When solving systems of equation with three variables we use the elimination method or the substitution method to make a system of two equations in two variables. Example Solve the systems of equations (this example is also shown in our video lesson) $$\left\{\begin{matrix} x+2y-z=4\\ 2x+y+z=-2\\ x+2y+z=2 \end{matrix}\right.$$ First
we add the first and second equation to make an equation with two variables, second we subtract the third equation from the second in order to get another equation with two variables. Now we have a system of two equations with two variables: $$\left\{\begin{matrix} 3x+3y = 2\\ x-y=-4 \end{matrix}\right.$$ We then multiply the second equation with 3 on both sides and add that to the first equation: $$6x=-10$$ $$x=\frac{-10}{6}$$ We plug this value into the 3x+3y=2
equation in order to determine our y-value: $$\begin{array}{lcl} \\ 3\cdot \frac{-10}{6}+3y&=&2\\ \\ -5+3y&=&2\\ 3y&=&7\\ \\ y&=&\frac{7}{3}\\ \end{array}$$ Last we plug our x- and y-value into any equation in first system in order to determine our z-value: $$\begin{array}{lcl} x+2y-z&=&4\\ \frac{-10}{6}+2\cdot \frac{7}{3}+z&=&2\\ 3+z&=&2\\ z&=&-1\\ \end{array}$$ Video lessonSolve the systems of equation in our example. Learning Outcomes
John received an inheritance of $12,000 that he divided into three parts and invested in three ways: in a money-market fund paying 3% annual interest; in municipal bonds paying 4% annual interest; and in mutual funds paying 7% annual interest. John invested $4,000 more in municipal funds than in municipal bonds. He earned $670 in interest the first year. How much did John invest in each type of fund? (credit: “Elembis,” Wikimedia Commons) Understanding the correct approach to setting up problems such as this one makes finding a solution a matter of following a pattern. We will solve this and similar problems involving three equations and three variables in this section. Doing so uses similar techniques as those used to solve systems of two equations in two variables. However, finding solutions to systems of three equations requires a bit more organization and a touch of visual gymnastics. Solve Systems of Three Equations in Three VariablesIn 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. A solution to a system of three equations in three variables [latex]\left(x,y,z\right),\text{}[/latex] is called an ordered triple. To find a solution, we can perform the following operations:
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. A General Note: Number of Possible SolutionsThe planes illustrate possible solution scenarios for three-by-three systems.
(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. Example: Determining Whether an Ordered Triple Is a Solution to a SystemDetermine whether the ordered triple [latex]\left(3,-2,1\right)[/latex] is a solution to the system. [latex]\begin{gathered}x+y+z=2 \\ 6x - 4y+5z=31 \\ 5x+2y+2z=13 \end{gathered}[/latex] How To: Given a linear system of three equations, solve for three unknowns.
Example: Solving a System of Three Equations in Three Variables by EliminationFind a solution to the following system: [latex]\begin{align}x - 2y+3z=9& &\text{(1)} \\ -x+3y-z=-6& &\text{(2)} \\ 2x - 5y+5z=17& &\text{(3)} \end{align}[/latex] Try ItSolve the system of equations in three variables. [latex]\begin{array}{l}2x+y - 2z=-1\hfill \\ 3x - 3y-z=5\hfill \\ x - 2y+3z=6\hfill \end{array}[/latex] In the following video, you 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. Example: Solving a Real-World Problem Using a System of Three Equations in Three VariablesIn 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? Try ItClassify Solutions to Systems in Three VariablesJust as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. The equations could represent three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. The process of elimination will result in a false statement, such as [latex]3=7[/latex] or some other contradiction. Example: Solving an Inconsistent System of Three Equations in Three VariablesSolve the following system. [latex]\begin{align}x - 3y+z=4 && \left(1\right) \\ -x+2y - 5z=3 && \left(2\right) \\ 5x - 13y+13z=8 && \left(3\right) \end{align}[/latex] Try ItSolve the system of three equations in three variables. [latex]\begin{array}{l}\text{ }x+y+z=2\hfill \\ \text{ }y - 3z=1\hfill \\ 2x+y+5z=0\hfill \end{array}[/latex] Expressing the Solution of a System of Dependent Equations Containing Three VariablesWe know from working with systems of equations in two variables that a dependent system 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. Or two of the equations could be the same and intersect the third on a line. Example: Finding the Solution to a Dependent System of EquationsFind the solution to the given system of three equations in three variables. [latex]\begin{align}2x+y - 3z=0 && \left(1\right)\\ 4x+2y - 6z=0 && \left(2\right)\\ x-y+z=0 && \left(3\right)\end{align}[/latex] Q & ADoes the generic solution to a dependent system always have to be written in terms of [latex]x?[/latex]No, you can write the generic solution in terms of any of the variables, but it is common to write it in terms of [latex]x[/latex] and if needed [latex]x[/latex] and [latex]y[/latex]. Try ItSolve the following system. [latex]\begin{gathered}x+y+z=7 \\ 3x - 2y-z=4 \\ x+6y+5z=24 \end{gathered}[/latex] Key Concepts
Glossarysolution set the set of all ordered pairs or triples that satisfy all equations in a system of equations How do you solve 3 equations with 3 variables in linear algebra?A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation.
|