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Math Help - Help Solving Systems of Equations in 3 Variables

  1. #1
    Junior Member
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    Help Solving Systems of Equations in 3 Variables

    Hi All,

    My issue is solving 3 systems of variables that are missing a variable in each equation.

    My 3 systems are:

    3p + 2r = 11, (1)
    q - 7r = 4, (2)
    p - q = 1 (3)

    What I'm seeing is - that there's no q in equation (1). That means that q has been eliminated from one equation. Since I need another equation with q eliminated, I wanted to use equations (2) and (3) to eliminate q.

    With that said, I proceeded with:

    Multiplied equation (2) by 1 and added it to equation (3)

    q - 7r = 4
    p - q =1
    ----------
    p - 7r = 5 (4)

    Next, I multiplied equation (4) by - 3 and added it to equation (1)

    3p + 2r = 11
    -3p + 21r = -15
    ----------------
    23r = -4
    r = -4/23 (fraction)

    Whenever, I get fraction that confuses me as to how to solve for the other variables. I plugged -4/23 back into equation (1) to solve for p and got an improper fraction. Is this correct? Or, can someone provide some assistance with how I should've proceeded in the first place? I'm missing something here and I'm not sure how I should proceed? Please help. TIA...
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lilrhino View Post
    Hi All,

    My issue is solving 3 systems of variables that are missing a variable in each equation.

    My 3 systems are:

    3p + 2r = 11, (1)
    q - 7r = 4, (2)
    p - q = 1 (3)

    What I'm seeing is - that there's no q in equation (1). That means that q has been eliminated from one equation. Since I need another equation with q eliminated, I wanted to use equations (2) and (3) to eliminate q.

    With that said, I proceeded with:

    Multiplied equation (2) by 1 and added it to equation (3)

    q - 7r = 4
    p - q =1
    ----------
    p - 7r = 5 (4)

    Next, I multiplied equation (4) by - 3 and added it to equation (1)

    3p + 2r = 11
    -3p + 21r = -15
    ----------------
    23r = -4
    r = -4/23 (fraction)

    Whenever, I get fraction that confuses me as to how to solve for the other variables. I plugged -4/23 back into equation (1) to solve for p and got an improper fraction. Is this correct? Or, can someone provide some assistance with how I should've proceeded in the first place? I'm missing something here and I'm not sure how I should proceed? Please help. TIA...
    You are doing great! Don't stop now on account of a few fractions:

    r = -4/23

    p - 7r = 5 ==> p - 7(-4/23) = 5

    p + 28/23 = 5

    p = 5 - 28/23

    p = 115/23 - 28/23 = 87/23

    And
    p - q = 1 ==> 87/23 - q = 1

    q = 87/23 - 1 = 87/23 - 23/23 = 64/23

    So your solution set is:
    p = 87/23
    q = 64/23
    r = -4/23

    -Dan
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  3. #3
    Junior Member
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    Quote Originally Posted by topsquark View Post
    You are doing great! Don't stop now on account of a few fractions:

    r = -4/23

    p - 7r = 5 ==> p - 7(-4/23) = 5

    p + 28/23 = 5

    p = 5 - 28/23

    p = 115/23 - 28/23 = 87/23

    And
    p - q = 1 ==> 87/23 - q = 1

    q = 87/23 - 1 = 87/23 - 23/23 = 64/23

    So your solution set is:
    p = 87/23
    q = 64/23
    r = -4/23

    -Dan
    Hi Dan, thanks...I got q = 64/23, but I stopped solving because I thought it was incorrect. Thanks a bunch!
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  4. #4
    rich.sisk
    Guest

    help me trying to so problem for a variable

    how do i solve 3x=wyz i need to solve
    for the variable x
    for the variable w
    for the variable y
    for the variable z
    plz need help trying to cheach up in class so i can pass any one is welcome


    Quote Originally Posted by lilrhino View Post
    Hi All,

    My issue is solving 3 systems of variables that are missing a variable in each equation.

    My 3 systems are:

    3p + 2r = 11, (1)
    q - 7r = 4, (2)
    p - q = 1 (3)

    What I'm seeing is - that there's no q in equation (1). That means that q has been eliminated from one equation. Since I need another equation with q eliminated, I wanted to use equations (2) and (3) to eliminate q.

    With that said, I proceeded with:

    Multiplied equation (2) by 1 and added it to equation (3)

    q - 7r = 4
    p - q =1
    ----------
    p - 7r = 5 (4)

    Next, I multiplied equation (4) by - 3 and added it to equation (1)

    3p + 2r = 11
    -3p + 21r = -15
    ----------------
    23r = -4
    r = -4/23 (fraction)

    Whenever, I get fraction that confuses me as to how to solve for the other variables. I plugged -4/23 back into equation (1) to solve for p and got an improper fraction. Is this correct? Or, can someone provide some assistance with how I should've proceeded in the first place? I'm missing something here and I'm not sure how I should proceed? Please help. TIA...
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  5. #5
    Super Member
    earboth's Avatar
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    Quote Originally Posted by rich.sisk View Post
    how do i solve 3x=wyz i need to solve
    for the variable x
    for the variable w
    for the variable y
    for the variable z
    plz need help trying to cheach up in class so i can pass any one is welcome
    Hello,

    one remark at first: If you have a new question please start a new thread. Otherwise you risk that nobody will notice that you need some help.

    Solving for a variable means that you collect all terms containing this variable at one side of the equation and all other terms on the other side. Presumely you have to get rid of some coefficients to isolate the variable:

    3x=wyz~\implies~\text{ divide by 3: }~\frac{3x}{3} =x= \frac13 \cdot w \cdot y \cdot z

    3x=wyz~\implies~\text{ divide by wz: }~y = \frac{3x}{ w  \cdot z}

    3x=wyz~\implies~\text{ divide by yz: }~w = \frac{3x}{ y \cdot z}

    3x=wyz~\implies~\text{ divide by wy: }~z = \frac{3x}{ w \cdot y}
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