{x-3y+4z=26 2x+y+z=3 -2x+3y-3z=-23} Solve the given system of equations. If the system has no​ solution, say that it is inconsistent. ?

5 Answers

Relevance
  • JOHN
    Lv 7
    2 weeks ago

    Hope the below is readable

    Attachment image
  • 2 weeks ago

    {x-3y+4z=26 2x+y+z=3 -2x+3y-3z=-23}

    Rewrite as: ❶: y+z=3-2x, ❷: 3y-3z=-23+2x, and ❸: x=26+3y-4z

    From ❶ and ❷ we get:

    ❹: y = ((3-2x)×-3-1×(-23+2x))/(1×-3-1×3) = (-7-2x)/3

    ❺: z = (1×(-23+2x)-(3-2x)×3)/(1×-3-1×3) = (16-4x)/3

    Substituting ❹ and ❺ into ❸ we get:

    ❻: x = 26+3(-7-2x)/3-4(16-4x)/3 = (10x-7)/3 = 1

    Substituting ❻ into ❹ and ❺ yields:

    ❼: y = (-7-2×1)/3 = -3 and ❽: z = (16-4×1)/3 = 4

    Solution point: (1, -3, 4)

  • 2 weeks ago

    Another approach, elimination:

    Add the 2nd & 3rd equations, obtaining:

    4y - 2z = -20.

    Double the 1st eqn and add it to the 3rd, obtaining:

    -3y + 5z = 29.

    Now multiply these two new equations so that the "y" coefficients will cancel one another out:

    12y - 6z = -60;

    -12y + 20z = 116.

    Adding these together gives 14z = 56, 

    so z = 4; back-substitution gives y = -3 and x = 1.

    Check by using any of the original equations, 

    for example:

    2x + y + z = 2 - 3 + 4 = 3,

    yes, that works.

    The solution by llaffer is wrong, does not check out.

  • 2 weeks ago

    x - 3y + 4z = 26 [eq 1]

    2x + y + z = 3 [eq 2]

    -2x + 3y - 3z = -23 [eq 3]

    Add eq 1 and 3:

    -x + z = 3 [eq 4]

    Triple eq 2 and add it to 1:

    7x + 7z = 35

    Divide by 7:

    x + z = 5 [eq 5]

    Add to eq 4:

    2z = 8

    z = 4

    Substitute in eq 5:

    x + 4 = 5

    x = 1

    Substitute in eq 2:

    2(1) + y + 4 = 3

    y = -3

    I'll leave the step of verifying to you.

    Answer:

    x = 1

    y = -3

    z = 4

  • How do you think about the answers? You can sign in to vote the answer.
  • 2 weeks ago

    x - 3y + 4z = 26 and 2x + y + z = 3 and -2x + 3y - 3z = -23

    Start by solving one equation for one variable then substitute it into the other two.  I'll solve the first equation for x in terms of y and z:

    x - 3y + 4z = 26

    x = 26 + 3y - 4z

    Substitute:

    2x + y + z = 3 and -2x + 3y - 3z = -23

    2(26 + 3y - 4z) + y + z = 3 and -2(26 + 3y - 4z) + 3y - 3z = -23

    52 + 6y - 8z + y + z = 3 and -52 - 6y + 8z + 3y - 3z = -23

    7y - 7z = -49 and -3y + 5z = 29

    Now solve the first equation for y in terms of z and substitute into the last:

    7y - 7z = -49

    y - z = -7

    y = z - 7

    -3y + 5z = 29

    -3(z - 7) + 5z = 29

    -3z + 21 + 5z = 29

    2z = 28

    z = 14

    Now that we have z, solve for y, then go back to x:

    y = z - 7

    y = 14 - 7

    y = 7

    x = 26 + 3y - 4z

    x = 26 + 3(7) - 4(14)

    x = 26 + 21 - 56

    x = -9

    The solution to this system of equations is:

    x = -9, y = 7, z = 14

Still have questions? Get your answers by asking now.