{"id":1220,"date":"2025-07-23T20:51:59","date_gmt":"2025-07-23T20:51:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1220"},"modified":"2026-03-20T16:11:55","modified_gmt":"2026-03-20T16:11:55","slug":"systems-of-linear-equations-in-three-variables-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/systems-of-linear-equations-in-three-variables-learn-it-2\/","title":{"raw":"Systems of Linear Equations in Three Variables: Learn It 2","rendered":"Systems of Linear Equations in Three Variables: Learn It 2"},"content":{"raw":"

Solve Systems of Three Equations in Three Variables<\/h2>\r\nSolving a system with three variables is very similar to solving one with two variables. It is important to keep track of your work as the addition of one more equation creates more steps in the solution process.\r\n\r\n
How To: Given a linear system of three equations, solve for three unknowns.\r\n<\/strong>\r\n
    \r\n \t
  1. Pick any pair of equations and solve for one variable.<\/li>\r\n \t
  2. Pick another pair of equations and solve for the same variable.<\/li>\r\n \t
  3. You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.<\/li>\r\n \t
  4. Back-substitute known variables into any one of the original equations and solve for the missing variable.<\/li>\r\n<\/ol>\r\n<\/section>
    We'll take the steps slowly in the following few examples. First, we'll look just at the last step in the process: back-substitution. Then, we'll look at an example that requires the addition (elimination) method to reach the first solution. Then we'll see some video examples that illustrate some of the different kinds of situations you may encounter when solving three-by-three systems. Finally, you'll have the opportunity to practice applying the complete process.<\/section>
    Determine whether the ordered triple [latex]\\left(3,-2,1\\right)[\/latex] is a solution to the system.\r\n

    [latex]\\begin{array}{l}\\text{ }x+y+z=2\\hfill \\\\ 6x - 4y+5z=31\\hfill \\\\ 5x+2y+2z=13\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"954288\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"954288\"]\r\n\r\nWe will check each equation by substituting in the values of the ordered triple for [latex]x,y[\/latex], and [latex]z[\/latex].\r\n

    [latex]\\begin{align} x+y+z=2\\\\ \\left(3\\right)+\\left(-2\\right)+\\left(1\\right)=2\\\\ \\text{True}\\end{align}\\hspace{5mm}[\/latex] [latex]\\hspace{5mm}\\begin{align} 6x - 4y+5z=31\\\\ 6\\left(3\\right)-4\\left(-2\\right)+5\\left(1\\right)=31\\\\ 18+8+5=31\\\\ \\text{True}\\end{align}\\hspace{5mm}[\/latex] [latex]\\hspace{5mm}\\begin{align}5x+2y+2z=13\\\\ 5\\left(3\\right)+2\\left(-2\\right)+2\\left(1\\right)=13\\\\ 15 - 4+2=13\\\\ \\text{True}\\end{align}[\/latex]<\/p>\r\nThe ordered triple [latex]\\left(3,-2,1\\right)[\/latex] is indeed a solution to the system.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

    [ohm_question hide_question_numbers=1]321618[\/ohm_question]<\/section>
    Solve the given system.
    [latex]\\displaystyle\\begin{cases}x-\\dfrac{1}{3}y+\\dfrac{1}{2}z=1\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y-\\dfrac{1}{2}z=4\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,z=-1\\end{cases}[\/latex]<\/center>\r\n[reveal-answer q=\"538379\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"538379\"]The third equation states that\u00a0[latex]z = \u22121[\/latex], so\u00a0we substitute this into the second equation to obtain a solution for\u00a0[latex]y[\/latex].\r\n
    [latex]\\begin{array}{l}y-\\dfrac{1}{2}(-1)=4\\\\y+\\dfrac{1}{2}=4\\\\y=4-\\dfrac{1}{2}\\\\y=\\dfrac{8}{2}-\\dfrac{1}{2}\\\\y=\\dfrac{7}{2}\\end{array}[\/latex]<\/center>\r\nNow we have two of our solutions, and we can substitute them both into the first equation to solve for\u00a0[latex]x[\/latex].\r\n
    [latex]\\begin{array}{l}x-\\dfrac{1}{3}\\left(\\dfrac{7}{2}\\right)+\\dfrac{1}{2}\\left(-1\\right)=1\\\\x-\\dfrac{7}{6}-\\dfrac{1}{2}=1\\\\x-\\dfrac{7}{6}-\\dfrac{3}{6}=1\\\\x-\\dfrac{10}{6}=1\\\\x=1+\\dfrac{10}{6}\\\\x=\\dfrac{6}{6}+\\dfrac{10}{6}\\\\x=\\dfrac{16}{6}=\\dfrac{8}{3}\\end{array}[\/latex]<\/center>Now we have our ordered triple; remember to place each variable solution in order.
    [latex](x,y,z)=\\left(\\dfrac{8}{3},\\dfrac{7}{2},-1\\right)[\/latex]<\/center>Analysis of the Solution<\/strong>Each of the lines in the system above represents a plane (think about a sheet of paper). If you imagine three sheets of notebook paper each representing a portion of these planes, you will start to see the complexities involved in how three such planes can intersect. Below is a sketch of the three planes. It turns out that any two of these planes intersect in a line, so our intersection point is where all three of these lines meet.\r\n\r\n[caption id=\"attachment_2377\" align=\"aligncenter\" width=\"300\"]\"Three Three planes intersecting.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
    [ohm_question hide_question_numbers=1]321619[\/ohm_question]<\/section>","rendered":"

    Solve Systems of Three Equations in Three Variables<\/h2>\n

    Solving a system with three variables is very similar to solving one with two variables. It is important to keep track of your work as the addition of one more equation creates more steps in the solution process.<\/p>\n

    How To: Given a linear system of three equations, solve for three unknowns.
    \n<\/strong><\/p>\n
      \n
    1. Pick any pair of equations and solve for one variable.<\/li>\n
    2. Pick another pair of equations and solve for the same variable.<\/li>\n
    3. You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.<\/li>\n
    4. Back-substitute known variables into any one of the original equations and solve for the missing variable.<\/li>\n<\/ol>\n<\/section>\n
      We’ll take the steps slowly in the following few examples. First, we’ll look just at the last step in the process: back-substitution. Then, we’ll look at an example that requires the addition (elimination) method to reach the first solution. Then we’ll see some video examples that illustrate some of the different kinds of situations you may encounter when solving three-by-three systems. Finally, you’ll have the opportunity to practice applying the complete process.<\/section>\n
      Determine whether the ordered triple [latex]\\left(3,-2,1\\right)[\/latex] is a solution to the system.<\/p>\n

      [latex]\\begin{array}{l}\\text{ }x+y+z=2\\hfill \\\\ 6x - 4y+5z=31\\hfill \\\\ 5x+2y+2z=13\\hfill \\end{array}[\/latex]<\/p>\n