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Systems of Linear Equations: Three Variables

Solve systems of three equations in three variables.

Systems of Linear Equations: Three Variables

Determine whether an ordered triple is a solution of a system of three linear equations with three variables (IA 4.4.1)
Solve a system of three linear equations with three variables (IA 4.4.2)

A linear equation with three variables where a, b, c, and d are real numbers and a, b, and c are not all 0, is of the form $ax+by+cz=d$ . The graph of a linear equation with three variables is a plane.

A system of linear equations with three variables is a set of linear equations with three variables. For example,

\[\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right\]

is a system of linear equations with three variables.

Solutions of a system of equations are the values of the variables that make all the equations true. A solution is represented by an ordered triple (x,y,z).

Example

Determine whether the ordered triples are solutions to the given system.

Try it.

$\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right$ at $(1,-3,2)$ and at $(4,-1,-5)$

Solution

We substitute 1, –3, 2 into all three equations:

$3x+y+z=2$	$x+2y+z=-3$	$3x+y+2z=4$
$\begin{array}{l}3(1)+(-3)+2=2 \\ 2=2\end{array}$ True	$\begin{array}{l}1+2(-3)+2=-3 \\ -3=-3\end{array}$ True	$\begin{array}{l}3(1)+(-3)+2(2)=4 \\ 4=4\end{array}$ True
Conclusion: Since (1,-3,2) is a solution to all three equations, then it is a solution to the system $\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right$

Next we substitute (4, –1, –5) into all three equations:

$3x+y+z=2$	$x+2y+z=-3$	$3x+y+2z=4$
$\begin{array}{l}3(4)+(-1)+(-5)=2 \\ 6=2\end{array}$ False	$\begin{array}{l}4+2(-1)+(-5)=-3 \\ -3=-3\end{array}$ True	$\begin{array}{l}3(4)+(-1)+2(-5)=4 \\ 1=4\end{array}$ False
Conclusion: Since (4,–1,–5) is not a solution to all three equations, then it is not a solution to the system $\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right$

Determine whether the ordered pairs are solutions to the given system.

\[\left\{\begin{array}{l}2x-6y+z=3 \\ 3x-4y-3z=2\ \text{at}\ (3,\ 1,\ 3)\ and at\ (4,\ 3,\ 7) \\ 2x+y-2z=3\end{array}\right\]

Try it.

At (3, 1, 3):

$2x-6y+z=3$	$3x-4y-3z=2$	$2x+3y-2z=3$
________________________	________________________	________________________
Conclusion: ________________________

Try it.

At (4, 3, 7):

$2x-6y+z=3$	$3x-4y-3z=2$	$2x+3y-2z=3$
________________________	________________________	________________________
Conclusion: ________________________

Condensed — the full section is in OpenStax College Algebra 2e.

Solving Systems of Three Equations in Three Variables

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 Gaussian elimination, named after the prolific German mathematician Karl Friedrich Gauss. 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 upper triangular form, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution $(x,y,z),$ which we call an ordered triple. A system in upper triangular form looks like the following:

\[\begin{array}{l}Ax+By+Cz=D \\ \ Ey+Fz=G \\ \ Hz=K\end{array}\]

The third equation can be solved for $z,$ and then we back-substitute to find $y$ and $x.$ To write the system in upper triangular form, we can perform the following operations:

Interchange the order of any two equations.
Multiply both sides of an equation by a nonzero constant.
Add a nonzero multiple of one equation to another equation.

The solution set to a three-by-three system is an ordered triple $\{(x,y,z)\}.$ 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.

Number of Possible Solutions

and illustrate possible solution scenarios for three-by-three systems.

Systems that have a single solution are those which, after elimination, result in a solution set consisting of an ordered triple $\{(x,y,z)\}.$ Graphically, the ordered triple defines a point that is the intersection of three planes in space.
Systems that have an infinite number of solutions are those which, after elimination, result in an expression that is always true, such as $0=0.$ Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space.
Systems that have no solution are those that, after elimination, result in a statement that is a contradiction, such as $3=0.$ Graphically, a system with no solution is represented by three planes with no point in common.

Condensed — the full section is in OpenStax College Algebra 2e.

Identifying Inconsistent Systems of Equations Containing Three Variables

Just 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 $3=7$ or some other contradiction.

Example

Try it.

Solve the following system.

\[\begin{array}{ll}\ x-3y+z=4 & (1) \\ \ -x+2y-5z=3 & (2) \\ 5x-13y+13z=8 & (3)\end{array}\]

Solution

Looking at the coefficients of $x,$ we can see that we can eliminate $x$ by adding equation (1) to equation (2).

\[\frac{\begin{array}{l}x-3y+z=4\ (1) \\ -x+2y-5z=3\ (2)\end{array}}{\ -y-4z=7\ (4)}\]

Next, we multiply equation (1) by $-5$ and add it to equation (3).

\[\begin{array}{lllllll}\underset{______________________________________}{\begin{array}{llll}-5x+15y-5z=-20 & & & (1)\ \text{multiplied}\ \text{by}\ -5 \\ 5x-13y+13z=8 & & & (3)\end{array}} \\ \begin{array}{llll}\ 2y+8z=-12 & & & (5)\end{array}\end{array}\]

Then, we multiply equation (4) by 2 and add it to equation (5).

\[\begin{array}{l}\underset{_______________________________________}{\begin{array}{l}-2y-8z=14(4)\ \text{multiplied}\ \text{by}\ 2 \\ 2y+8z=-12\ (5)\end{array}} \\ 0=2\end{array}\]

The final equation $0=2$ is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.

Expressing the Solution of a System of Dependent Equations Containing Three Variables

We 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

Try it.

Find the solution to the given system of three equations in three variables.

\[\begin{array}{ll}\ 2x+y-3z=0 & (1) \\ 4x+2y-6z=0 & (2) \\ x-y+z=0 & (3)\end{array}\]

Solution

First, we can multiply equation (1) by $-2$ and add it to equation (2).

\[\begin{array}{l}\underset{____________________________________________}{\begin{array}{l}-4x-2y+6z=0\ \text{equation }(1)\ \text{multiplied}\ \text{by}\ -2 \\ \text{}\text{}\text{}\text{}\ 4x+2y-6z=0\ \ (2)\end{array}} \\ 0=0\end{array}\]

We do not need to proceed any further. The result we get is an identity, $0=0,$ which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by $-2,$ and adding it to equation (1). We then perform the same steps as above and find the same result, $0=0.$

When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have

\[\begin{array}{l}\underset{_____________}{\begin{array}{l}\begin{array}{l} \\ 2x+y-3z=0\end{array} \\ \ x-y+z=0\end{array}} \\ \ 3x-2z=0\end{array}\]

We then solve the resulting equation for $z.$

\[\begin{array}{l}3x-2z=0 \\ \ z=\frac{3}{2}x\end{array}\]

We back-substitute the expression for $z$ into one of the equations and solve for $y.$

\[\begin{array}{l}2x+y-3(\frac{3}{2}x)=0 \\ 2x+y-\frac{9}{2}x=0 \\ \ y=\frac{9}{2}x-2x \\ \ y=\frac{5}{2}x\end{array}\]

So the general solution is $(x,\frac{5}{2}x,\frac{3}{2}x).$ In this solution, $x$ can be any real number. The values of $y$ and $z$ are dependent on the value selected for $x.$

Condensed — the full section is in OpenStax College Algebra 2e.

Key Concepts

A solution set is an ordered triple $\{(x,y,z)\}$ that represents the intersection of three planes in space. See
.
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. See .
Systems of three equations in three variables are useful for solving many different types of real-world problems. See .
A system of equations in three variables is inconsistent if no solution exists. After performing elimination operations, the result is a contradiction. See .
Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location.
A system of equations in three variables is dependent if it has an infinite number of solutions. After performing elimination operations, the result is an identity. See .
Systems of equations in three variables that are dependent could result from three identical planes, three planes intersecting at a line, or two identical planes that intersect the third on a line.

Systems of Linear Equations: Three Variables

Determine whether an ordered triple is a solution of a system of three linear equations with three variables (IA 4.4.1)
Solve a system of three linear equations with three variables (IA 4.4.2)

A system of linear equations with three variables is a set of linear equations with three variables. For example,

\[\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right\]

is a system of linear equations with three variables.

Solutions of a system of equations are the values of the variables that make all the equations true. A solution is represented by an ordered triple (x,y,z).

Example

Determine whether the ordered triples are solutions to the given system.

Try it.

$\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right$ at $(1,-3,2)$ and at $(4,-1,-5)$

Solution

We substitute 1, –3, 2 into all three equations:

$3x+y+z=2$	$x+2y+z=-3$	$3x+y+2z=4$
$\begin{array}{l}3(1)+(-3)+2=2 \\ 2=2\end{array}$ True	$\begin{array}{l}1+2(-3)+2=-3 \\ -3=-3\end{array}$ True	$\begin{array}{l}3(1)+(-3)+2(2)=4 \\ 4=4\end{array}$ True
Conclusion: Since (1,-3,2) is a solution to all three equations, then it is a solution to the system $\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right$

Next we substitute (4, –1, –5) into all three equations:

$3x+y+z=2$	$x+2y+z=-3$	$3x+y+2z=4$
$\begin{array}{l}3(4)+(-1)+(-5)=2 \\ 6=2\end{array}$ False	$\begin{array}{l}4+2(-1)+(-5)=-3 \\ -3=-3\end{array}$ True	$\begin{array}{l}3(4)+(-1)+2(-5)=4 \\ 1=4\end{array}$ False
Conclusion: Since (4,–1,–5) is not a solution to all three equations, then it is not a solution to the system $\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right$

Determine whether the ordered pairs are solutions to the given system.

\[\left\{\begin{array}{l}2x-6y+z=3 \\ 3x-4y-3z=2\ \text{at}\ (3,\ 1,\ 3)\ and at\ (4,\ 3,\ 7) \\ 2x+y-2z=3\end{array}\right\]

Try it.

At (3, 1, 3):

$2x-6y+z=3$	$3x-4y-3z=2$	$2x+3y-2z=3$
________________________	________________________	________________________
Conclusion: ________________________

Try it.

At (4, 3, 7):

$2x-6y+z=3$	$3x-4y-3z=2$	$2x+3y-2z=3$
________________________	________________________	________________________
Conclusion: ________________________

Condensed — the full section is in OpenStax Precalculus 2e.

Solving Systems of Three Equations in Three Variables

\[\begin{array}{l}Ax+By+Cz=D \\ \ Ey+Fz=G \\ \ Hz=K\end{array}\]

The third equation can be solved for $z,$ and then we back-substitute to find $y$ and $x.$ To write the system in upper triangular form, we can perform the following operations:

Interchange the order of any two equations.
Multiply both sides of an equation by a nonzero constant.
Add a nonzero multiple of one equation to another equation.

Number of Possible Solutions

and illustrate possible solution scenarios for three-by-three systems.

Systems that have a single solution are those which, after elimination, result in a solution set consisting of an ordered triple $\{(x,y,z)\}.$ Graphically, the ordered triple defines a point that is the intersection of three planes in space.
Systems that have an infinite number of solutions are those which, after elimination, result in an expression that is always true, such as $0=0.$ Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space.
Systems that have no solution are those that, after elimination, result in a statement that is a contradiction, such as $3=0.$ Graphically, a system with no solution is represented by three planes with no point in common.

Condensed — the full section is in OpenStax Precalculus 2e.

Identifying Inconsistent Systems of Equations Containing Three Variables

Example

Try it.

Solve the following system.

\[\begin{array}{ll}\ x-3y+z=4 & (1) \\ \ -x+2y-5z=3 & (2) \\ 5x-13y+13z=8 & (3)\end{array}\]

Solution

Looking at the coefficients of $x,$ we can see that we can eliminate $x$ by adding equation (1) to equation (2).

\[\frac{\begin{array}{l}x-3y+z=4\ (1) \\ -x+2y-5z=3\ (2)\end{array}}{\ -y-4z=7\ (4)}\]

Next, we multiply equation (1) by $-5$ and add it to equation (3).

Then, we multiply equation (4) by 2 and add it to equation (5).

\[\begin{array}{l}\underset{_______________________________________}{\begin{array}{l}-2y-8z=14(4)\ \text{multiplied}\ \text{by}\ 2 \\ 2y+8z=-12\ (5)\end{array}} \\ 0=2\end{array}\]

The final equation $0=2$ is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.

Expressing the Solution of a System of Dependent Equations Containing Three Variables

Example

Try it.

Find the solution to the given system of three equations in three variables.

\[\begin{array}{ll}\ 2x+y-3z=0 & (1) \\ 4x+2y-6z=0 & (2) \\ x-y+z=0 & (3)\end{array}\]

Solution

First, we can multiply equation (1) by $-2$ and add it to equation (2).

When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have

\[\begin{array}{l}\underset{_____________}{\begin{array}{l}\begin{array}{l} \\ 2x+y-3z=0\end{array} \\ \ x-y+z=0\end{array}} \\ \ 3x-2z=0\end{array}\]

We then solve the resulting equation for $z.$

\[\begin{array}{l}3x-2z=0 \\ \ z=\frac{3}{2}x\end{array}\]

We back-substitute the expression for $z$ into one of the equations and solve for $y.$

\[\begin{array}{l}2x+y-3(\frac{3}{2}x)=0 \\ 2x+y-\frac{9}{2}x=0 \\ \ y=\frac{9}{2}x-2x \\ \ y=\frac{5}{2}x\end{array}\]

So the general solution is $(x,\frac{5}{2}x,\frac{3}{2}x).$ In this solution, $x$ can be any real number. The values of $y$ and $z$ are dependent on the value selected for $x.$

Condensed — the full section is in OpenStax Precalculus 2e.

Key Concepts

A solution set is an ordered triple $\{(x,y,z)\}$ that represents the intersection of three planes in space. See
.
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. See .
Systems of three equations in three variables are useful for solving many different types of real-world problems. See .
A system of equations in three variables is inconsistent if no solution exists. After performing elimination operations, the result is a contradiction. See .
Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location.
A system of equations in three variables is dependent if it has an infinite number of solutions. After performing elimination operations, the result is an identity. See .
Systems of equations in three variables that are dependent could result from three identical planes, three planes intersecting at a line, or two identical planes that intersect the third on a line.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

$\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right$ at $(1,-3,2)$ and at $(4,-1,-5)$

Bonisa impendulo

We substitute 1, –3, 2 into all three equations:

$3x+y+z=2$	$x+2y+z=-3$	$3x+y+2z=4$
$\begin{array}{l}3(1)+(-3)+2=2 \\ 2=2\end{array}$ True	$\begin{array}{l}1+2(-3)+2=-3 \\ -3=-3\end{array}$ True	$\begin{array}{l}3(1)+(-3)+2(2)=4 \\ 4=4\end{array}$ True
Conclusion: Since (1,-3,2) is a solution to all three equations, then it is a solution to the system $\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right$

Next we substitute (4, –1, –5) into all three equations:

$3x+y+z=2$	$x+2y+z=-3$	$3x+y+2z=4$
$\begin{array}{l}3(4)+(-1)+(-5)=2 \\ 6=2\end{array}$ False	$\begin{array}{l}4+2(-1)+(-5)=-3 \\ -3=-3\end{array}$ True	$\begin{array}{l}3(4)+(-1)+2(-5)=4 \\ 1=4\end{array}$ False
Conclusion: Since (4,–1,–5) is not a solution to all three equations, then it is not a solution to the system $\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right$

At (3, 1, 3):
$2x-6y+z=3$ $3x-4y-3z=2$ $2x+3y-2z=3$
________________________ ________________________ ________________________
Conclusion: ________________________
At (4, 3, 7):
$2x-6y+z=3$ $3x-4y-3z=2$ $2x+3y-2z=3$
________________________ ________________________ ________________________
Conclusion: ________________________

Solve the system of equations: $\{\begin{array}{l}x+2y-z=1 \\ 2x+7y+4z=11 \\ x+3y+z=4\end{array}.$

Bonisa impendulo

\[\{\begin{array}{l}x+2y-z=1\ (1) \\ 2x+7y+4z=11\ (2) \\ x+3y+z=4\ (3)\end{array}\]

Use equation (1) and (3) to eliminate x.

Use equation (1) and (2) to eliminate x again.

Use equation (4) and (5) to eliminate $y$ .

	There are infinitely many solutions.
Solve equation (4) for y.	Represent the solution showing how x and y are dependent on z. $\begin{array}{lll}y+2z & = & 3 \\ y & = & -2z+3\end{array}$
Use equation (1) to solve for x.	$\ x+2y-z=1$
Substitute $y=-2z+3.$	$\begin{array}{lll}x+2(-2z+3)-z & = & 1 \\ x-4z+6-z & = & 1 \\ x-5z+6 & = & 1 \\ x & = & 5z-5\end{array}$

The true statement $0=0$ tells us that this is a dependent system that has infinitely many solutions. The solutions are of the form $(x,y,z)$ where $x=5z-5;y=-2z+3$ and z is any real number.

Solve the system of linear equations with three variables.

\[\left\{\begin{array}{l}x+2y-3z=-1 \\ x-3y+z=1 \\ 2x-y-2z=2\end{array}\right\begin{array}{l}(1) \\ (2) \\ (3)\end{array}\]

Step 1	Write the equations in standard form. If any coefficients are fractions, clear them. ________________________________________
Step 2	Let's use equations (1) and (2) to eliminate $x$ ________________________________________
Step 3	Let's use equations (1) and (3) to eliminate $x$ ________________________________________
Step 4	Let's now use the new equations (4) and (5) to eliminate $y$ ________________________________________
Step 5
Step 6

Determine whether the ordered pair is a solution to the given system
$\left\{\begin{array}{l}y-10z=-8 \\ 2x-y=2 \\ x-5z=3\end{array}\right\begin{array}{l} \\ \text{at}\ (7,\ 12,\ 2)\ \text{and at}\ (2,\ 2,\ 1) \\ \end{array}$
Determine whether the ordered triple $(3,-2,1)$ is a solution to the system.
\[\begin{array}{l}\ x+y+z=2 \\ 6x-4y+5z=31 \\ 5x+2y+2z=13\end{array}\]

Bonisa impendulo

We will check each equation by substituting in the values of the ordered triple for $x,y,$ and $z.$
$\begin{array}{lllll}\begin{array}{l}x+y+z=2 \\ (3)+(-2)+(1)=2 \\ \text{True}\end{array} & \ & \begin{array}{l}\ 6x-4y+5z=31 \\ 6(3)-4(-2)+5(1)=31 \\ 18+8+5=31 \\ \text{True}\end{array} & \ & \begin{array}{l}\ 5x+2y+2z=13 \\ 5(3)+2(-2)+2(1)=13 \\ \ 15-4+2=13 \\ \text{True}\end{array}\end{array}$
The ordered triple $(3,-2,1)$ is indeed a solution to the system.
Find a solution to the following system:
\[\begin{array}{ll}\ \ x-2y+3z=9 & \text{(1)} \\ -x+3y-z=-6 & \text{(2)} \\ 2x-5y+5z=17 & \text{(3)}\end{array}\]

Bonisa impendulo

There will always be several choices as to where to begin, but the most obvious first step here is to eliminate $x$ by adding equations (1) and (2).
\[\frac{\begin{array}{ll}x-2y+3z=9 & \text{(1)} \\ \ -x+3y-z=-6 & \ \text{(2)}\end{array}}{\begin{array}{ll}\ \ y+2z=3 & \ \text{(4)}\end{array}}\]
The second step is multiplying equation (1) by $-2$ and adding the result to equation (3). These two steps will eliminate the variable $x.$
\[\begin{array}{llll}\underset{____________________________________}{\begin{array}{ll} & \\ -2x+4y-6z=-18 & (1)\ \text{multiplied}\ \text{by}\ -2 \\ 2x-5y+5z=17 & (3)\end{array}} \\ \ \ -y-z=-1\ (5)\end{array}\]
In equations (4) and (5), we have created a new two-by-two system. We can solve for $z$ by adding the two equations.
\[\frac{\begin{array}{l}\begin{array}{l} \\ y+2z=3\ \ (4)\end{array} \\ -y-z=-1\ (5)\end{array}}{\ z=2\ \ (6)}\]
Choosing one equation from each new system, we obtain the upper triangular form:
\[\begin{array}{ll}\ x-2y+3z=9\ & (1) \\ \ y+2z=3 & (4) \\ \ z=2 & (6)\end{array}\]
Next, we back-substitute $z=2$ into equation (4) and solve for $y.$
\[\begin{array}{l}y+2(2)=3 \\ y+4=3 \\ \ y=-1\end{array}\]
Finally, we can back-substitute $z=2$ and $y=-1$ into equation (1). This will yield the solution for $x.$
\[\begin{array}{l}x-2(-1)+3(2)=9 \\ \ x+2+6=9 \\ \ x=1\end{array}\]
The solution is the ordered triple $(1,-1,2).$ See .
In the problem posed at the beginning of the section, Jordi 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. Jordi 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?

Bonisa impendulo

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:
\[\begin{array}{l}x=\text{amount invested in money-market fund} \\ y=\text{amount invested in municipal bonds} \\ z=\text{amount invested in mutual funds}\end{array}\]
The first equation indicates that the sum of the three principal amounts is $12,000.
\[x+y+z=12,000\]
We form the second equation according to the information that Jordi invested $4,000 more in mutual funds than he invested in municipal bonds.
\[z=y+4,000\]
The third equation shows that the total amount of interest earned from each fund equals $670.
\[0.03x+0.04y+0.07z=670\]
Then, we write the three equations as a system.
\[\begin{array}{l}\ x+y+z=12,000 \\ \ -y+z=4,000 \\ 0.03x+0.04y+0.07z=670\end{array}\]
To make the calculations simpler, we can multiply the third equation by 100. Thus,
\[\begin{array}{ll}\ x+\ y+z\ =12,000 & (1) \\ \ -y+z\ =4,000 & (2) \\ 3x+4y+7z=67,000 & (3)\end{array}\]
Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up.
\[\begin{array}{l}\ x+\ y+\ z=12,000 \\ 3x+4y+7z=67,000 \\ \ -y\ +\ z=4,000\end{array}\]
Step 2. Multiply equation (1) by $-3$ and add to equation (2). Write the result as row 2.
\[\begin{array}{l}x+y+z\ =12,000 \\ y+4z=31,000 \\ \ -y+z\ =4,000\end{array}\]
Step 3. Add equation (2) to equation (3) and write the result as equation (3).
\[\begin{array}{l}x+y+\ z=12,000 \\ \ y+4z=31,000 \\ \ 5z\ =35,000\end{array}\]
Step 4. Solve for $z$ in equation (3). Back-substitute that value in equation (2) and solve for $y.$ Then, back-substitute the values for $z$ and $y$ into equation (1) and solve for $x.$
\[\begin{array}{l}\ 5z=35,000 \\ \ z=7,000 \\ \\ \\ \ y+4(7,000)=31,000 \\ \ y=3,000 \\ \\ \\ x+3,000+7,000=12,000 \\ \ x=2,000\end{array}\]
Jordi invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds.
Solve the system of equations in three variables.
\[\begin{array}{l}2x+y-2z=-1 \\ 3x-3y-z=5 \\ x-2y+3z=6\end{array}\]

Bonisa impendulo

$(1,-1,1)$
Solve the following system.
\[\begin{array}{ll}\ x-3y+z=4 & (1) \\ \ -x+2y-5z=3 & (2) \\ 5x-13y+13z=8 & (3)\end{array}\]

Bonisa impendulo

Looking at the coefficients of $x,$ we can see that we can eliminate $x$ by adding equation (1) to equation (2).
\[\frac{\begin{array}{l}x-3y+z=4\ (1) \\ -x+2y-5z=3\ (2)\end{array}}{\ -y-4z=7\ (4)}\]
Next, we multiply equation (1) by $-5$ and add it to equation (3).
\[\begin{array}{lllllll}\underset{______________________________________}{\begin{array}{llll}-5x+15y-5z=-20 & & & (1)\ \text{multiplied}\ \text{by}\ -5 \\ 5x-13y+13z=8 & & & (3)\end{array}} \\ \begin{array}{llll}\ 2y+8z=-12 & & & (5)\end{array}\end{array}\]
Then, we multiply equation (4) by 2 and add it to equation (5).
\[\begin{array}{l}\underset{_______________________________________}{\begin{array}{l}-2y-8z=14(4)\ \text{multiplied}\ \text{by}\ 2 \\ 2y+8z=-12\ (5)\end{array}} \\ 0=2\end{array}\]
The final equation $0=2$ is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.
Solve the system of three equations in three variables.
\[\begin{array}{l}\ x+y+z=2 \\ \ y-3z=1 \\ 2x+y+5z=0\end{array}\]

Bonisa impendulo

No solution.
Find the solution to the given system of three equations in three variables.
\[\begin{array}{ll}\ 2x+y-3z=0 & (1) \\ 4x+2y-6z=0 & (2) \\ x-y+z=0 & (3)\end{array}\]

Bonisa impendulo

First, we can multiply equation (1) by $-2$ and add it to equation (2).
\[\begin{array}{l}\underset{____________________________________________}{\begin{array}{l}-4x-2y+6z=0\ \text{equation }(1)\ \text{multiplied}\ \text{by}\ -2 \\ \text{}\text{}\text{}\text{}\ 4x+2y-6z=0\ \ (2)\end{array}} \\ 0=0\end{array}\]
We do not need to proceed any further. The result we get is an identity, $0=0,$ which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by $-2,$ and adding it to equation (1). We then perform the same steps as above and find the same result, $0=0.$
When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have
\[\begin{array}{l}\underset{_____________}{\begin{array}{l}\begin{array}{l} \\ 2x+y-3z=0\end{array} \\ \ x-y+z=0\end{array}} \\ \ 3x-2z=0\end{array}\]
We then solve the resulting equation for $z.$
\[\begin{array}{l}3x-2z=0 \\ \ z=\frac{3}{2}x\end{array}\]
We back-substitute the expression for $z$ into one of the equations and solve for $y.$
\[\begin{array}{l}2x+y-3(\frac{3}{2}x)=0 \\ 2x+y-\frac{9}{2}x=0 \\ \ y=\frac{9}{2}x-2x \\ \ y=\frac{5}{2}x\end{array}\]
So the general solution is $(x,\frac{5}{2}x,\frac{3}{2}x).$ In this solution, $x$ can be any real number. The values of $y$ and $z$ are dependent on the value selected for $x.$
Solve the following system.
\[\begin{array}{l}\ \ x+y+z=7 \\ 3x-2y-z=4 \\ x+6y+5z=24\end{array}\]

Bonisa impendulo

Infinite number of solutions of the form $(x,4x-11,-5x+18).$
Can a linear system of three equations have exactly two solutions? Explain why or why not

Bonisa impendulo

No, there can be only one, zero, or infinitely many solutions.
If a given ordered triple solves the system of equations, is that solution unique? If so, explain why. If not, give an example where it is not unique.
If a given ordered triple does not solve the system of equations, is there no solution? If so, explain why. If not, give an example.

Bonisa impendulo

Not necessarily. There could be zero, one, or infinitely many solutions. For example, $(0,0,0)$ is not a solution to the system below, but that does not mean that it has no solution.
$\begin{array}{l}\ 2x+3y-6z=1 \\ -4x-6y+12z=-2 \\ \ x+2y+5z=10\end{array}$
Using the method of addition, is there only one way to solve the system?
Can you explain whether there can be only one method to solve a linear system of equations? If yes, give an example of such a system of equations. If not, explain why not.

Bonisa impendulo

Every system of equations can be solved graphically, by substitution, and by addition. However, systems of three equations become very complex to solve graphically so other methods are usually preferable.
$\begin{array}{l}2x-6y+6z=-12 \\ \ x+4y+5z=-1 \\ -x+2y+3z=-1\ \end{array}$ and $(0,1,-1)$
$\begin{array}{l}\ 6x-y+3z=6 \\ 3x+5y+2z=0\ \\ \ x+y=0\end{array}$ and $(3,-3,-5)$

Bonisa impendulo

No
$\begin{array}{l}6x-7y+z=2 \\ -x-y+3z=4\ \\ \ 2x+y-z=1\end{array}$ and $(4,2,-6)$
$\begin{array}{l}x-y=0 \\ \ x-z=5 \\ x-y+z=-1\ \end{array}$ and $(4,4,-1)$

Bonisa impendulo

Yes
$\begin{array}{l}-x-y+2z=3 \\ \ 5x+8y-3z=4 \\ -x+3y-5z=-5\ \end{array}$ and $(4,1,-7)$
$\begin{array}{l}3x-4y+2z=-15 \\ \ 2x+4y+z=16 \\ \ 2x+3y+5z=20\end{array}$

Bonisa impendulo

$(-1,4,2)$
$\begin{array}{l}5x-2y+3z=20 \\ 2x-4y-3z=-9 \\ \ x+6y-8z=21\end{array}$
$\begin{array}{l}\ 5x+2y+4z=9 \\ -3x+2y+z=10 \\ \ 4x-3y+5z=-3\end{array}$

Bonisa impendulo

$(-\frac{85}{107},\frac{312}{107},\frac{191}{107})$
$\begin{array}{l}4x-3y+5z=31 \\ -x+2y+4z=20 \\ \ x+5y-2z=-29\end{array}$
$\begin{array}{l}\ 5x-2y+3z=4 \\ -4x+6y-7z=-1 \\ \ 3x+2y-z=4\end{array}$

Bonisa impendulo

$(1,\frac{1}{2},0)$
$\begin{array}{l}\ 4x+6y+9z=0 \\ -5x+2y-6z=3 \\ 7x-4y+3z=-3\end{array}$
$\begin{array}{l}\ 2x-y+3z=17 \\ -5x+4y-2z=-46 \\ \ 2y+5z=-7\end{array}$

Bonisa impendulo

$(4,-6,1)$
$\begin{array}{l}5x-6y+3z=50 \\ -x+4y=10 \\ \ 2x-z=10\end{array}$
$\begin{array}{l}\ 2x+3y-6z=1 \\ -4x-6y+12z=-2 \\ \ x+2y+5z=10\end{array}$

Bonisa impendulo

$(x,\frac{1}{27}(65-16x),\frac{x+28}{27})$
$\begin{array}{l}\ 4x+6y-2z=8 \\ \ 6x+9y-3z=12 \\ -2x-3y+z=-4\end{array}$
$\begin{array}{l}\ 2x+3y-4z=5 \\ -3x+2y+z=11 \\ -x+5y+3z=4\end{array}$

Bonisa impendulo

$(-\frac{45}{13},\frac{17}{13},-2)$
$\begin{array}{l}10x+2y-14z=8 \\ \ -x-2y-4z=-1 \\ -12x-6y+6z=-12\end{array}$
$\begin{array}{l}\ x+y+z=14 \\ \ 2y+3z=-14 \\ -16y-24z=-112\end{array}$

Bonisa impendulo

No solutions exist
$\begin{array}{l}\ 5x-3y+4z=-1 \\ -4x+2y-3z=0 \\ \ -x+5y+7z=-11\end{array}$
$\begin{array}{l}\ x+y+z=0 \\ \ 2x-y+3z=0 \\ \ x-z=0\end{array}$

Bonisa impendulo

$(0,0,0)$
$\begin{array}{l}3x+2y-5z=6 \\ 5x-4y+3z=-12 \\ 4x+5y-2z=15\end{array}$

Symbols used here

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

matrix
A rectangular array of numbers; a linear map.

\det A,\ |A|

determinant
Scaling factor of area/volume under A; zero means singular.

A^{-1},\ A^{T}

inverse, transpose
The matrix that undoes A; A with rows and columns swapped.

\lambda

lambda (eigenvalue)
The factor by which an eigenvector is stretched: Av = λv.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Systems of Linear Equations: Three Variables

Solve systems of three equations in three variables.
Identify inconsistent systems of equations containing three variables.
Express the solution of a system of dependent equations containing three variables.
Determine whether an ordered triple is a solution of a system of three linear equations with three variables (IA 4.4.1)
Solve a system of three linear equations with three variables (IA 4.4.2)
Write the equations in standard form. If any coefficients are fractions, clear them.
Eliminate the same variable from two equations.
Decide which variable you will eliminate.

Questions people ask

What does a determinant mean geometrically?

It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.

What is an eigenvector?

A direction the matrix does not turn — it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.

Why is matrix multiplication not commutative?

Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.

Zama ngokwakho

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

IiNkqubo Linear Algebra

Determinants Matrix inverse Eigenvalues and eigenvectors Matrix multiplication Row reduction Vector spaces, span and linear independence Orthogonality, projections and least squares Diagonalisation and matrix powers Linear transformations and change of basis

\(3x+y+z=2\)	\(x+2y+z=-3\)	\(3x+y+2z=4\)
\(\begin{array}{l}3(1)+(-3)+2=2 \\ 2=2\end{array}\) True	\(\begin{array}{l}1+2(-3)+2=-3 \\ -3=-3\end{array}\) True	\(\begin{array}{l}3(1)+(-3)+2(2)=4 \\ 4=4\end{array}\) True
Conclusion: Since (1,-3,2) is a solution to all three equations, then it is a solution to the system \(\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right\)

\(3x+y+z=2\)	\(x+2y+z=-3\)	\(3x+y+2z=4\)
\(\begin{array}{l}3(4)+(-1)+(-5)=2 \\ 6=2\end{array}\) False	\(\begin{array}{l}4+2(-1)+(-5)=-3 \\ -3=-3\end{array}\) True	\(\begin{array}{l}3(4)+(-1)+2(-5)=4 \\ 1=4\end{array}\) False
Conclusion: Since (4,–1,–5) is not a solution to all three equations, then it is not a solution to the system \(\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right\)

\(3x+y+z=2\)	\(x+2y+z=-3\)	\(3x+y+2z=4\)
\(\begin{array}{l}3(1)+(-3)+2=2 \\ 2=2\end{array}\) True	\(\begin{array}{l}1+2(-3)+2=-3 \\ -3=-3\end{array}\) True	\(\begin{array}{l}3(1)+(-3)+2(2)=4 \\ 4=4\end{array}\) True
Conclusion: Since (1,-3,2) is a solution to all three equations, then it is a solution to the system \(\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right\)

\(3x+y+z=2\)	\(x+2y+z=-3\)	\(3x+y+2z=4\)
\(\begin{array}{l}3(4)+(-1)+(-5)=2 \\ 6=2\end{array}\) False	\(\begin{array}{l}4+2(-1)+(-5)=-3 \\ -3=-3\end{array}\) True	\(\begin{array}{l}3(4)+(-1)+2(-5)=4 \\ 1=4\end{array}\) False
Conclusion: Since (4,–1,–5) is not a solution to all three equations, then it is not a solution to the system \(\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right\)

\(3x+y+z=2\)	\(x+2y+z=-3\)	\(3x+y+2z=4\)
\(\begin{array}{l}3(1)+(-3)+2=2 \\ 2=2\end{array}\) True	\(\begin{array}{l}1+2(-3)+2=-3 \\ -3=-3\end{array}\) True	\(\begin{array}{l}3(1)+(-3)+2(2)=4 \\ 4=4\end{array}\) True
Conclusion: Since (1,-3,2) is a solution to all three equations, then it is a solution to the system \(\left\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}\right\)

\(2x-6y+z=3\)	\(3x-4y-3z=2\)	\(2x+3y-2z=3\)
________________________	________________________	________________________
Conclusion: ________________________