Solving 3x3 Systems of Equations Algebraically
Description/Explanation/Highlights
Video Description
This video explains how to solve a 3×3 system of equations algebraically.
Steps and Key Points to Remember
To solve a 3×3 system of equations algebraically, follow these steps:
- To solve a 3×3 system, we will use elimination and substitution to solve for one variable at a time until all variables are solved using the following general process.
- Put the equation in order with variables in alphabetical order to the left and the constant to the right of the equal sign.
- Line them up vertically adding any missing variables with a zero for the coefficient.
- Use one of the equations twice and combine it with each of the others to eliminate the same variable in each system.
- Use the newly created equations to form a 2×2 system.
- Use elimination to eliminate an additional variable and solve for the remaining variable.
- Substitute that solution into one of the equations in the 2×2 system to solve for the other variable.
- Use the two solutions in one of the original equations in the 3×3 system to find the value of the final variable.
- To see this process, look at the following example.
\(x+2y+z=9\)
\(2x+y=-3z\)
\(3x-1=4y+5z\)
\(2x+y=-3z\)
\(3x-1=4y+5z\)
- Begin by putting the equations in order and lining them up.
\(x+2y+z=9 \;\;\;\;\rightarrow\;\;x+2y+\;\;\,z=9\)
\(2x+y=-3z \;\;\;\;\;\,\rightarrow2x+\;\:y+\,3z=0\)
\(3x-1=4y+5z\rightarrow \underline{3x-4y-\,5z=1}\)
- Pick one of the equations to use twice to eliminate the same variable with each of the other equations.
- I picked the first equation with the intent of eliminating the x since it has a coefficient of 1.
\( \;\;x+2y+\;\;\,z=9\)
\( \underline{2x+\;\:y+\,3z=0}\)
\( \;\;x+2y+\;\;\,z=9\)
\(\underline{3x-4y-\,5z=1}\)
- Multiply the first equation in the first system by -2 and the first equation in the second system by -3 to eliminate x in each equation. Note: You must eliminate the same variable in each system.
- Combine each system to create equations with two variables.
\( -2(\;\;x+2y+\;\;\,z=9)\)
\( \underline{\;\;\;\;\;\;\;2x+\;\:y+\,3z=0}\)
\( -3(\;\;x+2y+\;\;\,z=9)\)
\(\underline{\;\;\;\;\;\;\;3x-4y-\,5z=1}\)
\( -2x-4y-2z=-18\)
\( \underline{\;\;\;2x+\;\:y+\,3z=\;\;\;\;0}\)
\(\;\;\;\;\;\;\;\;-3y+\;\;\;z=-18\)
\( -3x-6y-3z=-27\)
\(\underline{\;\;\;3x-4y-\,5z=\;\;\;\;1}\)
\(\;\;\;\;\;\;\:-10y-\:8z=-26\)
\(\underline{\;\;\;3x-4y-\,5z=\;\;\;\;1}\)
\(\;\;\;\;\;\;\:-10y-\:8z=-26\)
- Make the two equations into a 2×2 system to combine and eliminate another variable.
\(-10y-\:8z=-26\)
\(\underline{\;\;-3y+\;\;\:z=-18}\)
\(\underline{\;\;-3y+\;\;\:z=-18}\)
- Multiplying the second equation by 8 and combining will eliminate the z.
- Divide both sides by the coefficient of y to find the final value of y.
\(\;\;\;\;-10y-\:8z=-26\;\longrightarrow\)
\(\underline{8(\;\;-3y+\;\;\:z=-18)}\longrightarrow\)
\(-10y-\:8z=\;\;-26\)
\(\underline{-24y+\:8z=-144}\)
\(\underline{-34y}\;\;\;\;\;\;\;\;\;=\underline{-170}\)
\(\;-34\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-34\)
\(\;\;\;\;\;\;\;\;\;\;y=5\)
\(\underline{-24y+\:8z=-144}\)
\(\underline{-34y}\;\;\;\;\;\;\;\;\;=\underline{-170}\)
\(\;-34\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-34\)
\(\;\;\;\;\;\;\;\;\;\;y=5\)
- Substitute the value of y into one of the 2-variable equations we found above. We will use \(-3y+z=-18\). Substituting 5 for y we get \(-3(5)+z=-18\rightarrow -15+z=-18\) and adding 15 to both sides will give us \(z=-3\).
- We now know the value of y and z so we can substitute these values into one of the original 3-variable equations to find x. Using \(x+2y+z=9\) and substituting 5 and -3 gives us \(x+2(5)+(-3)=9\) so \(x+7=9\) and \(x=2\).
- The solution to our 3×3 system is \((2, 5,-3)\).
Here are some key points to keep in mind when solving a 3×3 system of equations algebraically.
- Put the equations in order and line up one under the other to attempt to combine and eliminate one of the variables.
- Include any missing variables with a zero as a coefficient.
- Use one of the equations with both of the other equations to eliminate the same variable in both sets of equations
- By choosing carefully, you may be able to multiply only one of the equations by a number to eliminate the same variable in both parts of the system.
- After eliminating a variable, combine the two equations to form a 2×2 system and eliminate another variable to solve for the remaining variable.
- Work backward and substitute the solution into one of the 2-variable equations to find the value of the second variable and finally use both solutions in one of the original 3-variable equations to solve for the final variables.
- Keeping work organized is important in working mistake free through the process.
- To check, plug x, y, and z into the original equations and make sure that simplifying gets a true statement for each.
Video Highlights
- 00:00 Introduction
- 00:13 \(x+2y+z=9\), \(2x+y=-3z\) and \(3x-1=4y+5z\) example of solving a 3×3 system of equations algebraically
- 11:03 Conclusion
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