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Solve Systems of Equations with Three Variables

Determine whether an ordered triple is a solution of a system of three linear equations with three variables

Determine Whether an Ordered Triple is a Solution of a System of Three Linear Equations with Three Variables

In this section, we will extend our work of solving a system of linear equations. So far we have worked with systems of equations with two equations and two variables. Now we will work with systems of three equations with three variables. But first let's review what we already know about solving equations and systems involving up to two variables.

We learned earlier that the graph of a linear equation, $ax+by=c,$ is a line. Each point on the line, an ordered pair $(x,y),$ is a solution to the equation. For a system of two equations with two variables, we graph two lines. Then we can see that all the points that are solutions to each equation form a line. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions

We know when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Similarly, for a linear equation with three variables $ax+by+cz=d,$ every solution to the equation is an ordered triple, $(x,y,z)$, that makes the equation true.

All the points that are solutions to one equation form a plane in three-dimensional space. And, by finding what the planes have in common, we’ll find the solution to the system.

When we solve a system of three linear equations represented by a graph of three planes in space, there are three possible cases.

Example

Try it.

Determine whether the ordered triple is a solution to the system: $\{\begin{array}{l}x-y+z=2 \\ 2x-y-z=-6 \\ 2x+2y+z=-3\end{array}.$

ⓐ $(-2,-1,3)$ ⓑ $(-4,-3,4)$

Solution

ⓐ

ⓑ

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

Solve a System of Linear Equations with Three Variables

To solve a system of linear equations with three variables, we basically use the same techniques we used with systems that had two variables. We start with two pairs of equations and in each pair we eliminate the same variable. This will then give us a system of equations with only two variables and then we know how to solve that system!

Next, we use the values of the two variables we just found to go back to the original equation and find the third variable. We write our answer as an ordered triple and then check our results.

How to Solve a System of Equations With Three Variables by Elimination

Try it.

Solve the system by elimination: $\{\begin{array}{l}x-2y+z=3 \\ 2x+y+z=4 \\ 3x+4y+3z=-1\end{array}.$

Solution

The steps are summarized here.

Solve a system of linear equations with three variables.

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.
- Work with a pair of equations to eliminate the chosen variable.
- Multiply one or both equations so that the coefficients of that variable are opposites.
- Add the equations resulting from Step 2 to eliminate one variable
Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
The two new equations form a system of two equations with two variables. Solve this system.
Use the values of the two variables found in Step 4 to find the third variable.
Write the solution as an ordered triple.
Check that the ordered triple is a solution to all three original equations.

When we solve a system and end up with no variables and a false statement, we know there are no solutions and that the system is inconsistent. The next example shows a system of equations that is inconsistent.

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

Solve Applications using Systems of Linear Equations with Three Variables

Applications that are modeled by a systems of equations can be solved using the same techniques we used to solve the systems. Many of the application are just extensions to three variables of the types we have solved earlier.

Example

Try it.

The community college theater department sold three kinds of tickets to its latest play production. The adult tickets sold for $15, the student tickets for $10 and the child tickets for $8. The theater department was thrilled to have sold 250 tickets and brought in $2,825 in one night. The number of student tickets sold is twice the number of adult tickets sold. How many of each type did the department sell?

Solution

We will use a chart to organize the information.
Number of students is twice number of adults.
Rewrite the equation in standard form.	$\begin{array}{lll}y & = & 2x \\ 2x-y & = & 0\end{array}$

Use equations (1) and (2) to eliminate z.

Use (3) and (4) to eliminate $y.$

Solve for x.	$\ x\ =75$ adult tickets
Use equation (3) to find y.	$\ -2x+y=0$
Substitute $x=75.$	$\ \begin{array}{lll}-2(75)+y & = & 0 \\ -150+y & = & 0 \\ y & = & 150\ \text{student tickets}\end{array}$
Use equation (1) to find z.	$\ x+y+z=250$
Substitute in the values $x=75,\ y=150.$	$\ \begin{array}{lll}75+150+z & = & 250 \\ 225+z & = & 250 \\ z & = & 25\ \text{child tickets}\end{array}$
Write the solution.	The theater department sold 75 adult tickets, 150 student tickets, and 25 child tickets.

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

Key Concepts

Linear Equation in Three Variables: 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
\[ax+by+cz=d\]
Every solution to the equation is an ordered triple, $(x,y,z)$ that makes the equation true.
How to solve a system of linear equations with three variables.
1. Write the equations in standard form
  If any coefficients are fractions, clear them.
2. Eliminate the same variable from two equations.
  Decide which variable you will eliminate.
  Work with a pair of equations to eliminate the chosen variable.
  Multiply one or both equations so that the coefficients of that variable are opposites.
  Add the equations resulting from Step 2 to eliminate one variable
3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
4. The two new equations form a system of two equations with two variables. Solve this system.
5. Use the values of the two variables found in Step 4 to find the third variable.
6. Write the solution as an ordered triple.
7. Check that the ordered triple is a solution to all three original equations.

Solve Systems of Equations with Three Variables

Determine Whether an Ordered Triple is a Solution of a System of Three Linear Equations with Three Variables

In the following exercises, determine whether the ordered triple is a solution to the system.

Try it.

$\{\begin{array}{l}2x-6y+z=3 \\ 3x-4y-3z=2 \\ 2x+3y-2z=3\end{array}$

ⓐ $(3,1,3)$ ⓑ $(4,3,7)$

Try it.

$\{\begin{array}{l}-3x+\ \ y+z=-4 \\ -x+2y-2z=1 \\ 2x-\ \ y-z=-1\end{array}$

ⓐ $(-5,-7,4)$ ⓑ $(5,7,4)$

Solution

ⓐ no ⓑ yes

Try it.

$\{\begin{array}{l}y-10z=-8 \\ 2x-y=2 \\ x-5z=3\end{array}$

ⓐ $(7,12,2)$ ⓑ $(2,2,1)$

Try it.

$\{\begin{array}{l}x+3y-z=15 \\ y=\frac{2}{3}x-2 \\ x-3y+z=-2\end{array}$

ⓐ $(-6,5,\frac{1}{2})$ ⓑ $(5,\frac{4}{3},-3)$

Solution

ⓐ no ⓑ no

Solve a System of Linear Equations with Three Variables

In the following exercises, solve the system of equations.

Try it.

$\{\begin{array}{l}5x+2y+z=5 \\ -3x-y+2z=6 \\ 2x+3y-3z=5\end{array}$

Try it.

$\{\begin{array}{l}6x-5y+2z=3 \\ 2x+y-4z=5 \\ 3x-3y+z=-1\end{array}$

Solution

$(4,5,2)$

Try it.

$\{\begin{array}{l}2x-5y+3z=8 \\ 3x-y+4z=7 \\ x+3y+2z=-3\end{array}$

Try it.

$\{\begin{array}{l}5x-3y+2z=-5 \\ 2x-y-z=4 \\ 3x-2y+2z=-7\end{array}$

Solution

$(7,12,-2)$

Try it.

$\{\begin{array}{l}3x-5y+4z=5 \\ 5x+2y+z=0 \\ 2x+3y-2z=3\end{array}$

Try it.

$\{\begin{array}{l}4x-3y+z=7 \\ 2x-5y-4z=3 \\ 3x-2y-2z=-7\end{array}$

Solution

$(-3,-5,4)$

Try it.

$\{\begin{array}{l}3x+8y+2z=-5 \\ 2x+5y-3z=0 \\ x+2y-2z=-1\end{array}$

Try it.

$\{\begin{array}{l}11x+9y+2z=-9 \\ 7x+5y+3z=-7 \\ 4x+3y+z=-3\end{array}$

Solution

$(2,-3,-2)$

Try it.

$\{\begin{array}{l}\frac{1}{3}x-y-z=1 \\ x+\frac{5}{2}y+z=-2 \\ 2x+2y+\frac{1}{2}z=-4\end{array}$

Try it.

$\{\begin{array}{l}x+\frac{1}{2}y+\frac{1}{2}z=0 \\ \frac{1}{5}x-\frac{1}{5}y+z=0 \\ \frac{1}{3}x-\frac{1}{3}y+2z=-1\end{array}$

Solution

$(6,-9,-3)$

Try it.

$\{\begin{array}{l}x+\frac{1}{3}y-2z=-1 \\ \frac{1}{3}x+y+\frac{1}{2}z=0 \\ \frac{1}{2}x+\frac{1}{3}y-\frac{1}{2}z=-1\end{array}$

Try it.

$\{\begin{array}{l}\frac{1}{3}x-y+\frac{1}{2}z=4 \\ \frac{2}{3}x+\frac{5}{2}y-4z=0 \\ x-\frac{1}{2}y+\frac{3}{2}z=2\end{array}$

Solution

$(3,-4,-2)$

Try it.

$\{\begin{array}{l}x+2z=0 \\ 4y+3z=-2 \\ 2x-5y=3\end{array}$

Try it.

$\{\begin{array}{l}2x+5y=4 \\ 3y-z=3 \\ 4x+3z=-3\end{array}$

Solution

$(-3,2,3)$

Try it.

$\{\begin{array}{l}2y+3z=-1 \\ 5x+3y=-6 \\ 7x+z=1\end{array}$

Try it.

$\{\begin{array}{l}3x-z=-3 \\ 5y+2z=-6 \\ 4x+3y=-8\end{array}$

Solution

$(-2,0,-3)$

Try it.

$\{\begin{array}{l}4x-3y+2z=0 \\ -2x+3y-7z=1 \\ 2x-2y+3z=6\end{array}$

Try it.

$\{\begin{array}{l}x-2y+2z=1 \\ -2x+y-z=2 \\ x-y+z=5\end{array}$

Solution

no solution

Try it.

$\{\begin{array}{l}2x+3y+z=12 \\ x+y+z=9 \\ 3x+4y+2z=20\end{array}$

Try it.

$\{\begin{array}{l}x+4y+z=-8 \\ 4x-y+3z=9 \\ 2x+7y+z=0\end{array}$

Solution

$x=\frac{203}{16};y=\frac{-25}{16};z=\frac{-231}{16};$

Try it.

$\{\begin{array}{l}x+2y+z=4 \\ x+y-2z=3 \\ -2x-3y+z=-7\end{array}$

Try it.

$\{\begin{array}{l}x+y-2z=3 \\ -2x-3y+z=-7 \\ x+2y+z=4\end{array}$

Solution

$(x,y,z)$ where $x=5z+2;y=-3z+1;z$ is any real number

Try it.

$\{\begin{array}{l}x+y-3z=-1 \\ y-z=0 \\ -x+2y=1\end{array}$

Try it.

$\{\begin{array}{l}x-2y+3z=1 \\ x+y-3z=7 \\ 3x-4y+5z=7\end{array}$

Solution

$(x,y,z)$ where $x=5z-2;y=4z-3;z$ is any real number

Solve Applications using Systems of Linear Equations with Three Variables

In the following exercises, solve the given problem.

Try it.

The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is twice the measure of the first angle. The third angle is twelve more than the second. Find the measures of the three angles.

Try it.

The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is three times the measure of the first angle. The third angle is fifteen more than the second. Find the measures of the three angles.

Solution

45 degrees, 60 degrees, 75 degrees

Try it.

After watching a major musical production at the theater, the patrons can purchase souvenirs. If a family purchases 4 t-shirts, the video, and 1 stuffed animal, their total is $135.

A couple buys 2 t-shirts, the video, and 3 stuffed animals for their nieces and spends $115. Another couple buys 2 t-shirts, the video, and 1 stuffed animal and their total is $85. What is the cost of each item?

Try it.

The church youth group is selling snacks to raise money to attend their convention. Amy sold 2 pounds of candy, 3 boxes of cookies and 1 can of popcorn for a total sales of $65. Brian sold 4 pounds of candy, 6 boxes of cookies and 3 cans of popcorn for a total sales of $140. Paulina sold 8 pounds of candy, 8 boxes of cookies and 5 cans of popcorn for a total sales of $250. What is the cost of each item?

Solution

$20, $5, $10

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

Practice (40)

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

Evaluate $5x-2y+3z$ when $x=-2,$ $y=-4,$ and $z=3.$
If you missed this problem, review .

விடை தெரியப்படுத்து

$7$
Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. $\{\begin{array}{l}-2x+y=-11 \\ x+3y=9\end{array}.$
If you missed this problem, review .

விடை தெரியப்படுத்து

conditional; one solution
Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. $\{\begin{array}{l}7x+8y=4 \\ 3x-5y=27\end{array}.$
If you missed this problem, review .

விடை தெரியப்படுத்து

conditional; one solution
Determine whether the ordered triple is a solution to the system: $\{\begin{array}{l}x-y+z=2 \\ 2x-y-z=-6 \\ 2x+2y+z=-3\end{array}.$
ⓐ $(-2,-1,3)$ ⓑ $(-4,-3,4)$

விடை தெரியப்படுத்து

ⓐ

ⓑ
Determine whether the ordered triple is a solution to the system: $\{\begin{array}{l}3x+y+z=2 \\ x+2y+z=-3 \\ 3x+y+2z=4\end{array}.$
ⓐ $(1,-3,2)$ ⓑ $(4,-1,-5)$

விடை தெரியப்படுத்து

ⓐ yes ⓑ no
Determine whether the ordered triple is a solution to the system: $\{\begin{array}{l}x-3y+z=-5 \\ -3x-y-z=1 \\ 2x-2y+3z=1\end{array}.$
ⓐ $(2,-2,3)$ ⓑ $(-2,2,3)$

விடை தெரியப்படுத்து

ⓐ no ⓑ yes
Solve the system by elimination: $\{\begin{array}{l}x-2y+z=3 \\ 2x+y+z=4 \\ 3x+4y+3z=-1\end{array}.$

விடை தெரியப்படுத்து
Solve the system by elimination: $\{\begin{array}{l}3x+y-z=2 \\ 2x-3y-2z=1 \\ 4x-y-3z=0\end{array}.$

விடை தெரியப்படுத்து

$(2,-1,3)$
Solve the system by elimination: $\{\begin{array}{l}4x+y+z=-1 \\ -2x-2y+z=2 \\ 2x+3y-z=1\end{array}.$

விடை தெரியப்படுத்து

$(-2,3,4)$
Solve: $\{\begin{array}{l}3x-4z=0 \\ 3y+2z=-3 \\ 2x+3y=-5\end{array}.$

விடை தெரியப்படுத்து

\[\{\begin{array}{l}3x-4z=0\ (1) \\ 3y+2z=-3\ (2) \\ 2x+3y=-5\ (3)\end{array}\]
We can eliminate $z$ from equations (1) and (2) by multiplying equation (2) by 2 and then adding the resulting equations.

Notice that equations (3) and (4) both have the variables $x$ and $y$. We will solve this new system for $x$ and $y$.

To solve for y, we substitute $x=-4$ into equation (3).

We now have $x=-4$ and $y=1.$ We need to solve for z. We can substitute $x=-4$ into equation (1) to find z.

We write the solution as an ordered triple. $\ (-4,1,-3)$
We check that the solution makes all three equations true.
$\begin{array}{lllllllllllllllllllllllll}\begin{array}{lll}3x-4z & = & 0(1) \\ 3(-4)-4(-3) & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array} & & & \begin{array}{lll}3y+2z & = & -3(2) \\ 3(1)+2(-3) & \overset{?}{=} & -3 \\ -3 & = & -3✓\end{array} & & & \begin{array}{lllllll} \\ \\ \begin{array}{lll}2x+3y & = & -5(3) \\ 2(-4)+3(1) & \overset{?}{=} & -5 \\ -5 & = & -5✓\end{array} \\ \text{The solution is}\ (-4,1,-3).\end{array}\end{array}$
Solve: $\{\begin{array}{l}3x-4z=-1 \\ 2y+3z=2 \\ 2x+3y=6\end{array}.$

விடை தெரியப்படுத்து

$(-3,4,-2)$
Solve: $\{\begin{array}{l}4x-3z=-5 \\ 3y+2z=7 \\ 3x+4y=6\end{array}.$

விடை தெரியப்படுத்து

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

விடை தெரியப்படுத்து

\[\{\begin{array}{l}x+2y-3z=-1\ (1) \\ x-3y+z=1\ (2) \\ 2x-y-2z=2\ (3)\end{array}\]
Use equation (1) and (2) to eliminate z.

Use (2) and (3) to eliminate $z$ again.

Use (4) and (5) to eliminate a variable.

There is no solution.
We are left with a false statement and this tells us the system is inconsistent and has no solution.
Solve the system of equations: $\{\begin{array}{l}x+2y+6z=5 \\ -x+y-2z=3 \\ x-4y-2z=1\end{array}.$

விடை தெரியப்படுத்து

no solution
Solve the system of equations: $\{\begin{array}{l}2x-2y+3z=6 \\ 4x-3y+2z=0 \\ -2x+3y-7z=1\end{array}.$

விடை தெரியப்படுத்து

no solution

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

விடை தெரியப்படுத்து

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

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 by equations: $\{\begin{array}{l}x+y-z=0 \\ 2x+4y-2z=6 \\ 3x+6y-3z=9\end{array}.$

விடை தெரியப்படுத்து

infinitely many solutions$(x,3,z)$ where $x=z-3;y=3;z$ is any real number
Solve the system by equations: $\{\begin{array}{l}x-y-z=1 \\ -x+2y-3z=-4 \\ 3x-2y-7z=0\end{array}.$

விடை தெரியப்படுத்து

infinitely many solutions $(x,y,z)$ where$x=5z-2;y=4z-3;z$ is any real number

விடை தெரியப்படுத்து

We will use a chart to organize the information.
Number of students is twice number of adults.
Rewrite the equation in standard form.	$\begin{array}{lll}y & = & 2x \\ 2x-y & = & 0\end{array}$

Use equations (1) and (2) to eliminate z.

Use (3) and (4) to eliminate $y.$

Solve for x.	$\ x\ =75$ adult tickets
Use equation (3) to find y.	$\ -2x+y=0$
Substitute $x=75.$	$\ \begin{array}{lll}-2(75)+y & = & 0 \\ -150+y & = & 0 \\ y & = & 150\ \text{student tickets}\end{array}$
Use equation (1) to find z.	$\ x+y+z=250$
Substitute in the values $x=75,\ y=150.$	$\ \begin{array}{lll}75+150+z & = & 250 \\ 225+z & = & 250 \\ z & = & 25\ \text{child tickets}\end{array}$
Write the solution.	The theater department sold 75 adult tickets, 150 student tickets, and 25 child tickets.

The community college fine arts department sold three kinds of tickets to its latest dance presentation. The adult tickets sold for $20, the student tickets for $12 and the child tickets for $10.The fine arts department was thrilled to have sold 350 tickets and brought in $4,650 in one night. The number of child tickets sold is the same as the number of adult tickets sold. How many of each type did the department sell?

விடை தெரியப்படுத்து

The fine arts department sold 75 adult tickets, 200 student tickets, and 75 child tickets.
The community college soccer team sold three kinds of tickets to its latest game. The adult tickets sold for $10, the student tickets for $8 and the child tickets for $5. The soccer team was thrilled to have sold 600 tickets and brought in $4,900 for one game. The number of adult tickets is twice the number of child tickets. How many of each type did the soccer team sell?

விடை தெரியப்படுத்து

The soccer team sold 200 adult tickets, 300 student tickets, and 100 child tickets.
$\{\begin{array}{l}2x-6y+z=3 \\ 3x-4y-3z=2 \\ 2x+3y-2z=3\end{array}$
ⓐ $(3,1,3)$ ⓑ $(4,3,7)$
$\{\begin{array}{l}-3x+\ \ y+z=-4 \\ -x+2y-2z=1 \\ 2x-\ \ y-z=-1\end{array}$
ⓐ $(-5,-7,4)$ ⓑ $(5,7,4)$

விடை தெரியப்படுத்து

ⓐ no ⓑ yes
$\{\begin{array}{l}y-10z=-8 \\ 2x-y=2 \\ x-5z=3\end{array}$
ⓐ $(7,12,2)$ ⓑ $(2,2,1)$
$\{\begin{array}{l}x+3y-z=15 \\ y=\frac{2}{3}x-2 \\ x-3y+z=-2\end{array}$
ⓐ $(-6,5,\frac{1}{2})$ ⓑ $(5,\frac{4}{3},-3)$

விடை தெரியப்படுத்து

ⓐ no ⓑ no
$\{\begin{array}{l}5x+2y+z=5 \\ -3x-y+2z=6 \\ 2x+3y-3z=5\end{array}$
$\{\begin{array}{l}6x-5y+2z=3 \\ 2x+y-4z=5 \\ 3x-3y+z=-1\end{array}$

விடை தெரியப்படுத்து

$(4,5,2)$
$\{\begin{array}{l}2x-5y+3z=8 \\ 3x-y+4z=7 \\ x+3y+2z=-3\end{array}$
$\{\begin{array}{l}5x-3y+2z=-5 \\ 2x-y-z=4 \\ 3x-2y+2z=-7\end{array}$

விடை தெரியப்படுத்து

$(7,12,-2)$
$\{\begin{array}{l}3x-5y+4z=5 \\ 5x+2y+z=0 \\ 2x+3y-2z=3\end{array}$
$\{\begin{array}{l}4x-3y+z=7 \\ 2x-5y-4z=3 \\ 3x-2y-2z=-7\end{array}$

விடை தெரியப்படுத்து

$(-3,-5,4)$
$\{\begin{array}{l}3x+8y+2z=-5 \\ 2x+5y-3z=0 \\ x+2y-2z=-1\end{array}$
$\{\begin{array}{l}11x+9y+2z=-9 \\ 7x+5y+3z=-7 \\ 4x+3y+z=-3\end{array}$

விடை தெரியப்படுத்து

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

விடை தெரியப்படுத்து

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

விடை தெரியப்படுத்து

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

விடை தெரியப்படுத்து

$(-3,2,3)$
$\{\begin{array}{l}2y+3z=-1 \\ 5x+3y=-6 \\ 7x+z=1\end{array}$

Symbols used here

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

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

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Systems of Equations with Three Variables

Determine whether an ordered triple is a solution of a system of three linear equations with three variables
Solve a system of linear equations with three variables
Solve applications using systems of linear equations with three variables
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.
Work with a pair of equations to eliminate the chosen variable.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

உங்களை முயற்சிக்கவும்

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

மேலும் Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

	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 \(y=-2z+3.\)	\(\begin{array}{lll}x+2(-2z+3)-z & = & 1 \\ x-4z+6-z & = & 1 \\ x-5z+6 & = & 1 \\ x & = & 5z-5\end{array}\)