SYSTEMS OF EQUATIONS in TWO VARIABLES

A system of equations is a collection of two or more equations with the same set of unknowns. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation in the system.


The equations in the system can be linear or non-linear. This tutorial reviews systems of linear equations.


A problem can be expressed in narrative form or the problem can be expressed in algebraic form.



Let's start with an example stated in narrative form. We'll convert it to an equivalent equation in algebraic form, and then we will solve it.



Problem 2.1f:


Solve for x and y in the following system of equations.

\begin{eqnarray*}&& \\
\frac{x+3}{4}+\frac{y-1}{3} &=&1 \\
&& \\
2x-y &=&12 \\
&& \\
&&
\end{eqnarray*}




Answer: x=5 and y=-2



Solutions:


A system of linear equations can be solved four different ways:

        Substitution

        Elimination

        Matrices

        Graphing



The Method of Substitution:


The method of substitution involves several steps:



\begin{eqnarray*}&& \\
\left( 1\right) &:&\frac{x+3}{4}+\frac{y-1}{3}=1 \\
&& \\
(2) &:&2x-y=12 \\
&& \\
&&
\end{eqnarray*}


Step 1:


Solve for y in equation (2).

\begin{eqnarray*}(2) &:&2x-y=12 \\
&& \\
-y &=&12-2x \\
&& \\
y &=&2x-12 \\
&&
\end{eqnarray*}



Step 2:


Substitute this value for y in equation (1). This will change equation (1) to an equation with just one variable, x.

\begin{eqnarray*}\left( 1\right) &:&\frac{x+3}{4}+\frac{y-1}{3}=1 \\
&& \\
\...
...frac{x+3}{4}+\frac{\left( 2x-12\right) -1}{3}=1 \\
&& \\
&&
\end{eqnarray*}



Step 3:


Solve for x in the translated equation (1).

\begin{eqnarray*}&& \\
\frac{x+3}{4}+\frac{\left( 2x-12\right) -1}{3} &=&1 \\ ...
...=&12 \\
&& \\
11x &=&55 \\
&& \\
x &=&5 \\
&& \\
&&
\end{eqnarray*}



Step 4:


Substitute this value of x in equation (2) and solve for y.

\begin{eqnarray*}&& \\
(2) &:&2\left( 5\right) -y=12 \\
&& \\
10-y &=&12 \\
&& \\
y &=&-2 \\
&& \\
&& \\
&&
\end{eqnarray*}


Step 5:


Check your answers by substituting the values of x and y in each of the original equations. If, after the substitution, the left side of the equation equals the right side of the equation, you know that your answers are correct.




\begin{eqnarray*}&& \\
\left( 1\right) &:&\frac{\left( 5\right) +3}{4}+\frac{\...
...
(2) &:&2\left( 5\right) -\left( -2\right) =12 \\
&& \\
&&
\end{eqnarray*}


The Method of Elimination:


The process of substitution involves several steps:


In a two-variable problem rewrite the equations so that when the equations are added, one of the variables is eliminated, and then solve for the remaining variable.




\begin{eqnarray*}&& \\
\left( 1\right) &:&\frac{x+3}{4}+\frac{y-1}{3}=1 \\
&& \\
(2) &:&2x-y=12 \\
&& \\
&&
\end{eqnarray*}


Step 1:


Simplify equation $\left( 1\right) $.

\begin{eqnarray*}&& \\
\frac{x+3}{4}+\frac{y-1}{3} &=&1 \\
&& \\
12\left[ ...
...&& \\
3x+9+4y-4 &=&12 \\
&& \\
3x+4y &=&7 \\
&& \\
&&
\end{eqnarray*}



\begin{eqnarray*}&& \\
\left( 1\right) &:&3x+4y=7 \\
&& \\
(2) &:&2x-y=12 \\
&& \\
&&
\end{eqnarray*}


Multiply equation (2) by 4 and add it to equation (1) to form equation (3) with just one variable.

\begin{eqnarray*}&& \\
\left( 1\right) &:&3x+4y=7 \\
&&+ \\
(2) &:&8x-4y=48 \\
&& \\
(3) &:&11x=55 \\
&&
\end{eqnarray*}


Step 2:


Solve for x.

\begin{eqnarray*}&& \\
11x &=&55 \\
&& \\
x &=&5 \\
&& \\
&&
\end{eqnarray*}


Step 2:


Substitute x=5 in equation (1) and solve for y.

\begin{eqnarray*}&& \\
\left( 1\right) &:&3x+4y=7 \\
&& \\
3\left( 5\right) +4y &=&7 \\
&& \\
4y &=&-8 \\
&& \\
y &=&-2 \\
&&
\end{eqnarray*}


Step 3:


Check your answers.





The Method of Matrices:


This method is essentially a shortcut for the method of elimination.


Rewrite equations (1) and (2) without the variables and operators. The left column contains the coefficients of the x's, the middle column contains the coefficients of the y's, and the right column contains the constants.



$
\begin{array}{r}
(1) \\
\\
(2)
\end{array}
\left[
\begin{array}{rrrr...
... 7 \\
& & & & \vert & & \\
2 & & -1 & & \vert & & 12
\end{array}
\right] $





The objective is to reorganize the original matrix into one that looks like


$
\begin{array}{r}
(1) \\
\\
(2)
\end{array}
\left[
\begin{array}{rrrr...
...
& & & & \vert & & \\
0 & & 1 & & \vert & & b
\end{array}
\right] \bigskip $
where a and b are the solutions to the system.





Step 1.


Manipulate the matrix so that the number in cell 11 (row 1-col 1) is 1. To accomplish this, multiply Row 1 by $\frac{1}{3}.$



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


$
\begin{array}{r}
(1) \\
\\
(2)
\end{array}
\left[
\begin{array}{rrrr...
...3} \\
& & & & \vert & & \\
2 & & -1 & & \vert & & 12
\end{array}
\right] $


Step 2:


Manipulate the matrix so that the number is cell 21 (row 2-col 1) is 0. Do this by adding -2 times Row 1 to Row 2 to form a new Row 2.

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


$
\begin{array}{r}
(1) \\
\\
(2)
\end{array}
\left[
\begin{array}{rrrr...
... & & \\
0 & & -\frac{11}{3} & & \vert & & \frac{22}{3}
\end{array}
\right] $




Step 3:


Manipulate the matrix so that the cell 22 is 1. Do this by multiplying row 2 by $-\frac{3}{11}$.



$
\begin{array}{r}
(1) \\
\\
(2)
\end{array}
\left[
\begin{array}{rrrr...
...{3} \\
& & & & \vert & & \\
0 & & 1 & & \vert & & -2
\end{array}
\right] $




Step 4: Manipulate the matrix so that cell 12 is 0. Do this by adding

\begin{eqnarray*}&& \\
-\frac{4}{3}[Row\ 2]+[Row\ 1] &=&[New\ Row\ 1] \\
&&
\end{eqnarray*}



$
\begin{array}{r}
(1) \\
\\
(2)
\end{array}
\left[
\begin{array}{rrrr...
...& 5 \\
& & & & \vert & & \\
0 & & 1 & & \vert & & -2
\end{array}
\right] $




You can read the answers off the matrix as x=5, and y=-2.






The method of Graphing:


In this method solve for y in each equation and graph both. The point of intersection is the solution.


If you would like to go back to the problem page, click on Problem.

If you would like to return to the beginning of the two by two system of equations, click on Example.




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Author: Nancy Marcus

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