{"id":1219,"date":"2025-07-23T20:53:29","date_gmt":"2025-07-23T20:53:29","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1219"},"modified":"2026-03-20T16:15:39","modified_gmt":"2026-03-20T16:15:39","slug":"systems-of-linear-equations-in-three-variables-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/systems-of-linear-equations-in-three-variables-learn-it-3\/","title":{"raw":"Systems of Linear Equations in Three Variables: Learn It 3","rendered":"Systems of Linear Equations in Three Variables: Learn It 3"},"content":{"raw":"

Solving a System of Three Equations in Three Variables by Elimination<\/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: Solve a system of linear equations with three variables by elimination<\/strong>\r\n
    \r\n \t
  1. Write the equations in standard form. If any coefficients are fractions, clear them.<\/li>\r\n \t
  2. Eliminate the same variable from two equations.\r\n
      \r\n \t
    1. Decide which variable you will eliminate.<\/li>\r\n \t
    2. Work with a pair of equations to eliminate the chosen variable.<\/li>\r\n \t
    3. Multiply one or both equations so that the coefficients of that variable are opposites.<\/li>\r\n \t
    4. Add the equations resulting from Step 2 to eliminate one variable.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
    5. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.<\/li>\r\n \t
    6. The two new equations form a system of two equations with two variables. Solve this system.<\/li>\r\n \t
    7. Use the values of the two variables found in Step 4 to find the third variable.<\/li>\r\n \t
    8. Write the solution as an ordered triple.<\/li>\r\n \t
    9. Check that the ordered triple is a solution to all three original equations<\/li>\r\n<\/ol>\r\n<\/section>
      Find a solution to the following system:\r\n
      [latex]\\begin{array}{lll}x-y+z=5& &\\text{(1)}\\\\-2y+z=6& &\\text{(2)}\\\\2y-2z=-12& &\\text{(3)}\\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"223787\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"223787\"]\r\n\r\nWe labeled the equations this time to be able to keep track of things a little more easily. The most obvious first step here is to eliminate [latex]y[\/latex] by adding equations (2) and (3).\r\n
      [latex]\\begin{array}{lll}\\,\\,\\,\\,\\,\\,\\,\\,\\,-2y+z=6\\,\\,\\,\\,(2)\\\\\\underline{\\,\\,\\,\\,2y-2z=-12}\\,\\,\\,\\,(3)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-z=-6\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,z=6\\end{array}[\/latex]<\/div>\r\n \r\n\r\nNow we can substitute the value for\u00a0[latex]z[\/latex] that we obtained into equation [latex](2)[\/latex].\r\n

      [latex]\\begin{array}{rrr}-2y+(6)=6\\\\-2y=6-6\\\\-2y=0\\\\\\,\\,\\,\\,y=0\\end{array}[\/latex]<\/p>\r\nBe careful here not to get confused with a solution of\u00a0[latex]y = 0[\/latex] and an inconsistent solution. \u00a0It is ok for variables to equal\u00a0[latex]0[\/latex].\r\n\r\nNow we can substitute\u00a0[latex]z = 6[\/latex] and\u00a0[latex]y = 0[\/latex] back into the first equation.\r\n

      [latex]\\begin{array}{rrr}x-y+z=5\\\\x-0+6=5\\\\x+6=5\\\\x=5-6\\\\x=-1\\end{array}[\/latex]<\/p>\r\nSolution: [latex]{(x,y,z)}={(-1,0,6)}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

      Solving three-by-three systems involves both creativity and careful, well-organized work. It will take some practice before it begins to feel natural.<\/section>
      Find a solution to the following system:\r\n

      [latex]\\begin{align}x - 2y+3z=9& &\\text{(1)} \\\\ -x+3y-z=-6& &\\text{(2)} \\\\ 2x - 5y+5z=17& &\\text{(3)} \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"462104\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"462104\"]\r\n\r\nThere will always be several choices as to where to begin, but the most obvious first step here is to eliminate [latex]x[\/latex] by adding equations (1) and (2).\r\n

      [latex]\\begin{align}x - 2y+3z&=9\\\\ -x+3y-z&=-6 \\\\ \\hline y+2z&=3 \\end{align}[\/latex][latex]\\hspace{5mm}\\begin{gathered}\\text{(1})\\\\ \\text{(2)}\\\\ \\text{(4)}\\end{gathered}[\/latex]<\/p>\r\nThe second step is multiplying equation (1) by [latex]-2[\/latex] and adding the result to equation (3). These two steps will eliminate the variable [latex]x[\/latex].\r\n

      [latex]\\begin{align}\u22122x+4y\u22126z&=\u221218 \\\\ 2x\u22125y+5z&=17 \\\\ \\hline \u2212y\u2212z&=\u22121\\end{align}[\/latex][latex]\\hspace{5mm}\\begin{align}&\\text{(2) multiplied by }\u22122\\\\&\\left(3\\right)\\\\&(5)\\end{align}[\/latex]<\/p>\r\nIn equations (4) and (5), we have created a new two-by-two system. We can solve for [latex]z[\/latex] by adding the two equations.\r\n

      [latex]\\begin{align}y+2z&=3 \\\\ -y-z&=-1 \\\\ \\hline z&=2 \\end{align}[\/latex][latex]\\hspace{5mm}\\begin{align}(4)\\\\(5)\\\\(6)\\end{align}[\/latex]<\/p>\r\nChoosing one equation from each new system, we obtain the upper triangular form:\r\n

      [latex]\\begin{align}x - 2y+3z&=9 && \\left(1\\right) \\\\ y+2z&=3 && \\left(4\\right) \\\\ z&=2 && \\left(6\\right) \\end{align}[\/latex]<\/p>\r\nNext, we back-substitute [latex]z=2[\/latex] into equation (4) and solve for [latex]y[\/latex].\r\n

      [latex]\\begin{align}y+2\\left(2\\right)&=3 \\\\ y+4&=3 \\\\ y&=-1 \\end{align}[\/latex]<\/p>\r\nFinally, we can back-substitute [latex]z=2[\/latex] and [latex]y=-1[\/latex] into equation (1). This will yield the solution for [latex]x[\/latex].\r\n

      [latex]\\begin{align} x - 2\\left(-1\\right)+3\\left(2\\right)&=9\\\\ x+2+6&=9\\\\ x&=1\\end{align}[\/latex]<\/p>\r\nThe solution is the ordered triple [latex]\\left(1,-1,2\\right)[\/latex].\r\n\r\n\"Three\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

      [ohm_question hide_question_numbers=1]321620[\/ohm_question]<\/section>
      [ohm_question hide_question_numbers=1]321621[\/ohm_question]<\/section>","rendered":"

      Solving a System of Three Equations in Three Variables by Elimination<\/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: Solve a system of linear equations with three variables by elimination<\/strong><\/p>\n
        \n
      1. Write the equations in standard form. If any coefficients are fractions, clear them.<\/li>\n
      2. Eliminate the same variable from two equations.\n
          \n
        1. Decide which variable you will eliminate.<\/li>\n
        2. Work with a pair of equations to eliminate the chosen variable.<\/li>\n
        3. Multiply one or both equations so that the coefficients of that variable are opposites.<\/li>\n
        4. Add the equations resulting from Step 2 to eliminate one variable.<\/li>\n<\/ol>\n<\/li>\n
        5. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.<\/li>\n
        6. The two new equations form a system of two equations with two variables. Solve this system.<\/li>\n
        7. Use the values of the two variables found in Step 4 to find the third variable.<\/li>\n
        8. Write the solution as an ordered triple.<\/li>\n
        9. Check that the ordered triple is a solution to all three original equations<\/li>\n<\/ol>\n<\/section>\n
          Find a solution to the following system:<\/p>\n
          [latex]\\begin{array}{lll}x-y+z=5& &\\text{(1)}\\\\-2y+z=6& &\\text{(2)}\\\\2y-2z=-12& &\\text{(3)}\\end{array}[\/latex]<\/div>\n