Solution to the System With 3 Variables
Solving Systems of Three Variables
Learning Objective(s)
· Solve a system of equations when no multiplication is necessary to eliminate a variable.
· Solve a system of equations when multiplication is necessary to eliminate a variable.
· Solve application problems that require the use of this method.
· Recognize systems that have no solution or an infinite number of solutions.
Introduction
Equations can have more than one or two variables. You are going to look at equations with three variables. Equations with one variable graph on a line. Equations with two variables graph on a plane. Equations with three variables graph in a 3-dimensional space.
Equations with one variable require only one equation to have a unique (one) solution. Equations with two variables require two equations to have a unique solution (one ordered pair). So it should not be a surprise that equations with three variables require a system of three equations to have a unique solution (one ordered triplet).
Solving A System of Three Variables
Just as when solving a system of two equations, there are three possible outcomes for the solution of a system of three variables. Let's look at this visually, although you will not be graphing these equations.
Case 1: There is one solution. In order for three equations with three variables to have one solution, the planes must intersect in a single point.
Case 2: There is no solution. The three planes do not have any points in common. (Note that two of the equations may have points in common with each other, but not all three.) Below are examples of some of the ways this can happen.
Case 3: There are an infinite number of solutions. This occurs when the three planes intersect in a line. And this can also occur when the three equations graph as the same plane.
Let's start by looking at Case 1, where the system has a unique (one) solution. This is the case that you are usually most interested in.
Here is a system of linear equations. There are three variables and three equations.
| 3x | + | 4y | – | z | = | 8 |
| 5x | – | 2y | + | z | = | 4 |
| 2x | – | 2y | + | z | = | 1 |
You know how to solve a system with two equations and two variables. For the first step, use the elimination method to remove one of the variables. In this case, z can be eliminated by adding the first and second equations.
| 3x | + | 4y | – | z | = | 8 |
| 5x | – | 2y | + | z | = | 4 |
| 8x | + | 2y | = | 12 |
To solve the system, though, you need two equations using two variables. Adding the first and third equations in the original system will also give an equation with x and y but not z.
| 3x | + | 4y | – | z | = | 8 |
| 2x | – | 2y | + | z | = | 1 |
| 5x | + | 2y | = | 9 |
Now you have a system of two equations and two variables.
Solve the system using elimination again. In this case, you can eliminate y by adding the opposite of the second equation:
| 8x | + | 2y | = | 12 | ||
| −5x | + | −2y | = | −9 | ||
| 3x | = | 3 |
Solve the resulting equation for the remaining variable.
Now you use one of the equations in the two-variable system to find y.
| 5x + 2y = 9 |
| 5(1) + 2y = 9 |
| 5 + 2y = 9 |
| 2y = 4 |
| y = 2 |
Finally, use any equation from the first system, along with the values already found, to solve for the last variable.
| 2x – 2y + z = 1 |
| 2(1) – 2(2) + z = 1 |
| 2 – 4 + z = 1 |
| −2 + z = 1 |
| z = 3 |
Be sure to check your answer. With this many steps, there are a lot of places to make a simple error!
| 3x + 4y – z = 8 | 5x – 2y + z = 4 | 2x – 2y + z = 1 | ||
| 3(1) + 4(2) – 3 = 8 | 5(1) – 2(2) + 3 = 4 | 2(1) – 2(2) + 3 = 1 | ||
| 3 + 8 – 3 = 8 | 5 – 4 + 3 = 4 | 2 – 4 + 3 = 1 | ||
| 11 – 3 = 8 8 = 8 | 1 + 3 = 4 4 = 4 | −2 + 3 = 1 1=1 | ||
| TRUE | TRUE | TRUE |
Since x = 1, y = 2, and z = 3 is a solution for all three equations, it's the solution for the system of equations. Just as two values can be written as an ordered pair, three values can be written as an ordered triplet: (x, y, z) = (1, 2, 3).
Solving a system of three variables
1. Choose two equations and use them to eliminate one variable.
2. Choose another pair of equations and use them to eliminate the same variable.
3. Use the resulting pair of equations from steps 1 and 2 to eliminate one of the two remaining variables.
4. Solve the final equation for the remaining variable.
5. Find the value of the second variable. Do this by using one of the resulting equations from steps 1 and 2 and the value of the found variable from step 4.
6. Find the value of the third variable. Do this by using one of the original equations and the values of the found variables from steps 4 and 5.
7. Check your answer in all three equations !
| Example | ||||||||||||||||||||||||
| Problem | Solve for f , g , and h . f + g + h = 13 f – h = −2 − 2f + g = 3 | |||||||||||||||||||||||
| Step 1: Choose two equations and eliminate a variable. The first two equations can be added to eliminate h. | |||||||||||||||||||||||
| 2f + g = 11 − 2f + g = 3 | Step 2: The third equation has no h variable, so there's nothing to eliminate! You have a system of two equations and two variables. | |||||||||||||||||||||||
| Step 3: Eliminate a second variable. These equations can be added to eliminate f. | |||||||||||||||||||||||
| 2g = 14 g = 7 | Step 4: Solve the resulting equation for the remaining variable. | |||||||||||||||||||||||
| 2f + g = 11 2f + 7 = 11 2f = 4 f = 2 | Step 5: Use that value and one of the equations from the system in step 3 that involves just two variables, one of which was g that you already know. Solve for the second variable. | |||||||||||||||||||||||
| f + g + h = 13 2 + 7 + h = 13 9 + h = 13 h = 4 | Step 6: Use the two found values and one of the original equations that had all three variables to solve for the third variable. | |||||||||||||||||||||||
| f + g + h = 13 2 + 7 + 4 = 13 9 + 4 = 13 13 = 13 TRUE f – h = −2 2 – 4 = −2 − 2= −2 TRUE − 2f + g = 3 − 2(2) + 7 = 3 − 4 + 7 = 3 3 = 3 TRUE | Step 7: Check your answer . | |||||||||||||||||||||||
| Answer | The solution is (f, g, h) = (2, 7, 4). | |||||||||||||||||||||||
As with systems of two equations with two variables, you may need to add the opposite of one of the equations or even multiply one of the equations before adding in order to eliminate one of the variables.
| Example | |||||||||||||||||||||||||||||||||||||||||||||||
| Problem | Solve for x, y, and z. 3x – 2y + z = 12 x + 3y + z = −4 2x + 2y – 4z = 6 | ||||||||||||||||||||||||||||||||||||||||||||||
| Step 1: First, choose two equations and eliminate a variable. Multiply the second equation by − 1, and then add it to the first equation. This will eliminate z. | ||||||||||||||||||||||||||||||||||||||||||||||
| Step 2: Next, combine the third equation and one of the first two to eliminate z again. However, the third equation has a coefficient of − 4 on z while the coefficients in the first two equations are both 1. So, multiply the second equation by 4 and add. | ||||||||||||||||||||||||||||||||||||||||||||||
| Step 3: Eliminate a second variable using the equations from steps 1 and 2. Again, they cannot be added as they are. Look at the coefficients on x. If you multiply the equation from step 1 by − 3, the x terms will have the same coefficient. | ||||||||||||||||||||||||||||||||||||||||||||||
| Multiply and then add. Be careful of the signs! | ||||||||||||||||||||||||||||||||||||||||||||||
| 29y = −58 y = −2 | Step 4: Solve the resulting equation for the remaining variable. | ||||||||||||||||||||||||||||||||||||||||||||||
| 2x – 5y = 16 2x – 5(−2) = 16 2x + 10 = 16 2x = 6 x = 3 | Step 5: Use that value and one of the equations from the system in step 3, that involves just two variables, one of which was y. Solve for the second variable. | ||||||||||||||||||||||||||||||||||||||||||||||
| x + 3y + z = −4 3 + 3(−2) + z = −4 3 + (−6) + z = −4 − 3 + z = −4 z = −1 | Step 6: Use the two found values and one of the original equations to solve for the third variable. | ||||||||||||||||||||||||||||||||||||||||||||||
| 3x – 2y + z = 12 3(3) – 2(−2) + (−1) = 12 9 + 4 – 1 = 12 13 – 1 = 12 12 = 12 TRUE x + 3y + z = −4 3 + 3(−2) + (−1) = −4 3 + (−6) + (−1) = −4 − 3 + (−1) = −4 − 4 = −4 TRUE 2x + 2y – 4z = 6 2(3) + 2(−2) – 4(−1) = 6 6 + (−4) + 4 = 6 2 + 4 = 6 6 = 6 TRUE | Step 7: Check your answer . | ||||||||||||||||||||||||||||||||||||||||||||||
| Answer | The solution is (x, y, z) = (3, −2, −1). | ||||||||||||||||||||||||||||||||||||||||||||||
These systems can be helpful for solving real-world problems.
| Example | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Problem | Andrea sells photographs at art fairs. She prices the photos according to size: small photos cost $10, medium photos cost $15, and large photos cost $40. She usually sells as many small photos as medium and large photos combined. She also sells twice as many medium photos as large. A booth at the art fair costs $300. If her sales go as usual, how many of each size photo must she sell to pay for the booth? | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| S = number of small photos sold M = number of medium photos sold L = number of large photos sold | To set up the system, first choose the variables. In this case the unknown values are the number of small, medium, and large photos. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 10S = money received for small photos 15M = money received for medium photos 40L = money received for large photos 10S + 15M + 40L = 300 | The total of her sales must be $300 to pay for the booth. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| S = M + L M = 2L | The number of small photos is the same as the total of medium and large photos. She sells twice as many medium photos as large photos. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 10S + 15M + 40L = 300 S – M – L = 0 M – 2L = 0 | To make things easier, rewrite the equations to be in the same format, with all variables on the left side of the equal sign and only a constant number on the right. Now solve the system. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Step 1: First choose two equations and eliminate a variable. Since one equation has no S variable, it may be helpful to use the other two equations and eliminate the S variable from them. Multiply the second equation by − 10 and add. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| M – 2L = 0 | Step 2: The second equation for our two-variable system will be the remaining equation (that has no S variable). | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Step 3: Eliminate a second variable using the equations from steps 1 and 2. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| While you could multiply the second equation by 25 to eliminate L, the numbers will stay nicer if you divide the first equation by 25. Don't forget to be careful of the signs! | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 2M = 12 M = 6 | Step 4: Solve the resulting equation for the remaining variable. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| M = 2L 6 = 2L 3 = L | Step 5: Use that value and one of the equations containing just two variables, one of those variables being L that you already know, to solve for the second variable. It's best to use one of the original equations—in case an error was made in multiplication. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| S = M + L S = 6 + 3 S = 9 | Step 6: Use the two found values and one of the original equations to solve for the third variable. You can even use one of the equations before you rewrote it for the system. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| She usually sells as many small photos as medium and large photos combined. Medium and large photos combined = 6 + 3 = 9, which is the number of small photos. She also sells twice as many medium photos as large. Medium photos is 6, which is twice the number of large photos (3). A booth at the art fair costs $300 . Andrea receives $10(9) or $90 for the 9 small photos, $15(6) or $90 for the 6 medium photos, and $40(3) or $120 for the large photos. $90 + $90 + $120 = $300. | Step 7: Check your answer . With application problems, it's sometimes easier (and better) to use the original wording of the problem rather than the equations you write. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Answer | If Andrea sells 9 small photos, 6 medium photos, and 3 large photos, she'll receive exactly the amount of money needed to pay for the booth. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
In the solution to this system, what is the value of x?
7x − 4y + 3z = 28
3x + 3y – z = 19
3x + 2y + z = 16
A) 5
B) 16
C) -31
D) 1
Show/Hide Answer
A) 5
Correct. Eliminate z by adding the last two equations together to get 6x + 5y = 35. Now multiply the second equation by 3 and add to the first equation to get 16x + 5y = 85. This creates a smaller system of two equations and two variables: 6x + 5y = 35 and 16x + 5y = 85. Multiply 6x + 5y = 35 by −1 to create −6x – 5y = −35 and now add this to 16x + 5y = 85. This eliminates y, giving 10x = 50, so x = 5.
B) 16
Incorrect. Eliminate z by adding the last two equations together, to get 6x + 5y = 35. Now multiply the second equation by 3 and add to the first equation to get 16x + 5y = 85. This creates a smaller system of two equations and two variables: 6x + 5y = 35 and 16x + 5y = 85. Multiply 6x + 5y = 35 by −1 to create −6x - 5y = −35 and now add this to 16x + 5y = 85. This eliminates y, giving 10x = 50, so x = 5.
C) -31
Incorrect. Eliminate z by adding the last two equations together, to get 6x + 5y = 35. Now multiply the second equation by 3 and add to the first equation to get 16x + 5y = 85. This creates a smaller system of two equations and two variables: 6x + 5y = 35 and 16x + 5y = 85. Multiply 6x + 5y = 35 by −1 to create −6x – 5y = −35 and now add this to 16x + 5y = 85. This eliminates y, giving 10x = 50, so x = 5.
D) 1
Incorrect. Eliminate z by adding the last two equations together, to get 6x + 5y = 35. Now multiply the second equation by 3 and add to the first equation to get 16x + 5y = 85. This creates a smaller system of two equations and two variables: 6x + 5y = 35 and 16x + 5y = 85. Multiply 6x + 5y = 35 by −1 to create −6x – 5y = −35 and now add this to 16x + 5y = 85. This eliminates y, giving 10x = 50, so x = 5.
Systems with No Solutions or an Infinite Number of Solutions
Now let's look at Case 2 (no solution) and Case 3 (an infinite number of solutions).
Since you will not graph these equations, as it is difficult to graph in three dimensions on a 2-dimensional sheet of paper, you will look at what happens when you try to solve systems with no solutions or an infinite number of solutions.
Let's look at a system that has no solutions.
5x – 2y + z = 3
4x – 4y – 8z = 2
− x + y + 2z = − 3
Suppose you wanted to solve this system, and you started with the last two equations. Multiply the last by 4 and add to eliminate x.
| 4x | – | 4y | – | 8z | = | 2 |
| 4(− x | + | y | + | 2z) | = | 4(−3) |
| 4x | – | 4y | – | 8z | = | 2 |
| − 4x | + | 4y | + | 8z | = | − 12 |
| 0 | = | − 10 |
In this case, the result is a false statement. This means there are no solutions to the two equations and therefore there can be no solutions for the system of three equations. If this occurs for any two of the three equations, then there is no solution for the system of equations.
Now let's look at a system that has an infinite number of solutions.
x – 2y + z = 3
− 3x + 6y – 3z = −9
4x – 8y + 4z = 12
For the first step, you would choose two equations and combine them to eliminate a variable. You can eliminate x by multiplying the first equation by 3 and adding to the second equation.
| 3(x | – | 2y | + | z ) | = | 3(3) |
| − 3x | + | 6y | – | 3z | = | − 9 |
| 3x | – | 6y | + | 3z | = | 9 |
| − 3x | + | 6y | – | 3z | = | − 9 |
| 0 | = | 0 |
Notice that when the two equations are added, all variables are eliminated! The final equation is a true statement: 0 = 0.
When this happens, it's because the two equations are equivalent. These two equations would graph as the same plane. But in order for the solution to the system of three equations to be infinite, you need to continue to check with the third equation.
Since the first two equations are equivalent, the system of equations could be written with only two equations. Continue as before. Multiply the first equation by − 4 and add the third equation.
− 4( x – 2y +z) =−4(3)
4x – 8y + 4z = −12
− 4x + 8y – 4z = −12
4x – 8y + 4z = 12
0 = 0
Again, the final equation is the true statement 0 = 0. So the third equation is the same plane as the first two. Now you can confirm that there are an infinite number of solutions—all of the points that are on the plane that these three equations each describe.
This is one type of situation where there are an infinite number of solutions. There are others, which you will not examine at this time.
| Example | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Problem | How many solutions does the following system of equations have? x + y + z = 2 2x + 2y + 2z = 4 −3x – 3y – 3z = −6 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Multiply the first equation by − 2 and then add that resulting equation to the second equation. 0 = 0 is a true statement, which leads us to believe that you may have an infinite number of solutions. This outcome indicates that the first pair of equations is really the same equation. The values of x, y, and z that will make the first equation work will also work for the second. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Now add the third equation with the first. Again, the result is another true statement. The first and third equations are the same. So you have three equations that will all graph as the same plane. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Answer | There are an infinite number of solutions to this system. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Example | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Problem | Solve the following system of equations. x – y – 2z = 4 4x – 4y – z = 2 −x + y + 2z = −3 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Compare the coefficients on the x terms. Multiply the first equation by − 4 and then add that resulting equation to the second equation. Notice that a false statement is produced: 0 = − 14. This means that there is no solution to this system of equations; you do not have to complete any further steps. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Answer | The system has no solutions. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
How many solutions does this system have?
6x + 4y + 2z = 32
3x – 3y – z = 19
3x + 2y + z = 32
A) No solutions
B) One
C) An infinite number of solutions
Show/Hide Answer
A) No solutions
Correct. Multiply the last equation by −2 to get −6x – 4y – 2z = −64. If you add this equation to the first one, you will get 0 = −32, a false statement. This means that this system has no solutions.
B) One
Incorrect. If you multiply the last equation by −2 and then add it to the first equation, you get 0 = −32, a false statement. This system has no solutions.
C) An infinite number of solutions
Incorrect. If you multiply the last equation by −2 and then add it to the first equation, you get 0 = −32, a false statement. This system has no solutions.
Summary
Combining equations is a powerful tool for solving a system of equations, including systems with three equations and three variables. Sometimes, you must multiply one of the equations before you add so that you can eliminate a variable. You continue the process of combining equation and eliminating variables until you have found the value of all of the variables. Occasionally this process leads to all of the variables being eliminated (eliminated not solved for). When all the variables are eliminated by such combinations of combining equations, if it leads to a false statement, then the system will have no solutions. When all the variables are eliminated by such combinations of combining equations, if one of the resulting equations is true, the system may have an infinite number of solutions. However, all the equations must be compared and found to true for there to be an infinite number of solutions, not just two of the three equations.
Solution to the System With 3 Variables
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