Systems with three unknowns via addition method

Often you can solve systems of equations with three unknowns with a similar procedure - almost like a cooking recipe. In this article, you'll learn a way that may not always be the fastest, but is failproof, on the other hand. Other (possilby faster) methods for solving are the Gaussian elimination and Cramer's rule.

Before you can see an example of how the recipe works, you will first get to know the general procedure.

Solving a system of equations with three unknowns requires a lot of concentration. Work clearly and check off all steps one after the other:

1. Reduce the system to two unknown variables a) Use the addition method on two of the equations to eliminate one variable. b) Apply the addition method to the third equation and any of the two above to eliminate the same variable.

2. Solve the resulting system of equations with two unknown variables a) Eliminate one of the variables using the addition method and find the other $\Rightarrow$ you know the first variable b) Plug in the now determined variable into one of the two equations

3. Find the last unknown in the original equation system $\Rightarrow$ You know all variables.

Step-by-step example

As an example, consider this system of equations:

$\begin{array}{ll}I)& 2x+3y-z=5\\ II)& x+y+z=6\\ III)& -3x-4y+3z=-5\end{array}$

1. reduce to a system of equations with two variables

Now try to eliminate all $x$ in equation $II$ and $III$ by adding them to equation $I$. This way you get two new equations containing only the variables $y$ and $z$. The first equation contains only two variables. This will give you two new equations containing only the variables $y$ and $z$. (You can, of course, use any other variable you like. And course, you can eliminate any other variable in any other equation).

1a) First time addition method

$\begin{array}{lrl}I)& 2x+3y-z& =5\\ II)& x+y+z& =6& |\cdot (-2)\end{array}$

Multiply equation $II$ by $-2$ so that when added to equation $I$, the $x$'s are omitted.

$\begin{array}{llrl}& I)& \phantom{-}2x+3y-z& =5\\ +& II)& -2x-2y-2z& =-12\end{array}$

Use the addition method: Calculate $I+II.$ We may call the resulating equation $A$.

$\begin{array}{llrl}& I)& \phantom{-}2x+3y-z& =5\\ +& II)& -2x-2y-2z& =-12\\ & A)& \phantom{-2x+}y-3z& =-7\end{array}$

1b) Second time addition method

$\begin{array}{lrlr}I)& 2x+3y-z& =5& |\cdot 3\\ III)& -3x-4y+3z& =-5& |\cdot 2\end{array}$

To eliminate all $x$ again, multiply equation $I$ by $3$ and equation $II$ by $2$ to get the same coefficient in front of the $x$. The opposite sign is the prerequisite for the addition method.

$\begin{array}{clrl}& I)& 6x+9y-3z& =15\\ +& III)& -6x-8y+6z& =-10\end{array}$

Use the addition method: Calculate $I+III$. We may call the resulating equation $B$.

$\begin{array}{clrl}& I)& 6x+9y-3z& =15\\ +& III)& -6x-8y+6z& =-10\\ & B)& y+3z& =5\end{array}$

The equations $A$ and $B$ form a system of equations with two unknowns:

$\begin{array}{lrl}A)& y-3z& =-7\\ B)& y+3z& =5\end{array}$

2. Solve the system of equations with two variables

In this article you will use the addition method again to eliminate the variable $z$. Of course you can use any other solution method. Of course, you can use any other solution method or eliminate $y$ as well.

2a) Find the first variable.

Note that the addition method is used throughout this article. You can also solve the system of equations with any other method!

Since both equations contain $3z$ with different signs, you can start directly with the addition method and calculate $A+B$ to eliminate the variable $y$.

$\begin{array}{lrl}A)& y-3z& =-7\\ B)& y+3z& =5\end{array}$

Now solve the resulting equation for $y$.

$\begin{array}{lrl}A)& y-3z& =-7\\ B)& y+3z& =5\\ & 2y\phantom{+3z}& =-2\end{array}$

Divide by 2.

$2y=-2|:2$

$y=-1$

You have found out the first variable!

2b) Find the second variable

Use the system of equations with two variables and your result $y=-1$ to find $z$. (You can substitute this into equation $A$ as well as into equation $B$.)

$y=-1$

Substitute into equation $A$.

Solve for $z$ . First, add $1$.

$-1-3z=-7$ $|+1$

Divide by $-3$.

$-3z=-6$ $|:(-3)$

$z=2$

You have now determined two of the three variables. Return to the original system of equations.

3. Determine the last variable

With $y=-1$ and $z=2$ you have two of the three variables. To find the last variable, you can plug $y$ and $z$ into each of the three equations $I$,$II$ and $III$. Here we plug it into equation $II$.

Plug in both unknowns.

$x-1+2=6$

Calculate on the left side.

$x+1=6$

Subtract $1$.

$x=5$

You have determined all three unknowns! The solution set is $𝕃=\{5;-1;2\}$.