{"id":2208,"date":"2016-11-03T18:56:50","date_gmt":"2016-11-03T18:56:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2208"},"modified":"2017-04-04T19:43:18","modified_gmt":"2017-04-04T19:43:18","slug":"solve-systems-of-three-equations-in-three-variables","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/solve-systems-of-three-equations-in-three-variables\/","title":{"raw":"Solve Systems of Three Equations in Three Variables","rendered":"Solve Systems of Three Equations in Three Variables"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Determine whether an ordered triple is a solution to a system of three equations<\/li>\r\n \t<li>Use back substitution to find a solution to a system of three equations<\/li>\r\n \t<li>Write the equations for a system given a scenario, and solve<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn order to solve systems of equations in three variables, known as three-by-three systems, the primary goal is to eliminate one variable at a time to achieve back-substitution. \u00a0A solution to a system of three equations in three variables [latex]\\left(x,y,z\\right),\\text{}[\/latex] is\u00a0called an <strong>ordered triple<\/strong>.\r\n\r\nTo find a solution, we can perform the following operations:\r\n<ol>\r\n \t<li>Interchange the order of any two equations.<\/li>\r\n \t<li>Multiply both sides of an equation by a nonzero constant.<\/li>\r\n \t<li>Add a nonzero multiple of one equation to another equation.<\/li>\r\n<\/ol>\r\nGraphically, the ordered triple defines the point that is the intersection of three planes in space. You can visualize such an intersection by imagining any corner in a rectangular room. A corner is defined by three planes: two adjoining walls and the floor (or ceiling). Any point where two walls and the floor meet represents the intersection of three planes.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Number of Possible Solutions<\/h3>\r\nThe planes\u00a0illustrate possible solution scenarios for three-by-three systems.\r\n<ul>\r\n \t<li>Systems that have a single solution are those which, after elimination, result in a <strong>solution set<\/strong> consisting of an ordered triple [latex]\\left\\{\\left(x,y,z\\right)\\right\\}[\/latex]. Graphically, the ordered triple defines a point that is the intersection of three planes in space.<\/li>\r\n \t<li>Systems that have an infinite number of solutions are those which, after elimination, result in an expression that is always true, such as [latex]0=0[\/latex]. Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space.<\/li>\r\n \t<li>Systems that have no solution are those that, after elimination, result in a statement that is a contradiction, such as [latex]3=0[\/latex]. Graphically, a system with no solution is represented by three planes with no point in common.<\/li>\r\n<\/ul>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03185113\/CNX_Precalc_Figure_09_02_006n2.jpg\" alt=\"\" width=\"487\" height=\"238\" data-media-type=\"image\/jpg\" \/> (a)Three planes intersect at a single point, representing a three-by-three system with a single solution. (b) Three planes intersect in a line, representing a three-by-three system with infinite solutions.[\/caption]\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03185116\/CNX_Precalc_Figure_09_02_007n2.jpg\" width=\"487\" height=\"188\" data-media-type=\"image\/jpg\" \/>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Determining Whether an Ordered Triple Is a Solution to a System<\/h3>\r\nDetermine whether the ordered triple [latex]\\left(3,-2,1\\right)[\/latex] is a solution to the system.\r\n<p style=\"text-align: center;\">[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\"]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<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\\begin{array}{r}\\hfill x+y+z=2\\\\ \\hfill \\left(3\\right)+\\left(-2\\right)+\\left(1\\right)=2\\\\ \\hfill \\text{True}\\end{array}&amp; &amp; \\begin{array}{r}\\hfill \\text{}6x - 4y+5z=31\\\\ \\hfill 6\\left(3\\right)-4\\left(-2\\right)+5\\left(1\\right)=31\\\\ \\hfill 18+8+5=31\\\\ \\hfill \\text{True}\\end{array}&amp; &amp; \\begin{array}{r}\\hfill \\text{}5x+2y+2z=13\\\\ \\hfill 5\\left(3\\right)+2\\left(-2\\right)+2\\left(1\\right)=13\\\\ \\hfill \\text{}15 - 4+2=13\\\\ \\hfill \\text{True}\\end{array}\\end{array}[\/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<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a linear system of three equations, solve for three unknowns.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Pick any pair of equations and solve for one variable.<\/li>\r\n \t<li>Pick another pair of equations and solve for the same variable.<\/li>\r\n \t<li>You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.<\/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\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a System of Three Equations in Three Variables by Elimination<\/h3>\r\nFind a solution to the following system:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\text{ }x - 2y+3z=9\\hfill &amp; \\text{(1)}\\hfill \\\\ \\text{ }-x+3y-z=-6\\hfill &amp; \\text{(2)}\\hfill \\\\ 2x - 5y+5z=17\\hfill &amp; \\text{(3)}\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"462104\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"462104\"]\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<p style=\"text-align: center;\">[latex]\\frac{\\begin{array}{ll}\\text{ }\\text{}x - 2y+3z=9\\hfill &amp; \\text{(1)}\\hfill \\\\ \\text{ }-x+3y-z=-6\\hfill &amp; \\text{ (2)}\\hfill \\end{array}}{\\begin{array}{ll}\\text{ }\\text{}\\text{}y+2z=3\\hfill &amp; \\text{ (3)}\\hfill \\end{array}}[\/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<p style=\"text-align: center;\">[latex]\\begin{array} \\hfill\u22122x+4y\u22126z=\u221218 \\hfill&amp; \\left(1\\right)\\text{ multiplied by }\u22122 \\\\ 2x\u22125y+5z=17 \\hfill&amp; \\left(3\\right) \\\\ \\text{_____________________________} \\\\ \\hfill\u2212y\u2212z=\u22121\\left(5\\right)\\hfill&amp; \\end{array}[\/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<p style=\"text-align: center;\">[latex]\\frac{\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\text{}y+2z=3\\text{ }\\left(4\\right)\\hfill \\end{array}\\hfill \\\\ -y-z=-1\\text{ }\\left(5\\right)\\hfill \\end{array}}{z=2\\text{ }\\left(6\\right)}[\/latex]<\/p>\r\nChoosing one equation from each new system, we obtain the upper triangular form:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\text{}\\text{}x - 2y+3z=9\\text{ }\\hfill &amp; \\left(1\\right)\\hfill \\\\ \\text{ }y+2z=3\\hfill &amp; \\left(4\\right)\\hfill \\\\ \\text{ }z=2\\hfill &amp; \\left(6\\right)\\hfill \\end{array}[\/latex]<\/p>\r\nNext, we back-substitute [latex]z=2[\/latex] into equation (4) and solve for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y+2\\left(2\\right)=3\\hfill \\\\ \\text{ }y+4=3\\hfill \\\\ \\text{ }y=-1\\hfill \\end{array}[\/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<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill x - 2\\left(-1\\right)+3\\left(2\\right)=9\\\\ \\hfill \\text{ }x+2+6=9\\\\ \\hfill \\text{ }x=1\\end{array}[\/latex]<\/p>\r\nThe solution is the ordered triple [latex]\\left(1,-1,2\\right)[\/latex].\r\n\r\n<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 1, negative 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=negative 2.\" width=\"487\" height=\"324\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the system of equations in three variables.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+y - 2z=-1\\hfill \\\\ 3x - 3y-z=5\\hfill \\\\ x - 2y+3z=6\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"232738\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"232738\"][latex]\\left(1,-1,1\\right)[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=23765&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\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\nhttps:\/\/youtu.be\/wIE8KSpb-E8\r\n\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Real-World Problem Using a System of Three Equations in Three Variables<\/h3>\r\nIn the problem posed at the beginning of the section, John invested his inheritance of $12,000 in three different funds: part in a money-market fund paying 3% interest annually; part in municipal bonds paying 4% annually; and the rest in mutual funds paying 7% annually. John invested $4,000 more in mutual funds than he invested in municipal bonds. The total interest earned in one year was $670. How much did he invest in each type of fund?\r\n\r\n[reveal-answer q=\"348091\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"348091\"]\r\nTo solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\text{amount invested in money-market fund}\\hfill \\\\ y=\\text{amount invested in municipal bonds}\\hfill \\\\ z=\\text{amount invested in mutual funds}\\hfill \\end{array}[\/latex]<\/p>\r\nThe first equation indicates that the sum of the three principal amounts is $12,000.\r\n<p style=\"text-align: center;\">[latex]x+y+z=12,000[\/latex]<\/p>\r\nWe form the second equation according to the information that John invested $4,000 more in mutual funds than he invested in municipal bonds.\r\n<p style=\"text-align: center;\">[latex]z=y+4,000[\/latex]<\/p>\r\nThe third equation shows that the total amount of interest earned from each fund equals $670.\r\n<p style=\"text-align: center;\">[latex]0.03x+0.04y+0.07z=670[\/latex]<\/p>\r\nThen, we write the three equations as a system.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+y+z=12,000\\hfill \\\\ \\text{ }-y+z=4,000\\hfill \\\\ 0.03x+0.04y+0.07z=670\\hfill \\end{array}[\/latex]<\/p>\r\nTo make the calculations simpler, we can multiply the third equation by 100. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\text{ }x+\\text{ }y+z\\text{ }=12,000\\hfill &amp; \\left(1\\right)\\hfill \\\\ \\text{ }-y+z\\text{ }=4,000\\hfill &amp; \\left(2\\right)\\hfill \\\\ 3x+4y+7z=67,000\\hfill &amp; \\left(3\\right)\\hfill \\end{array}[\/latex]<\/p>\r\n<strong>Step 1.<\/strong> Interchange equation (2) and equation (3) so that the two equations with three variables will line up.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+\\text{ }y +\\text{ }z=12,000\\hfill \\\\ 3x+4y +7z=67,000\\hfill \\\\ \\text{ }-y\\text{ }+\\text{ }z=4,000\\hfill \\end{array}[\/latex]<\/p>\r\n<strong>Step 2.<\/strong> Multiply equation (1) by [latex]-3[\/latex] and add to equation (2). Write the result as row 2.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+y+z\\text{ }=12,000\\hfill \\\\ \\text{ }y+4z=31,000\\hfill \\\\ \\text{ }-y+z\\text{ }=4,000\\hfill \\end{array}[\/latex]<\/p>\r\n<strong>Step 3.<\/strong> Add equation (2) to equation (3) and write the result as equation (3).\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+y+\\text{ }z=12,000\\hfill \\\\ \\text{ }y+4z=31,000\\hfill \\\\ \\text{ }5z\\text{ }=35,000\\hfill \\end{array}[\/latex]<\/p>\r\n<strong>Step 4.<\/strong> Solve for [latex]z[\/latex] in equation (3). Back-substitute that value in equation (2) and solve for [latex]y[\/latex]. Then, back-substitute the values for [latex]z[\/latex] and [latex]y[\/latex] into equation (1) and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }5z=35,000\\hfill \\\\ \\text{ }z=7,000\\hfill \\\\ \\hfill \\\\ \\hfill \\\\ \\text{ }y+4\\left(7,000\\right)=31,000\\hfill \\\\ \\text{ }y=3,000\\hfill \\\\ \\hfill \\\\ \\hfill \\\\ x+3,000+7,000=12,000\\hfill \\\\ \\text{ }x=2,000\\hfill \\end{array}[\/latex]<\/p>\r\nJohn invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=19353&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Determine whether an ordered triple is a solution to a system of three equations<\/li>\n<li>Use back substitution to find a solution to a system of three equations<\/li>\n<li>Write the equations for a system given a scenario, and solve<\/li>\n<\/ul>\n<\/div>\n<p>In order to solve systems of equations in three variables, known as three-by-three systems, the primary goal is to eliminate one variable at a time to achieve back-substitution. \u00a0A solution to a system of three equations in three variables [latex]\\left(x,y,z\\right),\\text{}[\/latex] is\u00a0called an <strong>ordered triple<\/strong>.<\/p>\n<p>To find a solution, we can perform the following operations:<\/p>\n<ol>\n<li>Interchange the order of any two equations.<\/li>\n<li>Multiply both sides of an equation by a nonzero constant.<\/li>\n<li>Add a nonzero multiple of one equation to another equation.<\/li>\n<\/ol>\n<p>Graphically, the ordered triple defines the point that is the intersection of three planes in space. You can visualize such an intersection by imagining any corner in a rectangular room. A corner is defined by three planes: two adjoining walls and the floor (or ceiling). Any point where two walls and the floor meet represents the intersection of three planes.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Number of Possible Solutions<\/h3>\n<p>The planes\u00a0illustrate possible solution scenarios for three-by-three systems.<\/p>\n<ul>\n<li>Systems that have a single solution are those which, after elimination, result in a <strong>solution set<\/strong> consisting of an ordered triple [latex]\\left\\{\\left(x,y,z\\right)\\right\\}[\/latex]. Graphically, the ordered triple defines a point that is the intersection of three planes in space.<\/li>\n<li>Systems that have an infinite number of solutions are those which, after elimination, result in an expression that is always true, such as [latex]0=0[\/latex]. Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space.<\/li>\n<li>Systems that have no solution are those that, after elimination, result in a statement that is a contradiction, such as [latex]3=0[\/latex]. Graphically, a system with no solution is represented by three planes with no point in common.<\/li>\n<\/ul>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03185113\/CNX_Precalc_Figure_09_02_006n2.jpg\" alt=\"\" width=\"487\" height=\"238\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">(a)Three planes intersect at a single point, representing a three-by-three system with a single solution. (b) Three planes intersect in a line, representing a three-by-three system with infinite solutions.<\/p>\n<\/div>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03185116\/CNX_Precalc_Figure_09_02_007n2.jpg\" width=\"487\" height=\"188\" data-media-type=\"image\/jpg\" alt=\"image\" \/><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Determining Whether an Ordered Triple Is a Solution to a System<\/h3>\n<p>Determine whether the ordered triple [latex]\\left(3,-2,1\\right)[\/latex] is a solution to the system.<\/p>\n<p style=\"text-align: center;\">[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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q954288\">Solution<\/span><\/p>\n<div id=\"q954288\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will check each equation by substituting in the values of the ordered triple for [latex]x,y[\/latex], and [latex]z[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\\begin{array}{r}\\hfill x+y+z=2\\\\ \\hfill \\left(3\\right)+\\left(-2\\right)+\\left(1\\right)=2\\\\ \\hfill \\text{True}\\end{array}& & \\begin{array}{r}\\hfill \\text{}6x - 4y+5z=31\\\\ \\hfill 6\\left(3\\right)-4\\left(-2\\right)+5\\left(1\\right)=31\\\\ \\hfill 18+8+5=31\\\\ \\hfill \\text{True}\\end{array}& & \\begin{array}{r}\\hfill \\text{}5x+2y+2z=13\\\\ \\hfill 5\\left(3\\right)+2\\left(-2\\right)+2\\left(1\\right)=13\\\\ \\hfill \\text{}15 - 4+2=13\\\\ \\hfill \\text{True}\\end{array}\\end{array}[\/latex]<\/p>\n<p>The ordered triple [latex]\\left(3,-2,1\\right)[\/latex] is indeed a solution to the system.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a linear system of three equations, solve for three unknowns.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Pick any pair of equations and solve for one variable.<\/li>\n<li>Pick another pair of equations and solve for the same variable.<\/li>\n<li>You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.<\/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<div class=\"textbox exercises\">\n<h3>Example: Solving a System of Three Equations in Three Variables by Elimination<\/h3>\n<p>Find a solution to the following system:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\text{ }x - 2y+3z=9\\hfill & \\text{(1)}\\hfill \\\\ \\text{ }-x+3y-z=-6\\hfill & \\text{(2)}\\hfill \\\\ 2x - 5y+5z=17\\hfill & \\text{(3)}\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q462104\">Solution<\/span><\/p>\n<div id=\"q462104\" class=\"hidden-answer\" style=\"display: none\">\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).<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\begin{array}{ll}\\text{ }\\text{}x - 2y+3z=9\\hfill & \\text{(1)}\\hfill \\\\ \\text{ }-x+3y-z=-6\\hfill & \\text{ (2)}\\hfill \\end{array}}{\\begin{array}{ll}\\text{ }\\text{}\\text{}y+2z=3\\hfill & \\text{ (3)}\\hfill \\end{array}}[\/latex]<\/p>\n<p>The 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].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array} \\hfill\u22122x+4y\u22126z=\u221218 \\hfill& \\left(1\\right)\\text{ multiplied by }\u22122 \\\\ 2x\u22125y+5z=17 \\hfill& \\left(3\\right) \\\\ \\text{_____________________________} \\\\ \\hfill\u2212y\u2212z=\u22121\\left(5\\right)\\hfill& \\end{array}[\/latex]<\/p>\n<p>In 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.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\text{}y+2z=3\\text{ }\\left(4\\right)\\hfill \\end{array}\\hfill \\\\ -y-z=-1\\text{ }\\left(5\\right)\\hfill \\end{array}}{z=2\\text{ }\\left(6\\right)}[\/latex]<\/p>\n<p>Choosing one equation from each new system, we obtain the upper triangular form:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\text{}\\text{}x - 2y+3z=9\\text{ }\\hfill & \\left(1\\right)\\hfill \\\\ \\text{ }y+2z=3\\hfill & \\left(4\\right)\\hfill \\\\ \\text{ }z=2\\hfill & \\left(6\\right)\\hfill \\end{array}[\/latex]<\/p>\n<p>Next, we back-substitute [latex]z=2[\/latex] into equation (4) and solve for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y+2\\left(2\\right)=3\\hfill \\\\ \\text{ }y+4=3\\hfill \\\\ \\text{ }y=-1\\hfill \\end{array}[\/latex]<\/p>\n<p>Finally, 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].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill x - 2\\left(-1\\right)+3\\left(2\\right)=9\\\\ \\hfill \\text{ }x+2+6=9\\\\ \\hfill \\text{ }x=1\\end{array}[\/latex]<\/p>\n<p>The solution is the ordered triple [latex]\\left(1,-1,2\\right)[\/latex].<\/p>\n<p><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 1, negative 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=negative 2.\" width=\"487\" height=\"324\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the system of equations in three variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+y - 2z=-1\\hfill \\\\ 3x - 3y-z=5\\hfill \\\\ x - 2y+3z=6\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q232738\">Solution<\/span><\/p>\n<div id=\"q232738\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(1,-1,1\\right)[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=23765&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\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\" id=\"oembed-1\" title=\"Systems of Equations in Three Variables:  Part 1 of 2\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/wIE8KSpb-E8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Real-World Problem Using a System of Three Equations in Three Variables<\/h3>\n<p>In the problem posed at the beginning of the section, John invested his inheritance of $12,000 in three different funds: part in a money-market fund paying 3% interest annually; part in municipal bonds paying 4% annually; and the rest in mutual funds paying 7% annually. John invested $4,000 more in mutual funds than he invested in municipal bonds. The total interest earned in one year was $670. How much did he invest in each type of fund?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348091\">Solution<\/span><\/p>\n<div id=\"q348091\" class=\"hidden-answer\" style=\"display: none\">\nTo solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\text{amount invested in money-market fund}\\hfill \\\\ y=\\text{amount invested in municipal bonds}\\hfill \\\\ z=\\text{amount invested in mutual funds}\\hfill \\end{array}[\/latex]<\/p>\n<p>The first equation indicates that the sum of the three principal amounts is $12,000.<\/p>\n<p style=\"text-align: center;\">[latex]x+y+z=12,000[\/latex]<\/p>\n<p>We form the second equation according to the information that John invested $4,000 more in mutual funds than he invested in municipal bonds.<\/p>\n<p style=\"text-align: center;\">[latex]z=y+4,000[\/latex]<\/p>\n<p>The third equation shows that the total amount of interest earned from each fund equals $670.<\/p>\n<p style=\"text-align: center;\">[latex]0.03x+0.04y+0.07z=670[\/latex]<\/p>\n<p>Then, we write the three equations as a system.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+y+z=12,000\\hfill \\\\ \\text{ }-y+z=4,000\\hfill \\\\ 0.03x+0.04y+0.07z=670\\hfill \\end{array}[\/latex]<\/p>\n<p>To make the calculations simpler, we can multiply the third equation by 100. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\text{ }x+\\text{ }y+z\\text{ }=12,000\\hfill & \\left(1\\right)\\hfill \\\\ \\text{ }-y+z\\text{ }=4,000\\hfill & \\left(2\\right)\\hfill \\\\ 3x+4y+7z=67,000\\hfill & \\left(3\\right)\\hfill \\end{array}[\/latex]<\/p>\n<p><strong>Step 1.<\/strong> Interchange equation (2) and equation (3) so that the two equations with three variables will line up.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+\\text{ }y +\\text{ }z=12,000\\hfill \\\\ 3x+4y +7z=67,000\\hfill \\\\ \\text{ }-y\\text{ }+\\text{ }z=4,000\\hfill \\end{array}[\/latex]<\/p>\n<p><strong>Step 2.<\/strong> Multiply equation (1) by [latex]-3[\/latex] and add to equation (2). Write the result as row 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+y+z\\text{ }=12,000\\hfill \\\\ \\text{ }y+4z=31,000\\hfill \\\\ \\text{ }-y+z\\text{ }=4,000\\hfill \\end{array}[\/latex]<\/p>\n<p><strong>Step 3.<\/strong> Add equation (2) to equation (3) and write the result as equation (3).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+y+\\text{ }z=12,000\\hfill \\\\ \\text{ }y+4z=31,000\\hfill \\\\ \\text{ }5z\\text{ }=35,000\\hfill \\end{array}[\/latex]<\/p>\n<p><strong>Step 4.<\/strong> Solve for [latex]z[\/latex] in equation (3). Back-substitute that value in equation (2) and solve for [latex]y[\/latex]. Then, back-substitute the values for [latex]z[\/latex] and [latex]y[\/latex] into equation (1) and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }5z=35,000\\hfill \\\\ \\text{ }z=7,000\\hfill \\\\ \\hfill \\\\ \\hfill \\\\ \\text{ }y+4\\left(7,000\\right)=31,000\\hfill \\\\ \\text{ }y=3,000\\hfill \\\\ \\hfill \\\\ \\hfill \\\\ x+3,000+7,000=12,000\\hfill \\\\ \\text{ }x=2,000\\hfill \\end{array}[\/latex]<\/p>\n<p>John invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=19353&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/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-2208\">\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>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><\/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>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/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><\/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":21,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"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\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Systems of Equations in Three Variables: Part 1 of 2 \",\"author\":\"Sousa, James 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