SYSTEMS OF EQUATIONS in THREE VARIABLES

It is often desirable or even necessary to use more than one variable to model a situation in a field such as business, science, psychology, engineering, education, and sociology, to name a few. When this is the case, we write and solve a system of equations in order to answer questions about the situation.



If a system of linear equations has at least one solution, it is consistent. If the system has no solutions, it is inconsistent. If the system has an infinity number of solutions, it is dependent. Otherwise it is independent.



A linear equation in three variables is an equation equivalent to the equation

\begin{eqnarray*}&& \\
Ax+By+Cz+D &=&0 \\
&&
\end{eqnarray*}


where A, B, C, and D are real numbers and A, B, C, and D are not all 0. This is the equation of a plane.




Example 5:


Joe Avalos has taken three exams in precalculus this semester. His average grade is 84. The highest grade is 18 points more than the average of the two lowers grades, and his lowest grade is 15 points less than the average of the two higher grades. What did he make on each exam?



There are three unknowns:

\begin{eqnarray*}&& \\
1 &:&\text{ The lowest grade} \\
2 &:&\text{ The middle grade} \\
3 &:&\text{ The highest grade} \\
&&
\end{eqnarray*}


Let's rewrite the paragraph that asks the question we are to answer.


The first sentence can be rewritten as

\begin{eqnarray*}&& \\
\frac{\lbrack \text{The lowest grade}]+[\text{The middle grade}]+[\text{The
highest grade}]}{3} &=&84 \\
&&
\end{eqnarray*}


The second sentence can be rewritten as

\begin{eqnarray*}&& \\
\lbrack \text{The highest grade}]-\frac{[\text{The lowest grade}]+[\text{The
middle grade}]}{2} &=&18 \\
&&
\end{eqnarray*}


The third sentence can be rewritten as

\begin{eqnarray*}&& \\
\frac{\lbrack \text{The middle grade}]+[\text{The highest grade}]}{2}-[\text{%
The lowest grade}] &=&15 \\
&&
\end{eqnarray*}


.



It is going to get boring if we keep repeating the phrases

\begin{eqnarray*}&& \\
1 &:&\text{ The lowest grade} \\
2 &:&\text{ The middle grade} \\
3 &:&\text{ The highest grade} \\
&&
\end{eqnarray*}


Let's create a shortcut by letting symbols represent these phrases. Let

\begin{eqnarray*}&& \\
L &:&\text{ The lowest grade} \\
M &:&\text{ The middle grade} \\
H &:&\text{ The highest grade} \\
&&
\end{eqnarray*}


in the three sentences, and then rewrite them.



The first sentence

\begin{eqnarray*}&& \\
\frac{\lbrack \text{The lowest grade}]+[\text{The middle grade}]+[\text{The
highest grade}]}{3} &=&84 \\
&&
\end{eqnarray*}


can now be written in the algebraic form

\begin{eqnarray*}&& \\
\frac{L+M+H}{3} &=&84 \\
&& \\
L+M+H &=&252 \\
&& \\
&&
\end{eqnarray*}


The second sentence

\begin{eqnarray*}&& \\
\lbrack \text{The highest grade}]-\frac{[\text{The lowest grade}]+[\text{The
middle grade}]}{2} &=&18 \\
&&
\end{eqnarray*}


can now be written in the algebraic form

\begin{eqnarray*}&& \\
H-\frac{L+M}{2} &=&18 \\
&& \\
2H-L-M &=&36 \\
&&
\end{eqnarray*}


The third sentence

\begin{eqnarray*}&& \\
\frac{\lbrack \text{The middle grade}]+[\text{The highest grade}]}{2}-[\text{%
The lowest grade}] &=&15 \\
&&
\end{eqnarray*}


can now be written in the algebraic form

\begin{eqnarray*}&& \\
\frac{M+H}{2}-L &=&15 \\
&& \\
M+H-2L &=&30 \\
&&
\end{eqnarray*}




We have converted the problem from one described by words to one that is described by three equations.

       
L+M+H = 252 (1)
       
2H-L-M = 36 (2)
       
M+H-2L = 30 (3)
       

We are going to show you how to solve this system of equations three different ways:


1)        Substitution, 2)        Elimination 3)        Matrices




SUBSTITUTION:


The process of substitution involves several steps:


Step 1:        Solve for one of the variables in one of the equations. It makes no difference which equation and which variable you choose. Let's solve for L in equation (1).

\begin{eqnarray*}&& \\
L+M+H &=&252 \\
&& \\
L &=&-M-H+252 \\
&& \\
&&
\end{eqnarray*}


Step 2:        Substitute this value for L in equations (2) and (3). This will change equations (1) and (2) to equations in the two variables M and H. Call the changed equations (4) and (5), respectively..
       
2H-L-M = 36  
       
$\displaystyle 2H-\left( -M-H+252\right) -M$ = 36  
       
2H+M+H-252-M = 36  
       
3H-252 = 36  
       
3H = 288  
       
H = 96 (4)
       
       
$\displaystyle M+H-2\left( -M-H+252\right)$ = 30  
       
M+H+2M+2H-504 = 30  
       
3M+3H = 534  
       
M+H = 178 (5)
       
       

Step 3:        Substitute H=96 in equation (5) and solve for M.

\begin{eqnarray*}&& \\
M+H &=&178 \\
&& \\
M+96 &=&178 \\
&& \\
M &=&82 \\
&& \\
&&
\end{eqnarray*}


Step 4:        Substitute 96 for H and 82 for M in equation (1) and solve for L.

\begin{eqnarray*}&& \\
L+M+H &=&252 \\
&& \\
L+\left( 82\right) +\left( 96\right) &=&252 \\
&& \\
L &=&74 \\
&& \\
&&
\end{eqnarray*}


The solution: Joe's lowest grade was 74, his middle grade was 82, and his highest grade was 96. Step 5:        Check the solutions:

\begin{eqnarray*}&& \\
\frac{74+82+96}{3}84 &\rightarrow &Yes \\
&& \\
96-...
...+96}{2}-74 &=&15\rightarrow Yes \\
&& \\
&& \\
&& \\
&&
\end{eqnarray*}


ELIMINATION:


The process of elimination involves several steps: First you reduce three equations to two equations with two variables, and then to one equation with one variable.



Step 1:        Decide which variable you will eliminate. It makes no difference which one you choose. Let us eliminate M first..

\begin{eqnarray*}&& \\
\begin{array}{l}
(1) \\
\\
(2) \\
\\
(3)
\end...
...M & + & H & = & 30
\end{array}
\end{array}
& \\
&& \\
&&
\end{eqnarray*}


Step 2:        Add equations (1) and (2) to form equation (4).

\begin{eqnarray*}&& \\
\begin{array}{l}
(1) \\
\\
(2) \\
\\
(4)
\end...
...& & & 3H & = & 288
\end{array}
\end{array}
& \\
&& \\
&&
\end{eqnarray*}


Step 3:        Add equations (2) and (3) to form equation (5).

\begin{eqnarray*}&& \\
\begin{array}{l}
(2) \\
\\
(3) \\
\\
(5)
\end...
... & & & 3H & = & 66
\end{array}
\end{array}
& \\
&& \\
&&
\end{eqnarray*}


Step 4:        We now have two equations with two variables.



\begin{eqnarray*}(4) &:&3H=288 \\
&& \\
H &=&96 \\
&& \\
(5) &:&-3L+3H=66 \\
&& \\
-3L+3\left( 96\right) &=&66 \\
&& \\
L &=&74
\end{eqnarray*}


Step 5:        Substitute 74 for L and 96 for Hin equation (1) and solve for M.

\begin{eqnarray*}&& \\
L+H+M &=&252 \\
&& \\
74+M+96 &=&252 \\
&& \\
M &=&82 \\
&& \\
&&
\end{eqnarray*}


The solution is: $L=74,\ M=82,$ and H=96. In terms of the original problems, Joe's lowest grade was 74, his middle grade was 82, and his highest grade was 96.

MATRICES:


The process of using matrices is essentially a shortcut of the process of elimination. Each row of the matrix represents an equation and each column represents coefficients of one of the variables.


Step 1: Create a three-row by four-column matrix using coefficients and the constant of each equation.

\begin{eqnarray*}&& \\
&&
\begin{array}{r}
(1) \\
\\
(2) \\
\\
(3)
...
... \\
-2 & & 1 & & 1 & \vert & 30
\end{array}
\right] \\
&&
\end{eqnarray*}



The vertical lines in the matrix stands for the equal signs between both sides of each equation. The first column contains the coefficients of L, the second column contains the coefficients of M, the third column contains the coefficients of H, and the last column contains the constants.



We want to convert the original matrix

\begin{eqnarray*}&& \\
&&
\begin{array}{r}
(1) \\
\\
(2) \\
\\
(3)
...
... \\
-2 & & 1 & & 1 & \vert & 30
\end{array}
\right] \\
&&
\end{eqnarray*}


to the following matrix.

\begin{eqnarray*}&& \\
&&
\begin{array}{r}
(1) \\
\\
(2) \\
\\
(3)
...
... & \\
0 & & 0 & & 1 & \vert & c
\end{array}
\right] \\
&&
\end{eqnarray*}


Because then you can read the matrix as L=a, M=b, and H=c..



Step 2:        We work with column 1 first. The number 1 is already in cell 11(Row1-Col 1). Add Row 1 to Row 2 to form a new Row 2, and add 2 times Row 1 to Row 3 to form a new Row 3..

\begin{eqnarray*}&& \\
\left[ Row\ 1\right] +\left[ Row\ 2\right] &=&\left[ Ne...
...ght] +\left[ Row\ 3\right] &=&\left[ New\ Row\ 3\right] \\
&&
\end{eqnarray*}



\begin{eqnarray*}&&
\begin{array}{r}
(1) \\
\\
(2) \\
\\
(3)
\end{arr...
... \\
0 & & 3 & & 3 & \vert & 534
\end{array}
\right] \\
&&
\end{eqnarray*}


Step 3:        We will now work with column 2. We want 1 in Cell 22, and we achieve this by interchaning Row 2 and Row 3.

\begin{eqnarray*}&& \\
\left[ Row\ 2\right] &\Longleftrightarrow &\left[ Row\ 3\right] \\
&& \\
&&
\end{eqnarray*}



\begin{eqnarray*}&&
\begin{array}{r}
(1) \\
\\
(2) \\
\\
(3)
\end{arr...
...& & 0 & & 3 & \vert & 288
\end{array}
\right] \\
&& \\
&&
\end{eqnarray*}


Step 4:        Multiply Row 2 by $\frac{1}{3}$ to form a new Row 2..

\begin{eqnarray*}&& \\
\frac{1}{3}\left[ Row\ 2\right] &=&\left[ New\ Row\ 2\right] \\
&& \\
&&
\end{eqnarray*}



\begin{eqnarray*}&&
\begin{array}{r}
(1) \\
\\
(2) \\
\\
(3)
\end{arr...
... \\
0 & & 0 & & 3 & \vert & 288
\end{array}
\right] \\
&&
\end{eqnarray*}


Step 5:        Let's now manipulate the matrix so that there are zeros in Cell 12 and Cell 32. We do this by adding -1 times Row 2 to Row 1 to form a new Row 1.

\begin{eqnarray*}&& \\
-1\left[ Row\ 2\right] +\left[ Row\ 1\right] &=&\left[ New\ Row\ 1\right] \\
&&
\end{eqnarray*}



\begin{eqnarray*}&&
\begin{array}{r}
(1) \\
\\
(2) \\
\\
(3)
\end{arr...
...& & 0 & & 3 & \vert & 288
\end{array}
\right] \\
&& \\
&&
\end{eqnarray*}


Step 6:        Let's now manipulate the matrix so that there is a 1 in Cell 33. We do this by multiplying Row 3 by $\frac{1}{3}.$

\begin{eqnarray*}&& \\
\frac{1}{3}\left[ Row\ 3\right] &=&\left[ New\ Row\ 3\right] \\
&&
\end{eqnarray*}



\begin{eqnarray*}&&
\begin{array}{r}
(1) \\
\\
(2) \\
\\
(3)
\end{arr...
... & & 0 & & 1 & \vert & 96
\end{array}
\right] \\
&& \\
&&
\end{eqnarray*}


Step 7:        Let's now manipulate the matrix so that there are zeros in Cell 13 and Cell 23. We do this by adding -1 times Row 3 to Row 2 for a new Row 2.

\begin{eqnarray*}&& \\
-1\left[ Row\ 3\right] +\left[ Row\ 2\right] &=&\left[ New\ Row\ 2\right] \\
&&
\end{eqnarray*}



\begin{eqnarray*}&&
\begin{array}{r}
(1) \\
\\
(2) \\
\\
(3)
\end{arr...
... & & 0 & & 1 & \vert & 96
\end{array}
\right] \\
&& \\
&&
\end{eqnarray*}


The solution is: $L=74,\ M=82,$ and H=96. In terms of the original problems, Joe's lowest grade was 74, his middle grade was 82, and his highest grade was 96.


If you would like to work a similar example, click on Example.


If you would like to test yourself by working some problem similar to this example, click on Problem.



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