y = 5 , z = - 2

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    Bosnian

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