{"id":1149,"date":"2022-03-19T00:07:43","date_gmt":"2022-03-19T00:07:43","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=1149"},"modified":"2025-11-17T00:52:28","modified_gmt":"2025-11-17T00:52:28","slug":"2-7-system-of-linear-equations-in-three-variables","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/2-7-system-of-linear-equations-in-three-variables\/","title":{"raw":"2.7.2: Solving Systems of Linear Equations in Three Variables","rendered":"2.7.2: Solving Systems of Linear Equations in Three Variables"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n<div class=\"wrapper\">\r\n<div id=\"wrap\">\r\n<div id=\"content\" role=\"main\">\r\n<div id=\"post-524\" class=\"standard post-524 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Outcomes<\/h1>\r\n<ul>\r\n \t<li>Use back substitution to find a solution to a system of linear equations in three variables<\/li>\r\n \t<li>Solve a system of linear equations in three variables<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe know how to solve a system of two linear equations in two variables. So our goal in solving a system of three linear equations in three variable is to simplify it to a system of two linear equations in two variables. In order to to this we need to eliminate a variable by working with two pairs of equations to eliminate the same variable in each pair. By doing this we will end up with two new equations in two variables that we already know how to solve. Solving this new system will result in determining the values of two of the variables. We can then find the remaining variable by substitution.\r\n<div class=\"textbox\">\r\n<h3>SOLVING a linear system of three equations<\/h3>\r\n<ol>\r\n \t<li>Pick any pair of equations and eliminate one variable using the elimination (addition) method.<\/li>\r\n \t<li>Pick another pair of equations and eliminate the same variable using the elimination (addition) method.<\/li>\r\n \t<li>Solve the resulting system of the two new equations.<\/li>\r\n \t<li>Back-substitute known variables into any one of the original equations and solve for the missing variable.<\/li>\r\n<\/ol>\r\n<\/div>\r\nSolving a system with three variables is very similar to solving one with two variables. It is important to keep track of our work as the addition of one more equation creates more steps in the solution process.\r\n\r\nWe\u2019ll take the steps slowly in the following few examples. First, we\u2019ll look just at the last step in the process: back-substitution. Then, we\u2019ll look at an example that requires the elimination (addition) method to reach the first solution. We have also included some video examples that illustrate some of the different kinds of situations we may encounter when solving three-by-three systems.\r\n\r\nIn the example that follows, we will solve the system by using back-substitution.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nSolve the system of linear equations.\r\n\r\n[latex]\\left\\lbrace\\begin{array}{rrrrrrr}x&amp;-&amp;\\dfrac{1}{3}y&amp;+&amp;\\dfrac{1}{2}z&amp;=&amp;1\\\\&amp;{}&amp;y&amp;-&amp;\\dfrac{1}{2}z&amp;=&amp;4\\\\&amp;{}&amp;{}&amp;{}&amp;{}z&amp;=&amp;-1\\end{array}\\right.[\/latex]\r\n<h4>Solution<\/h4>\r\nThe third equation states that\u00a0[latex]z = \u22121[\/latex], so\u00a0we substitute this into the second equation to obtain a value for\u00a0[latex]y[\/latex].\r\n\r\n[latex]\\begin{aligned}y-\\dfrac{1}{2}z&amp;=4\\\\y-\\dfrac{1}{2}(-1)&amp;=4\\\\y+\\dfrac{1}{2}&amp;=4\\\\y&amp;=4-\\dfrac{1}{2}\\\\y&amp;=\\dfrac{8}{2}-\\dfrac{1}{2}\\\\y&amp;=\\dfrac{7}{2}\\end{aligned}[\/latex]\r\n\r\nNow we have two of our values, and we can substitute them both into the first equation to solve for\u00a0[latex]x[\/latex].\r\n\r\n[latex]\\begin{aligned}x-\\dfrac{1}{3}y+\\dfrac{1}{2}z&amp;=1\\\\x-\\dfrac{1}{3}\\left(\\dfrac{7}{2}\\right)+\\dfrac{1}{2}\\left(-1\\right)&amp;=1\\\\x-\\dfrac{7}{6}-\\dfrac{1}{2}&amp;=1\\\\x-\\dfrac{7}{6}-\\dfrac{3}{6}&amp;=1\\\\x-\\dfrac{10}{6}&amp;=1\\\\x&amp;=1+\\dfrac{10}{6}\\\\x&amp;=\\dfrac{6}{6}+\\dfrac{10}{6}\\\\x&amp;=\\dfrac{16}{6}\\\\x&amp;=\\dfrac{8}{3}\\end{aligned}[\/latex]\r\n\r\nNow we have our ordered triple; remember to place each variable value in order.\r\n\r\n[latex](x,y,z)=\\left(\\dfrac{8}{3},\\dfrac{7}{2},-1\\right)[\/latex]\r\n\r\n&nbsp;\r\n<h4>Analysis of the Solution:<\/h4>\r\nEach of the lines in the system above represents a plane. If you imagine three stiff sheets of 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<div id=\"attachment_2377\" class=\"wp-caption aligncenter\" style=\"width: 310px;\">\r\n\r\n<img class=\"aligncenter wp-image-2377 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11200619\/Screen-Shot-2016-07-11-at-1.04.41-PM-300x237.png\" alt=\"Three Planes Intersecting at a point.\" width=\"300\" height=\"237\" \/>\r\n<p class=\"wp-caption-text\">Three planes intersecting.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nThe following video shows another example of using back-substitution to solve a linear system in three variables.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/HHIjTChrIxE?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-1.docx\">Transcript-2.7.2-1<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nSolve the system of linear equations.\r\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}2x&amp;+&amp;3y&amp;-&amp;z&amp;=&amp;-1\\\\&amp;&amp;2y&amp;+&amp;3z&amp;=&amp;-3\\\\&amp;&amp;&amp;&amp;z&amp;=&amp;1\\end{array}\\right.[\/latex]<\/p>\r\n[reveal-answer q=\"hjm232\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm232\"]\r\n\r\n[latex]\\left (\\frac{9}{2}, \u20133,1\\right )[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn the next example we\u2019ll need to use the elimination method to find our first solution.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nFind a solution to the following system:\r\n<div style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrrr}x&amp;-&amp;y&amp;+&amp;z&amp;=&amp;5\\,\\,\\,\\,&amp;(1)\\\\&amp;&amp;-2y&amp;+&amp;z&amp;=&amp;6\\,\\,\\,\\,&amp;(2)\\\\{}&amp;{}&amp;2y&amp;-&amp;2z&amp;=&amp;-12\\,\\,\\,\\,&amp;(3)\\end{array}\\right.[\/latex]<\/div>\r\n<h4>Solution<\/h4>\r\nWe labeled the equations this time to be able to keep track of things a little more easily.\r\n\r\nLooking at the equations we see that equations (2) and (3) have only two variables. So we will solve this system by adding them together to eliminate [latex]y[\/latex] and solve for [latex]z[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{rrrrrr}-2y&amp;+&amp;z&amp;=&amp;6\\,\\,\\,\\,&amp;(2)\\\\2y&amp;-&amp;2z&amp;=&amp;-12\\,\\,\\,\\,&amp;(3)\\\\\\hline &amp;&amp;-z&amp;=&amp;-6&amp;{}\\\\&amp;&amp;z&amp;=&amp;6&amp;{}\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNow we can substitute in\u00a0 6 for [latex]z[\/latex] in equation (2):\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}-2y+(6)&amp;=6\\\\-2y&amp;=6-6\\\\-2y&amp;=0\\\\y&amp;=0\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nNOTE: Be 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\n&nbsp;\r\n\r\nNow we can substitute\u00a0[latex]z = 6[\/latex] and\u00a0[latex]y = 0[\/latex] into equation (1) and solve for [latex]x[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x-y+z&amp;=5\\\\x-0+6&amp;=5\\\\x+6&amp;=5\\\\x&amp;=5-6\\\\x&amp;=-1\\end{aligned}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex](x,y,z)=(-1,0,6)[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following videos for more examples of the algebra that may be encountered when solving systems with three variables.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/r6htz3gaHZ0?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-2.docx\">Transcript-2.7.2-2<\/a>\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/3RbVSvvRyeI?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-3.docx\">Transcript-2.7.2-3<\/a>\r\n\r\nSolving three-by-three systems involves both creativity and careful, well-organized work. It may take some practice before it begins to feel natural.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nSolve the system of linear equations:\r\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}x&amp;-&amp; 2y&amp;+&amp;3z&amp;=&amp;9\\\\ -x&amp;+&amp;3y&amp;-&amp;z&amp;=&amp;-6\\\\ 2x&amp;-&amp;5y&amp;+&amp;5z&amp;=&amp;17 \\end{array}\\right.[\/latex][latex]\\hspace{5mm}\\begin{gathered}\\text{(1})\\\\ \\text{(2)}\\\\ \\text{(3)}\\end{gathered}[\/latex]<\/p>\r\n\r\n<h4>Solution<\/h4>\r\nThere will always be several choices as to where to begin. Look for the variable that can be eliminated most easily from two pairs of equations. Let's start by eliminating [latex]x[\/latex]:\r\n\r\nAdd equations (1) and (2):\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrrrr}x &amp;-&amp; 2y&amp;+&amp;3z&amp;=&amp;9\\\\ -x&amp;+&amp;3y&amp;-&amp;z&amp;=&amp;-6 \\\\ \\hline &amp;&amp;y&amp;+&amp;2z&amp;=&amp;3 \\end{array}[\/latex][latex]\\hspace{5mm}\\begin{gathered}\\text{(1})\\\\ \\text{(2)}\\\\ \\text{(4)}\\end{gathered}[\/latex]<\/p>\r\nThe result is a new equation in two variables that we've labeled as equation (4).\r\n\r\nNow choose another pair of the original equations to eliminate [latex]x[\/latex]:\r\n\r\nUsing equations (1) and (3):\r\n\r\nWe need to multiply equation (1) by [latex]-2[\/latex] and add the result to equation (3).\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrrrr}\u22122x&amp;+&amp;4y&amp;\u2212&amp;6z&amp;=&amp;\u221218 \\\\ 2x&amp;\u2212&amp;5y&amp;+&amp;5z&amp;=&amp;17 \\\\ \\hline &amp;\u2212&amp;y&amp;\u2212&amp;z&amp;=&amp;\u22121\\end{array}[\/latex][latex]\\hspace{5mm}\\begin{align}&amp;\\text{(2) multiplied by }\u22122\\\\&amp;\\left(3\\right)\\\\&amp;(5)\\end{align}[\/latex]<\/p>\r\nThe result is equation (5). Equations (4) and (5) now form a system of two linear equations in the two variables y and z:\r\n\r\nWe can solve for [latex]z[\/latex] by adding equations (4) and (5):\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrr}y&amp;+&amp;2z&amp;=&amp;3 \\\\ -y&amp;-&amp;z&amp;=&amp;-1 \\\\ \\hline &amp;&amp;z&amp;=&amp;2 \\end{array}[\/latex][latex]\\hspace{5mm}\\begin{align}(4)\\\\(5)\\\\(6)\\end{align}[\/latex]<\/p>\r\nNow we know the value of [latex]x[\/latex], we can substitute into either equation (4) or (5) to find [latex]y[\/latex].\r\n\r\nBack-substitute [latex]z=2[\/latex] into equation (4) and solve for [latex]y[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}y+2\\left(2\\right)&amp;=3 \\\\ y+4&amp;=3 \\\\ y&amp;=-1 \\end{align}[\/latex]<\/p>\r\nKnowing the values of [latex]z[\/latex] and [latex]y[\/latex] allows us to\u00a0back-substitute [latex]z=2[\/latex] and [latex]y=-1[\/latex] into equation (1). This will yield the solution for [latex]x[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{align} x - 2\\left(-1\\right)+3\\left(2\\right)&amp;=9\\\\ x+2+6&amp;=9\\\\ x&amp;=1\\end{align}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe solution is the ordered triple [latex]\\left(1,-1,2\\right)[\/latex].\r\n\r\n&nbsp;\r\n<div class=\"wp-nocaption aligncenter\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03185119\/CNX_Precalc_Figure_09_02_0082.jpg\" alt=\"Three planes intersect at a point (1,-1, 2). The light blue plane's equation is x=1. The dark blue plane's equation is z=2. The orange plane's equation is y=-2.\" width=\"487\" height=\"324\" \/><\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>TRY IT 2<\/h3>\r\n<iframe id=\"ohm23765\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23765&amp;theme=oea&amp;iframe_resize_id=ohm23765&amp;show_question_numbers\" width=\"100%\" height=\"300\" data-mce-fragment=\"1\"><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><\/iframe>\r\n\r\n<\/div>\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-TryIt2.docx\">Transcript-2.7.2-TryIt2<\/a>\r\n\r\nIn the following video,\u00a0you will see a visual representation of the three possible outcomes for solutions to a system of equations in three variables. There is also a worked example of solving a system using elimination.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/wIE8KSpb-E8?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-4.docx\">Transcript-2.7.2-4<\/a>\r\n<h3>Dependent and Inconsistent Systems<\/h3>\r\nWhen solving a system of equations algebraically, we know we have a dependent system when we get an equation that is an identity (e.g., 0 = 0). We also know that if we end up with a contradiction, the system is inconsistent, and there is no solution.\r\n<div class=\"textbox examples\">\r\n<h3>Example 4<\/h3>\r\nSolve the following system.\r\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}x&amp;-&amp;3y&amp;+&amp;z&amp;=&amp;4\\\\{}&amp;-&amp;y&amp;-&amp;4z&amp;=&amp;7\\\\{}&amp;&amp;2y&amp;+&amp;8z&amp;=&amp;-12\\end{array}\\right.[\/latex][latex]\\hspace{5mm}\\begin{gathered}\\text{(1})\\\\ \\text{(2)}\\\\ \\text{(3)}\\end{gathered}[\/latex]<\/p>\r\n\r\n<h4 class=\"lang-tex s-code-block\">Solution<\/h4>\r\nFirst notice that equations (2) and (3) have no [latex]x[\/latex]-term. So we can solve them to find [latex]y[\/latex] and [latex]z[\/latex].\r\n\r\nTo eliminate [latex]y[\/latex] using equations (2) and (3), we can multiply equation (2) by\u00a02 and add equations (2) and (3) together:\r\n\r\nFirst, multiply both sides of equation (2) by 2:\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrr}\\color{blue}{2}(-y&amp;-&amp;4z)&amp;=&amp;\\color{blue}{2}(7)\\\\-2y&amp;-&amp;8z\\,&amp;=&amp;14\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Next, add equations\u00a0[latex](2)[\/latex] and\u00a0[latex](3)[\/latex] together to eliminate y and solve for\u00a0[latex]z[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrr}-2y&amp;-&amp;8z&amp;=&amp;14\\\\2y&amp;+&amp;8z&amp;=&amp;-12\\\\ \\hline 0&amp;+&amp;0&amp;=&amp;2\\\\&amp;{}&amp;0&amp;=&amp;2\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We have a contradiction so the two planes represented by equations (2) and (3) are parallel.<\/p>\r\nThis means there is no solution to the system of equations.\r\n\r\n<\/div>\r\nThe next video shows another example of using elimination to solve a system in three variables that ends up being <em><strong>inconsistent<\/strong> <\/em>and having no solution.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/ryNQsWrUoJw?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-5.docx\">Transcript-2.7.2-5<\/a>\r\n\r\nWe know from working with systems of equations in two variables that a\u00a0<strong><em>dependent system<\/em>\u00a0<\/strong>of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions. The other possibility is that two of the equations could be the same and intersect the third on a line.\r\n<div class=\"textbox examples\">\r\n<h3>Example 5<\/h3>\r\nFind the solution to the given system of three equations in three variables.\r\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}2x&amp;+&amp;y&amp;-&amp;3z&amp;=&amp;0&amp; \\hfill \\left(1\\right)\\\\ 4x&amp;+&amp;2y&amp; - &amp;6z&amp;=&amp;0&amp; \\hfill \\left(2\\right)\\\\ x&amp;-&amp;y&amp;+&amp;z&amp;=&amp;0&amp; \\hfill \\left(3\\right)\\end{array}\\right.[\/latex]<\/p>\r\n\r\n<h4>Solution<\/h4>\r\nFirst, we decide which variable looks easiest to eliminate: [latex]y[\/latex]\r\n\r\nWe can multiply equation\u00a0[latex](1)[\/latex] by [latex]-2[\/latex] and add it to equation\u00a0[latex](2)[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrrl} \u22124x&amp;\u2212&amp;2y&amp;+&amp;6z&amp;=&amp;0 \\hfill&amp;(1)\\text{ multiplied by }\u22122 \\\\ 4x&amp;+&amp;2y&amp;\u2212&amp;6z&amp;=&amp;0\\hfill&amp;(2) \\end{array}[\/latex]<\/p>\r\nThe result we get when adding the two equations is an identity, [latex]0=0[\/latex] which tells us that the two planes represented by equations (1) and (2) are coincidental.\r\n\r\nWe now need to determine if the third plane is a) coincidental with the first two (in which case there are infinite solutions on the plane), or b) parallel to the coincident planes (in which case the system has no solution), or c) intersects with the two coincidental planes in a line (in which case there are infinite solutions on the line).\r\n\r\nSInce equations (1) and (2) are equivalent, we can use either one with equation (3) to help make our determination.\r\n\r\nLet's consider equations (1) and (3). If we add (1) and (3) together, the [latex]y[\/latex]-terms will cancel each other out:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrrrr}2x&amp;+&amp;y&amp;-&amp;3z&amp;=&amp;0&amp; \\hfill \\left(1\\right)\\\\ x&amp;-&amp;y&amp;+&amp;z&amp;=&amp;0&amp; \\hfill \\left(3\\right)\\\\ \\hline3x&amp;&amp;&amp;-&amp;2z&amp;=&amp;0&amp;\\end{array}[\/latex]<\/p>\r\nThis is a linear equation in the two variables {latex]x[\/latex] and [latex]z[\/latex], so the two coincident planes\u00a0intersect the third plane in a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations.\r\n\r\n<\/div>\r\nThe last video example shows a dependent system that has an infinite number of solutions. It is enough for this course to know why the system is dependent. The video goes on to discuss how to write all variables in term of a\u00a0<em><strong>parameter<\/strong><\/em>. Although that is not covered in this class, we will see it in College Algebra.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/mThiwW8nYAU?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-6.docx\">Transcript-2.7.2-6<\/a>\r\n\r\nIt is important to know why a system of linear equations in three variables is dependent. The infinite solutions to such a system do not just float around anywhere in space! Rather they are either all on a line or on a plane.\r\n\r\nThe infinite solutions will be on a line if two of the three equations are equivalent. Graphically, this means that two of the planes are identical and the third plane crosses them in a line.\r\n\r\nThe infinite solutions will be on a line if the three planes meet in a single line.\r\n\r\nThe infinite solutions will be on a plane if all three equations are equivalent. Graphically this means that all three planes coincide and it is this plane that is the solution set.\r\n\r\n<strong>IMPORTANT NOTE:<\/strong>\u00a0 When we get an identity from manipulating two equations, that does not mean that there are automatically an infinite number of solutions! It simply means that two of the three planes are coincident. We still need to determine whether the third plane is also coincident (resulting in infinite solutions on the three coincident planes), is parallel to the two coincident planes (resulting in no solution), or intersects the two coincident planes in a line (resulting in infinite solutions on the line of intersection).\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nSolve the system of linear equations:\r\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}2x&amp;-&amp;3y&amp;+&amp;4z&amp;=&amp;9\\\\x&amp;+&amp;5y&amp;-&amp;3z&amp;=&amp;12\\\\-4x&amp;+&amp;6y&amp;-&amp;8z&amp;=&amp;-18\\end{array}\\right.[\/latex]<\/p>\r\n[reveal-answer q=\"hjm796\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm796\"]\r\n\r\nThe system is dependent. The infinite solutions lie on the line formed by two coincident planes (equations (1) and (3)) intersecting with the third plane (equation (2)).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nSolve the system of linear equations:\r\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}2x&amp;-&amp;3y&amp;+&amp;4z&amp;=&amp;9\\\\2x&amp;-&amp;3y&amp;+&amp;4z&amp;=&amp;12\\\\-4x&amp;+&amp;6y&amp;-&amp;8z&amp;=&amp;-5\\end{array}\\right.[\/latex]<\/p>\r\n[reveal-answer q=\"hjm799\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm799\"]\r\n\r\nThe system is inconsistent. There are no solutions. All three planes are parallel.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<div class=\"wrapper\">\n<div id=\"wrap\">\n<div id=\"content\" role=\"main\">\n<div id=\"post-524\" class=\"standard post-524 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div class=\"textbox learning-objectives\">\n<h1>Learning Outcomes<\/h1>\n<ul>\n<li>Use back substitution to find a solution to a system of linear equations in three variables<\/li>\n<li>Solve a system of linear equations in three variables<\/li>\n<\/ul>\n<\/div>\n<p>We know how to solve a system of two linear equations in two variables. So our goal in solving a system of three linear equations in three variable is to simplify it to a system of two linear equations in two variables. In order to to this we need to eliminate a variable by working with two pairs of equations to eliminate the same variable in each pair. By doing this we will end up with two new equations in two variables that we already know how to solve. Solving this new system will result in determining the values of two of the variables. We can then find the remaining variable by substitution.<\/p>\n<div class=\"textbox\">\n<h3>SOLVING a linear system of three equations<\/h3>\n<ol>\n<li>Pick any pair of equations and eliminate one variable using the elimination (addition) method.<\/li>\n<li>Pick another pair of equations and eliminate the same variable using the elimination (addition) method.<\/li>\n<li>Solve the resulting system of the two new equations.<\/li>\n<li>Back-substitute known variables into any one of the original equations and solve for the missing variable.<\/li>\n<\/ol>\n<\/div>\n<p>Solving a system with three variables is very similar to solving one with two variables. It is important to keep track of our work as the addition of one more equation creates more steps in the solution process.<\/p>\n<p>We\u2019ll take the steps slowly in the following few examples. First, we\u2019ll look just at the last step in the process: back-substitution. Then, we\u2019ll look at an example that requires the elimination (addition) method to reach the first solution. We have also included some video examples that illustrate some of the different kinds of situations we may encounter when solving three-by-three systems.<\/p>\n<p>In the example that follows, we will solve the system by using back-substitution.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Solve the system of linear equations.<\/p>\n<p>[latex]\\left\\lbrace\\begin{array}{rrrrrrr}x&-&\\dfrac{1}{3}y&+&\\dfrac{1}{2}z&=&1\\\\&{}&y&-&\\dfrac{1}{2}z&=&4\\\\&{}&{}&{}&{}z&=&-1\\end{array}\\right.[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>The third equation states that\u00a0[latex]z = \u22121[\/latex], so\u00a0we substitute this into the second equation to obtain a value for\u00a0[latex]y[\/latex].<\/p>\n<p>[latex]\\begin{aligned}y-\\dfrac{1}{2}z&=4\\\\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{aligned}[\/latex]<\/p>\n<p>Now we have two of our values, and we can substitute them both into the first equation to solve for\u00a0[latex]x[\/latex].<\/p>\n<p>[latex]\\begin{aligned}x-\\dfrac{1}{3}y+\\dfrac{1}{2}z&=1\\\\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}\\\\x&=\\dfrac{8}{3}\\end{aligned}[\/latex]<\/p>\n<p>Now we have our ordered triple; remember to place each variable value in order.<\/p>\n<p>[latex](x,y,z)=\\left(\\dfrac{8}{3},\\dfrac{7}{2},-1\\right)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<h4>Analysis of the Solution:<\/h4>\n<p>Each of the lines in the system above represents a plane. If you imagine three stiff sheets of 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.<\/p>\n<div id=\"attachment_2377\" class=\"wp-caption aligncenter\" style=\"width: 310px;\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2377 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11200619\/Screen-Shot-2016-07-11-at-1.04.41-PM-300x237.png\" alt=\"Three Planes Intersecting at a point.\" width=\"300\" height=\"237\" \/><\/p>\n<p class=\"wp-caption-text\">Three planes intersecting.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows another example of using back-substitution to solve a linear system in three variables.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/HHIjTChrIxE?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-1.docx\">Transcript-2.7.2-1<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Solve the system of linear equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}2x&+&3y&-&z&=&-1\\\\&&2y&+&3z&=&-3\\\\&&&&z&=&1\\end{array}\\right.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm232\">Show Answer<\/span><\/p>\n<div id=\"qhjm232\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left (\\frac{9}{2}, \u20133,1\\right )[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In the next example we\u2019ll need to use the elimination method to find our first solution.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Find a solution to the following system:<\/p>\n<div style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrrr}x&-&y&+&z&=&5\\,\\,\\,\\,&(1)\\\\&&-2y&+&z&=&6\\,\\,\\,\\,&(2)\\\\{}&{}&2y&-&2z&=&-12\\,\\,\\,\\,&(3)\\end{array}\\right.[\/latex]<\/div>\n<h4>Solution<\/h4>\n<p>We labeled the equations this time to be able to keep track of things a little more easily.<\/p>\n<p>Looking at the equations we see that equations (2) and (3) have only two variables. So we will solve this system by adding them together to eliminate [latex]y[\/latex] and solve for [latex]z[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rrrrrr}-2y&+&z&=&6\\,\\,\\,\\,&(2)\\\\2y&-&2z&=&-12\\,\\,\\,\\,&(3)\\\\\\hline &&-z&=&-6&{}\\\\&&z&=&6&{}\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Now we can substitute in\u00a0 6 for [latex]z[\/latex] in equation (2):<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}-2y+(6)&=6\\\\-2y&=6-6\\\\-2y&=0\\\\y&=0\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>NOTE: Be 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].<\/p>\n<p>&nbsp;<\/p>\n<p>Now we can substitute\u00a0[latex]z = 6[\/latex] and\u00a0[latex]y = 0[\/latex] into equation (1) and solve for [latex]x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x-y+z&=5\\\\x-0+6&=5\\\\x+6&=5\\\\x&=5-6\\\\x&=-1\\end{aligned}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex](x,y,z)=(-1,0,6)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following videos for more examples of the algebra that may be encountered when solving systems with three variables.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/r6htz3gaHZ0?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-2.docx\">Transcript-2.7.2-2<\/a><\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/3RbVSvvRyeI?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-3.docx\">Transcript-2.7.2-3<\/a><\/p>\n<p>Solving three-by-three systems involves both creativity and careful, well-organized work. It may take some practice before it begins to feel natural.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Solve the system of linear equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}x&-& 2y&+&3z&=&9\\\\ -x&+&3y&-&z&=&-6\\\\ 2x&-&5y&+&5z&=&17 \\end{array}\\right.[\/latex][latex]\\hspace{5mm}\\begin{gathered}\\text{(1})\\\\ \\text{(2)}\\\\ \\text{(3)}\\end{gathered}[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>There will always be several choices as to where to begin. Look for the variable that can be eliminated most easily from two pairs of equations. Let&#8217;s start by eliminating [latex]x[\/latex]:<\/p>\n<p>Add equations (1) and (2):<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrrrr}x &-& 2y&+&3z&=&9\\\\ -x&+&3y&-&z&=&-6 \\\\ \\hline &&y&+&2z&=&3 \\end{array}[\/latex][latex]\\hspace{5mm}\\begin{gathered}\\text{(1})\\\\ \\text{(2)}\\\\ \\text{(4)}\\end{gathered}[\/latex]<\/p>\n<p>The result is a new equation in two variables that we&#8217;ve labeled as equation (4).<\/p>\n<p>Now choose another pair of the original equations to eliminate [latex]x[\/latex]:<\/p>\n<p>Using equations (1) and (3):<\/p>\n<p>We need to multiply equation (1) by [latex]-2[\/latex] and add the result to equation (3).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrrrr}\u22122x&+&4y&\u2212&6z&=&\u221218 \\\\ 2x&\u2212&5y&+&5z&=&17 \\\\ \\hline &\u2212&y&\u2212&z&=&\u22121\\end{array}[\/latex][latex]\\hspace{5mm}\\begin{align}&\\text{(2) multiplied by }\u22122\\\\&\\left(3\\right)\\\\&(5)\\end{align}[\/latex]<\/p>\n<p>The result is equation (5). Equations (4) and (5) now form a system of two linear equations in the two variables y and z:<\/p>\n<p>We can solve for [latex]z[\/latex] by adding equations (4) and (5):<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrr}y&+&2z&=&3 \\\\ -y&-&z&=&-1 \\\\ \\hline &&z&=&2 \\end{array}[\/latex][latex]\\hspace{5mm}\\begin{align}(4)\\\\(5)\\\\(6)\\end{align}[\/latex]<\/p>\n<p>Now we know the value of [latex]x[\/latex], we can substitute into either equation (4) or (5) to find [latex]y[\/latex].<\/p>\n<p>Back-substitute [latex]z=2[\/latex] into equation (4) and solve for [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}y+2\\left(2\\right)&=3 \\\\ y+4&=3 \\\\ y&=-1 \\end{align}[\/latex]<\/p>\n<p>Knowing the values of [latex]z[\/latex] and [latex]y[\/latex] allows us to\u00a0back-substitute [latex]z=2[\/latex] and [latex]y=-1[\/latex] into equation (1). This will yield the solution for [latex]x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} x - 2\\left(-1\\right)+3\\left(2\\right)&=9\\\\ x+2+6&=9\\\\ x&=1\\end{align}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The solution is the ordered triple [latex]\\left(1,-1,2\\right)[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<div class=\"wp-nocaption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03185119\/CNX_Precalc_Figure_09_02_0082.jpg\" alt=\"Three planes intersect at a point (1,-1, 2). The light blue plane's equation is x=1. The dark blue plane's equation is z=2. The orange plane's equation is y=-2.\" width=\"487\" height=\"324\" \/><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>TRY IT 2<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm23765\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23765&amp;theme=oea&amp;iframe_resize_id=ohm23765&amp;show_question_numbers\" width=\"100%\" height=\"300\" data-mce-fragment=\"1\"><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><\/iframe><\/p>\n<\/div>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-TryIt2.docx\">Transcript-2.7.2-TryIt2<\/a><\/p>\n<p>In the following video,\u00a0you will see a visual representation of the three possible outcomes for solutions to a system of equations in three variables. There is also a worked example of solving a system using elimination.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/wIE8KSpb-E8?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-4.docx\">Transcript-2.7.2-4<\/a><\/p>\n<h3>Dependent and Inconsistent Systems<\/h3>\n<p>When solving a system of equations algebraically, we know we have a dependent system when we get an equation that is an identity (e.g., 0 = 0). We also know that if we end up with a contradiction, the system is inconsistent, and there is no solution.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4<\/h3>\n<p>Solve the following system.<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}x&-&3y&+&z&=&4\\\\{}&-&y&-&4z&=&7\\\\{}&&2y&+&8z&=&-12\\end{array}\\right.[\/latex][latex]\\hspace{5mm}\\begin{gathered}\\text{(1})\\\\ \\text{(2)}\\\\ \\text{(3)}\\end{gathered}[\/latex]<\/p>\n<h4 class=\"lang-tex s-code-block\">Solution<\/h4>\n<p>First notice that equations (2) and (3) have no [latex]x[\/latex]-term. So we can solve them to find [latex]y[\/latex] and [latex]z[\/latex].<\/p>\n<p>To eliminate [latex]y[\/latex] using equations (2) and (3), we can multiply equation (2) by\u00a02 and add equations (2) and (3) together:<\/p>\n<p>First, multiply both sides of equation (2) by 2:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrr}\\color{blue}{2}(-y&-&4z)&=&\\color{blue}{2}(7)\\\\-2y&-&8z\\,&=&14\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Next, add equations\u00a0[latex](2)[\/latex] and\u00a0[latex](3)[\/latex] together to eliminate y and solve for\u00a0[latex]z[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrr}-2y&-&8z&=&14\\\\2y&+&8z&=&-12\\\\ \\hline 0&+&0&=&2\\\\&{}&0&=&2\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">We have a contradiction so the two planes represented by equations (2) and (3) are parallel.<\/p>\n<p>This means there is no solution to the system of equations.<\/p>\n<\/div>\n<p>The next video shows another example of using elimination to solve a system in three variables that ends up being <em><strong>inconsistent<\/strong> <\/em>and having no solution.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/ryNQsWrUoJw?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-5.docx\">Transcript-2.7.2-5<\/a><\/p>\n<p>We know from working with systems of equations in two variables that a\u00a0<strong><em>dependent system<\/em>\u00a0<\/strong>of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions. The other possibility is that two of the equations could be the same and intersect the third on a line.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 5<\/h3>\n<p>Find the solution to the given system of three equations in three variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}2x&+&y&-&3z&=&0& \\hfill \\left(1\\right)\\\\ 4x&+&2y& - &6z&=&0& \\hfill \\left(2\\right)\\\\ x&-&y&+&z&=&0& \\hfill \\left(3\\right)\\end{array}\\right.[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>First, we decide which variable looks easiest to eliminate: [latex]y[\/latex]<\/p>\n<p>We can multiply equation\u00a0[latex](1)[\/latex] by [latex]-2[\/latex] and add it to equation\u00a0[latex](2)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrrl} \u22124x&\u2212&2y&+&6z&=&0 \\hfill&(1)\\text{ multiplied by }\u22122 \\\\ 4x&+&2y&\u2212&6z&=&0\\hfill&(2) \\end{array}[\/latex]<\/p>\n<p>The result we get when adding the two equations is an identity, [latex]0=0[\/latex] which tells us that the two planes represented by equations (1) and (2) are coincidental.<\/p>\n<p>We now need to determine if the third plane is a) coincidental with the first two (in which case there are infinite solutions on the plane), or b) parallel to the coincident planes (in which case the system has no solution), or c) intersects with the two coincidental planes in a line (in which case there are infinite solutions on the line).<\/p>\n<p>SInce equations (1) and (2) are equivalent, we can use either one with equation (3) to help make our determination.<\/p>\n<p>Let&#8217;s consider equations (1) and (3). If we add (1) and (3) together, the [latex]y[\/latex]-terms will cancel each other out:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrrrrrr}2x&+&y&-&3z&=&0& \\hfill \\left(1\\right)\\\\ x&-&y&+&z&=&0& \\hfill \\left(3\\right)\\\\ \\hline3x&&&-&2z&=&0&\\end{array}[\/latex]<\/p>\n<p>This is a linear equation in the two variables {latex]x[\/latex] and [latex]z[\/latex], so the two coincident planes\u00a0intersect the third plane in a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations.<\/p>\n<\/div>\n<p>The last video example shows a dependent system that has an infinite number of solutions. It is enough for this course to know why the system is dependent. The video goes on to discuss how to write all variables in term of a\u00a0<em><strong>parameter<\/strong><\/em>. Although that is not covered in this class, we will see it in College Algebra.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/mThiwW8nYAU?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.7.2-6.docx\">Transcript-2.7.2-6<\/a><\/p>\n<p>It is important to know why a system of linear equations in three variables is dependent. The infinite solutions to such a system do not just float around anywhere in space! Rather they are either all on a line or on a plane.<\/p>\n<p>The infinite solutions will be on a line if two of the three equations are equivalent. Graphically, this means that two of the planes are identical and the third plane crosses them in a line.<\/p>\n<p>The infinite solutions will be on a line if the three planes meet in a single line.<\/p>\n<p>The infinite solutions will be on a plane if all three equations are equivalent. Graphically this means that all three planes coincide and it is this plane that is the solution set.<\/p>\n<p><strong>IMPORTANT NOTE:<\/strong>\u00a0 When we get an identity from manipulating two equations, that does not mean that there are automatically an infinite number of solutions! It simply means that two of the three planes are coincident. We still need to determine whether the third plane is also coincident (resulting in infinite solutions on the three coincident planes), is parallel to the two coincident planes (resulting in no solution), or intersects the two coincident planes in a line (resulting in infinite solutions on the line of intersection).<\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Solve the system of linear equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}2x&-&3y&+&4z&=&9\\\\x&+&5y&-&3z&=&12\\\\-4x&+&6y&-&8z&=&-18\\end{array}\\right.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm796\">Show Answer<\/span><\/p>\n<div id=\"qhjm796\" class=\"hidden-answer\" style=\"display: none\">\n<p>The system is dependent. The infinite solutions lie on the line formed by two coincident planes (equations (1) and (3)) intersecting with the third plane (equation (2)).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Solve the system of linear equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\lbrace\\begin{array}{rrrrrrr}2x&-&3y&+&4z&=&9\\\\2x&-&3y&+&4z&=&12\\\\-4x&+&6y&-&8z&=&-5\\end{array}\\right.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm799\">Show Answer<\/span><\/p>\n<div id=\"qhjm799\" class=\"hidden-answer\" style=\"display: none\">\n<p>The system is inconsistent. There are no solutions. All three planes are parallel.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1149\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Dependent and Inconsistent Systems; Try It: hjm232; hjm796; hjm799. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 23765. <strong>Authored by<\/strong>: Shahbazian,Roy. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Systems of Equations in Three Variables: Part 1 of 2 . <strong>Authored by<\/strong>: Sousa, James (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/wIE8KSpb-E8\">https:\/\/youtu.be\/wIE8KSpb-E8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 5: System of Three Equations with Three Unknowns Using Elimination (Infinite Solutions). <strong>Authored by<\/strong>: James Sousa (mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/mThiwW8nYAU\">https:\/\/youtu.be\/mThiwW8nYAU<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 4: System of Three Equations with Three Unknowns Using Elimination (No Solution). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"\"><\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":422608,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"Hazel McKenna\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 23765\",\"author\":\"Shahbazian,Roy\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Systems of Equations in Three Variables: Part 1 of 2 \",\"author\":\"Sousa, James (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/wIE8KSpb-E8\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Dependent and Inconsistent Systems; 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