Anonyme
Anonyme question posée dans Science & MathematicsMathematics · Il y a 1 mois

System of equations ?

Could you please explain or show me how to solve this system of equations problem 

4x-2y+z=5

2x+y-2z=4

x+3y-2z=6

11 réponses

Pertinence
  • Il y a 1 mois

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

    (2) : 2x + y - 2z = 4

    (3) : x + 3y - 2z = 6

    You calculate (2) - [2 * (3)] and you obtain the equation (4)

    (4) : (2x + y - 2z) - 2.(x + 3y - 2z) = 4 - (2 * 6)

    (4) : 2x + y - 2z - 2x - 6y + 4z = 4 - 12

    (4) : - 5y + 2z = - 8

    (4) : 2z = 5y - 8

    (4) : z = (5y - 8)/2

    You calculate (1) - [4 * (3)]

    (4x - 2y + z) - 4.(x + 3y - 2z) = 5 - (4 * 6)

    4x - 2y + z - 4x - 12y + 8z = 5 - 24

    - 14y + 9z = - 19

    9z = 14y - 19

    z = (14y - 19)/9 → recall (4): z = (5y - 8)/2

    (14y - 19)/9 = (5y - 8)/2

    2.(14y - 19) = 9.(5y - 8)

    28y - 38 = 45y - 72

    28y - 45y = - 72 + 38

    - 17y = - 34

    → y = 2

    Recall (4): z = (5y - 8)/2

    z = (10 - 8)/2

    → z = 1

    Recall (3) : x + 3y - 2z = 6

    x = 6 - 3y + 2z

    x = 6 - 6 + 2

    → x = 2

  • Il y a 1 mois

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

    2x+y-2z=4------(2)

    x+3y-2z=6------(3)

    The answer to this problem is

    x=2

    y=2

    z=1

    Method:

    (i) (3)-(2), you can obtain a new equation

    in x, y only by eliminating z, call it (4).

    (ii) 2*(1)+(2) or 2*(1)+(3), You obtain another

    new equation in x, y only, call it (5).

    (iii) from (4) & (5), you can solve  for the value

    of x, & y. Then use any of the (1), (2) & (3),

    you can find the solution of z with the help

    of the solutions of x & y.

  • Il y a 1 mois

    First rearrange the equations listing only the coefficients and according to the least leading coefficient, i.e,

    1+3-2=6

    2 +1 -2 =4

    4-2+1=5

    Now use the Gauss-Jordan method to reduce the system to Reduced Row Echelon form;

    It starts out like this

    1+3-2=6

    0 -5  2 =-8 ---> -2R1 +R2 Replaces R2

    0-14+9=-19-->-4R1 +R3 replaces R3

    Continue in this manner until you get the following rref form.

    1 0 0 2

    0 1 0 2

    0 0 1 1 

    Which means x=2, y=2, and z = 1. If you don't understand what I'm doing look up Gauss-Jordan method in any algebra book.

    If you have a TI-graphing calculator you can check your answer by entering the coefficients as a matrix, going to Matrix, Math and seledting rref( Entering the matrix and press Enter

  • sepia
    Lv 7
    Il y a 1 mois

    4x - 2y + z = 5

    2x + y - 2z = 4

    x + 3y - 2z = 6

    Solution:

    x = 2, y = 2, z = 1

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  • Philip
    Lv 6
    Il y a 1 mois

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

    2x+y-2z=4...(2).;

    x+3y-2z=6...(3).;

    Step 1. We pick any 2 eqns and eliminate 1 variable.

    [2(2)-(1)]---> 4y-5z =3...(4).

    Step 2. We pick another 2 eqns and eliminate the same variable as in step 1. 

    [2(3)-(2)]---> 5y-2z = 8...(5).

    Now we have 2 eqns in 2 unknowns. Using (4) & (5) we eliminate one variable.

    [4(5)-5(4)]---> -8z+25z = 32-15, ie., 17z = 17, ie., z = 1. Then (5)---> y = 2 and

    (1)---> 4x = 5+2y-z = 5+4-1 = 8, ie., x = 2. Then (x,y,z) = (2,2,1).

    Note that in multiplying and subtracting eqns we concentrate term by term which

    helps us avoid making mistakes.

  • Il y a 1 mois

    this picture should help

    Attachment image
  • Il y a 1 mois

    4x - 2y + z = 5

    2x + y - 2z = 4

    x + 3y - 2z = 6

    10x - 3y = 14

    x - 2y = -2

     

    20x - 6y = 28

    3x - 6y = -6

    17x = 34

    x = 2 

    y = 2

    z = 1

  • RR
    Lv 7
    Il y a 1 mois

    It's a simultaneous equation. You have to eliminate one unknown at a time and substitute. It's a bit long-winded but quite straight-forward. I will number the equations to help you follow:

    (i) 4x-2y+z=5

    (ii) 2x+y-2z=4

    (iii) x+3y-2z=6

    ___________________

    Eliminate z

    (ii) - (iii)

    (iv) x - 2y = -2

    ------

    (i) x 2

    (iv) 8x-4y+2z =10

    (iv) + (iii)

    (vi) 9x-y = 16

    _______________

    You now have a simultaneous equation with 2 unkowns

    (iv) x - 2y = -2

    (vi) 9x - y = 16

    ____________________

    Eliminate y and find x

    (vi) x 2

    (viii)18x - 2y = 32

    (viii) - (iv)

    17x = 34

    x = 2

    ___________________

    Substitute x = 2 into (iv)

    2 - 2y = -2

    2y = -4

    y = -2

    __________________

    Substitute x = 2 and y = -2 into (i)

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

    8 - 4 + z = 5

    z= 5 - 8 + 4

    z = 1

    _______________

    ANS x = 2, y = -2, z = 1 Phew!

  • Ian H
    Lv 7
    Il y a 1 mois

    2z = x + 3y – 6 = 2x + y – 4 = 4y - 8x + 10

    x – 2y = -2

    10x – 3y = 14

    6y = 3x + 6 = 20x – 28

    17x = 34, x = 2, y = (x + 2)/2 = 2

    z = 5 – 4x + 2y = 1

  • Amy
    Lv 7
    Il y a 1 mois

    Add or subtract two equations in the right ratio to eliminate one of the variables.

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

    10x - 3y = 14

    Add or subtract a different two equations in the right ratio to eliminate the same variable.

    (2x+y-2z) - (x+3y-2z) = 4 - 6

    x - 2y = -2

    Add or subtract your resulting equations in the right ratio to eliminate one of the variables.

    (10x - 3y) - 10(x - 2y) = 14 - 10(-2)

    17y = 34

    This gives you the value of the one remaining variable.

    y = 2

    Plug that into one of your two-variable equations to solve for the other.

    x - 2(2) = -2

    x = 2

    And finally plug both values into one of the original equations to solve for the last variable.

    4(2) - 2(2) + z = 5

    z = 1

    Verify your solution by plugging all the values into all the original equations.

    4(2) - 2(2) + 1 = 5

    2(2) + 2 - 2(1) = 4

    2 + 3(2) - 2(1) = 6

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