{"id":1745,"date":"2015-11-12T18:30:45","date_gmt":"2015-11-12T18:30:45","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=1745"},"modified":"2015-11-12T18:30:45","modified_gmt":"2015-11-12T18:30:45","slug":"solving-systems-of-three-equations-in-three-variables","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/solving-systems-of-three-equations-in-three-variables\/","title":{"raw":"Solving Systems of Three Equations in Three Variables","rendered":"Solving Systems of Three Equations in Three Variables"},"content":{"raw":"<p>In order to solve systems of equations in three variables, known as three-by-three systems, the primary tool we will be using is called <strong>Gaussian elimination<\/strong>, named after the prolific German mathematician Karl Friedrich <strong>Gauss<\/strong>. While there is no definitive order in which operations are to be performed, there are specific guidelines as to what type of moves can be made. We may number the equations to keep track of the steps we apply. The goal is to eliminate one variable at a time to achieve <strong>upper triangular form<\/strong>, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution [latex]\\left(x,y,z\\right),\\text{}[\/latex] which we call an <strong>ordered triple<\/strong>. A system in upper triangular form looks like the following:\n<\/p><div style=\"text-align: center;\">[latex]\\begin{array}{l}Ax+By+Cz=D\\hfill \\\\ \\text{ }Ey+Fz=G\\hfill \\\\ \\text{ }Hz=K\\hfill \\end{array}[\/latex]<\/div>\nThe third equation can be solved for [latex]z,\\text{}[\/latex] and then we back-substitute to find [latex]y[\/latex] and [latex]x[\/latex]. To write the system in upper triangular form, we can perform the following operations:\n<ol><li>Interchange the order of any two equations.<\/li>\n\t<li>Multiply both sides of an equation by a nonzero constant.<\/li>\n\t<li>Add a nonzero multiple of one equation to another equation.<\/li>\n<\/ol>\nThe <strong>solution set<\/strong> to a three-by-three system is an ordered triple [latex]\\left\\{\\left(x,y,z\\right)\\right\\}[\/latex]. 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.\n<div class=\"textbox\">\n<h3>A General Note: Number of Possible Solutions<\/h3>\nFigure 2 and Figure 3\u00a0illustrate possible solution scenarios for three-by-three systems.\n<ul><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\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>\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>\n<\/ul>\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202112\/CNX_Precalc_Figure_09_02_006n2.jpg\" alt=\"\" width=\"487\" height=\"238\" data-media-type=\"image\/jpg\"\/><b>Figure 2.<\/b> (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]\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202114\/CNX_Precalc_Figure_09_02_007n2.jpg\" alt=\"\" width=\"487\" height=\"188\" data-media-type=\"image\/jpg\"\/><b>Figure 3<\/b>[\/caption]\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Determining Whether an Ordered Triple Is a Solution to a System<\/h3>\nDetermine whether the ordered triple [latex]\\left(3,-2,1\\right)[\/latex] is a solution to the system.\n<div 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]<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\nWe will check each equation by substituting in the values of the ordered triple for [latex]x,y[\/latex], and [latex]z[\/latex].\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>\nThe ordered triple [latex]\\left(3,-2,1\\right)[\/latex] is indeed a solution to the system.\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a linear system of three equations, solve for three unknowns.<strong>\n<\/strong><\/h3>\n<ol><li>Pick any pair of equations and solve for one variable.<\/li>\n\t<li>Pick another pair of equations and solve for the same variable.<\/li>\n\t<li>You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.<\/li>\n\t<li>Back-substitute known variables into any one of the original equations and solve for the missing variable.<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Solving a System of Three Equations in Three Variables by Elimination<\/h3>\nFind a solution to the following system:\n<div 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]<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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).\n<div 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]<\/div>\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].\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>\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.\n<div 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]<\/div>\nChoosing one equation from each new system, we obtain the upper triangular form:\n<div 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]<\/div>\nNext, we back-substitute [latex]z=2[\/latex] into equation (4) and solve for [latex]y[\/latex].\n<div 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]<\/div>\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].\n<div 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]<\/div>\nThe solution is the ordered triple [latex]\\left(1,-1,2\\right)[\/latex].\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202115\/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\"\/><b>Figure 4<\/b>[\/caption]\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Solving a Real-World Problem Using a System of Three Equations in Three Variables<\/h3>\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?\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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:\n<div 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]<\/div>\nThe first equation indicates that the sum of the three principal amounts is $12,000.\n<div style=\"text-align: center;\">[latex]x+y+z=12,000[\/latex]<\/div>\nWe form the second equation according to the information that John invested $4,000 more in mutual funds than he invested in municipal bonds.\n<div style=\"text-align: center;\">[latex]z=y+4,000[\/latex]<\/div>\nThe third equation shows that the total amount of interest earned from each fund equals $670.\n<div style=\"text-align: center;\">[latex]0.03x+0.04y+0.07z=670[\/latex]<\/div>\nThen, we write the three equations as a system.\n<div 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]<\/div>\nTo make the calculations simpler, we can multiply the third equation by 100. Thus,\n<div 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]<\/div>\n<strong>Step 1.<\/strong> Interchange equation (2) and equation (3) so that the two equations with three variables will line up.\n<div 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]<\/div>\n<strong>Step 2.<\/strong> Multiply equation (1) by [latex]-3[\/latex] and add to equation (2). Write the result as row 2.\n<div 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]<\/div>\n<strong>Step 3.<\/strong> Add equation (2) to equation (3) and write the result as equation (3).\n<div 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]<\/div>\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].\n<div 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]<\/div>\nJohn invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds.\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 1<\/h3>\nSolve the system of equations in three variables.\n<div 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]<\/div>\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-18\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>","rendered":"<p>In order to solve systems of equations in three variables, known as three-by-three systems, the primary tool we will be using is called <strong>Gaussian elimination<\/strong>, named after the prolific German mathematician Karl Friedrich <strong>Gauss<\/strong>. While there is no definitive order in which operations are to be performed, there are specific guidelines as to what type of moves can be made. We may number the equations to keep track of the steps we apply. The goal is to eliminate one variable at a time to achieve <strong>upper triangular form<\/strong>, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution [latex]\\left(x,y,z\\right),\\text{}[\/latex] which we call an <strong>ordered triple<\/strong>. A system in upper triangular form looks like the following:\n<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}Ax+By+Cz=D\\hfill \\\\ \\text{ }Ey+Fz=G\\hfill \\\\ \\text{ }Hz=K\\hfill \\end{array}[\/latex]<\/div>\n<p>The third equation can be solved for [latex]z,\\text{}[\/latex] and then we back-substitute to find [latex]y[\/latex] and [latex]x[\/latex]. To write the system in upper triangular form, 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>The <strong>solution set<\/strong> to a three-by-three system is an ordered triple [latex]\\left\\{\\left(x,y,z\\right)\\right\\}[\/latex]. 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>Figure 2 and Figure 3\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-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202112\/CNX_Precalc_Figure_09_02_006n2.jpg\" alt=\"\" width=\"487\" height=\"238\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2.<\/b> (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<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202114\/CNX_Precalc_Figure_09_02_007n2.jpg\" alt=\"\" width=\"487\" height=\"188\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: 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<div 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]<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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 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 shaded\">\n<h3>Example 2: Solving a System of Three Equations in Three Variables by Elimination<\/h3>\n<p>Find a solution to the following system:<\/p>\n<div 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]<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<p>There 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<div 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]<\/div>\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<div 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]<\/div>\n<p>Choosing one equation from each new system, we obtain the upper triangular form:<\/p>\n<div 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]<\/div>\n<p>Next, we back-substitute [latex]z=2[\/latex] into equation (4) and solve for [latex]y[\/latex].<\/p>\n<div 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]<\/div>\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<div 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]<\/div>\n<p>The solution is the ordered triple [latex]\\left(1,-1,2\\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202115\/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<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: 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>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<p>To 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<div 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]<\/div>\n<p>The first equation indicates that the sum of the three principal amounts is $12,000.<\/p>\n<div style=\"text-align: center;\">[latex]x+y+z=12,000[\/latex]<\/div>\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<div style=\"text-align: center;\">[latex]z=y+4,000[\/latex]<\/div>\n<p>The third equation shows that the total amount of interest earned from each fund equals $670.<\/p>\n<div style=\"text-align: center;\">[latex]0.03x+0.04y+0.07z=670[\/latex]<\/div>\n<p>Then, we write the three equations as a system.<\/p>\n<div 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]<\/div>\n<p>To make the calculations simpler, we can multiply the third equation by 100. Thus,<\/p>\n<div 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]<\/div>\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<div 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]<\/div>\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<div 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]<\/div>\n<p><strong>Step 3.<\/strong> Add equation (2) to equation (3) and write the result as equation (3).<\/p>\n<div 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]<\/div>\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<div 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]<\/div>\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 class=\"bcc-box bcc-success\">\n<h3>Try It 1<\/h3>\n<p>Solve the system of equations in three variables.<\/p>\n<div 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]<\/div>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-18\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\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-1745\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/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":276,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1745","chapter","type-chapter","status-publish","hentry"],"part":1739,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1745","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1745\/revisions"}],"predecessor-version":[{"id":2266,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1745\/revisions\/2266"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1739"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1745\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/wp\/v2\/media?parent=1745"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1745"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1745"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/wp\/v2\/license?post=1745"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}