Anonymous
Anonymous asked in Education & ReferenceHomework Help · 8 years ago

solving linear systems in three variables?

use elimination to solve each system of equations, i dont know how to solve for three variables, can you guys help me out? these are the two systems i need help with. can you please show the steps too so i get an understanding of how to solve more of these, thank you for the help

1.

8x+3y-6z=4

x-2y-z=2

4x+y-2z=-4

2.

2x-y+3z= 7

5x-4y-2z= 3

3x+3y+2z= -8

2 Answers

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  • Pramod
    Lv 7
    8 years ago
    Best Answer

    Q1.

    8 x + 3 y - 6 z = 4 .............. (1)

    1 x - 2 y - 1 z = 2 .............. (2)

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

    Understanding steps : Step (i) Consider eqns (1) and (2). ASSUME all the numerals and z as constants. Re-write these eqns. as follows

    8 x + 3 y = 6 z + 4 .............. ( modified 1)

    1 x - 2 y = - 1 z + 2 .............. ( modified 2)

    Step (ii) use elimination method to find value of x and y in terms of z . We get these values as follows ( I am not showing the intermediate steps )

    .............(3) (z+2) - [ - 2 ( 6z+4) ]. . . . 15 z + 14

    => x = ----------------------------- ------ . = ------------ ...................... (4)

    . . . . . . . . . . . (3+16). . . . . . . . . . . . . .19

    Substitute this value of x in eqn (2). We get y = - (12+2z)/19 .............. (5)

    Step (iii) Follow the same method and again find value of x and y considering eqns (2) and (3)

    We get -----

    => x = (5 z - 6) / 9 and y = - (2z + 12) / 9 ................... (6)

    From eqns (4) and (6) we see that ---

    (15z+14) / 19 = x = (5 z - 6) / 9

    -------------------------------------------

    => z = -- 6 ......... Answer

    --------------------------------------------

    Now if we substitute value of z ( ie -- 6 ) in the given eqns. we get --

    First eqn => 8 x + 3 y = -- 32

    Sec. eqn => . . x - 2 y = -- 4

    Third eqn => 4x + y = -- 16

    We can now find x and y easily by taking any two eqns ( say eqn1 and 2 ) and solving for x and y

    by elimination method.

    Answer x = -- 4 , y = 0 and z = -- 6 ........ Answer Answer

    Q2. First step ---

    2 x - y + 3 z = 7 ................ (1)

    5 x - 4 y - 2 z = 3 ....................(2)

    3 x + 3 y + 2 z = - 8 ...............(3)

    Multiply eqn (1) by 2 and eqn (2) by 3 ---

    4 x - 2 y + 6 z = 14 ........modified (1)

    15 x - 12 y - 6 z = 9 ..............modified.(2)

    ----------------------------------------- Add the two eqns

    19 x - 14 y = 23 ........... (4)

    Second step : Considering eqns (2) and (3)

    5 x - 4 y - 2 z = 3

    3 x + 3 y + 2 z = - 8

    ------------------------------------- Add the two eqns

    8 x - y = - 5 ............... (5)

    Final step: Subtract 14 times the eqn (5) from eqn (4)

    ..19 x - 14 y = 23 ........... (4)

    ±112 x ∓ 14 y = ∓ 70 ............... (5)

    -------------------------------------------------------------------- Subtract ( dont forget to change sign )

    -- 93 x + 0 (y) = 93

    => x = -- 1 ............... Answer

    Again multiply eqn (4) by 8 and eqn (5) by 19

    ..152 x - 112 y = 184 ........... (4)

    ±152 x ∓ 19 y = ∓ 95 ............... (5)

    ---------------------------------------- Subtract

    ...... . . . . . -- 93 y = 279

    => y = - (279/93) = - 3 .......... Answer

    ----------------- -------------------------------------

    Ans: x = -- 1 , y = - 3 and z = + 2

    ----------------- -------------------------------------

    >>>>>>>>>>>>>>>>>>

    .

  • 8 years ago

    They want you to use the elimination method:

    1. 8x + 3y - 6z = 4 ... [eqn 1]

    x - 2y - z = 2 ... [eqn 2]

    4x + y - 2z = -4 ... [eqn 3]

    eliminating z to get an equation in terms of x and y:

    [eqn 2] * 6 → 6x - 12y - 6z = 12 ... [eqn 4]

    [eqn 1] - [eqn 4] → 2x + 15y = -8 ......................... [eqn 5] ... [be careful with signs]

    eliminating z to get a second equation in x and y:

    [eqn 2] * 2 → 2x - 4y - 2z = 4 ... [eqn 6]

    [eqn 3] - [eqn 6] → 2x + 5y = -8 ...........................[eqn 7]

    using the two equations in x and y to eliminate x:

    [eqn 5] - [eqn 7] → 10y = 0

    y = 0

    solving for x:

    subs y = 0 into [eqn 7] → 2x = -8

    x = -4

    solving for z:

    subs x = -4, y = 0 into [eqn 2] → -4 - 0 - z = 2

    z = -6

    so the solution is (-4, 0, -6)

    Second one is done the same way ... except I'd start by eliminating y:

    2x - y + 3z = 7 ... [eqn 1]

    5x - 4y - 2z = 3 ... [eqn 2]

    3x + 3y + 2z = -8 ... [eqn 3]

    [eqn 1] * 4 → 8x - 4y + 12z = 28 ... [eqn 4]

    [eqn 2] - [eqn 4] → -3x -14z = -25 ............................. [eqn 5]

    [eqn 1] * 3 → 6x - 3y + 9z = 21 ... [eqn 6]

    [eqn 3] + [eqn 6] → 9x + 11z = 13 ............................. [eqn 7] ... [NOTE: The equations had to be added to eliminate the y terms]

    [eqn 5] * 3 → -9x - 42z = -75 ... [eqn 8]

    [eqn 7] + [eqn 8] → -31z = -62

    z = 2

    subs z = 2 into [eqn 7] → 9x + 22 = 13

    9x = -9

    x = -1

    subs x = -1, z = 2 into [eqn 1] → -2 - y + 6 = 7

    y = -3

    so the solution is (-1, -3, 2)

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