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 2.1b:


An airplane flying into a head wind travels the 1800-mile flying distance between two cities in 3 hours and 36 minutes. On the return flight, the same distance is traveled in 3 hours. Find the ground speed of the plane and the speed of the wind, assuming that both remain constant. [Ground speed is the speed of the plane if there were no wind.]



Answer: The ground speed of the plane is 550 miles per hour and the wind speed is 50 miles per hour.



Solutions:


The relationship between distance, speed, and time is distance equals speed times time. Let the symbol d represent distance, the symbol r represent speed (or rate), and the symbol t represent the time. The relationship between the three can then be expressed algebraically.

\begin{eqnarray*}&& \\

d &=&rt \\

&& \\

&&or \\

&& \\

r &=&\frac{d}{t} \\

&& \\

&&

\end{eqnarray*}


The net speed has two components: ground speed of the plane and wind speed (velocity), where ground speed represents the speed of the plane without wind. When the plane is flying into a head wind (wind pushing against plane), the net speed of the plane is the ground speed of the plane minus the wind speed. When the plane is flying with a tail wind (wind pushing plane), the net speed of the plane is the ground speed of the plane plus the wind speed. For this problem, we will assume that the plane and the wind are in line. When you get into the wind pushing against the plane at, say a thirty-degree angle, you solve it with vectors.



When the plane's speed is helped by the wind, the net speed increases and the time to reach destination decreases. When the plane's is hindered by the wind, the net speed decreases and the time to reach destination increases. Therefore, we know that the plane had a tail wind when the time is 3 hours, and the plane had a head wind when the time is 3 hours and 36 minutes.



Let's rewrite the problem.


With head wind: distance = (plane ground speed - wind speed) $\times $ time or


1,800= (plane speed - wind speed) $\times $ 3 hours and 36 minutes




With tail wind: distance = (plane speed + wind speed) $\times $ time or


1,800= (plane speed + wind speed) $\times $ 3 hours




Let the symbol x represent the phrase plane ground speed and let the symbol y represent the phrase wind speed, convert 3 hours and 36 minutes to 3.6 hours, and rewrite the two equations in algebraic form.

x - y = 3.6
x + y = 3.0


\begin{eqnarray*}&& \\

(1) &:&1,800=\left( x-y\right) 3.6 \\

&& \\

&& \\

(2) &:&1,800=\left( x+y\right) 3 \\

&& \\

&&

\end{eqnarray*}



We have converted a narrative statement of the problem to an equivalent algebraic statement of the problem. Let's solve this system of equations.



     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:


Step 1:


Solve for x in equation (1).

\begin{eqnarray*}(1) &:&1,800=3.6x-3.6y \\

&& \\

3.6x &=&1,800+3.6y \\

&& \\

x &=&500+y

\end{eqnarray*}



Step 2:


Substitute this value for x in equation (2). This will change equation (2) to an equation with just one variable, y.

\begin{eqnarray*}(2) &:&1,800=\left( x+y\right) 3 \\

&& \\

(2) &:&1,800=\left( 500+y\right) 3+3y \\

&& \\

&&

\end{eqnarray*}



Step 3:


Solve for y in the translated equation (2).

\begin{eqnarray*}&& \\

1,800 &=&\left( 500+y\right) 3+3y \\

&& \\

1,800 &=&1500+6y \\

&& \\

300 &=&6y \\

&& \\

y &=&50 \\

&&

\end{eqnarray*}



Step 4:


Substitute this value of y in equation (1) and solve for x.

\begin{eqnarray*}&& \\

(1) &:&1,800=\left( x-y\right) 3.6 \\

&& \\

1,800 &...

...\\

&& \\

1980 &=&3.6x \\

&& \\

x &=&550 \\

&& \\

&&

\end{eqnarray*}


The ground speed of the plane was 500 miles per hour and the wind speed was 50 miles per hour.


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*}(1) &:&\left( 550-50\right) 3.6=1,800 \\

&& \\

(2) &:&\left( 550+50\right) 3=1,800 \\

&& \\

&& \\

&&

\end{eqnarray*}


The Method of Elimination:


The process of substitution involves several steps:


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




\begin{eqnarray*}&& \\

(1) &:&1,800=\left( x-y\right) 3.6 \\

&& \\

(2) &:&1,800=\left( x+y\right) 3 \\

&& \\

&&

\end{eqnarray*}


Step 1:


Multiply equation (1) by 3, multiply equation (2) by 3.6, and add the two equations to form equation (3) with just one variable.

\begin{eqnarray*}&& \\

(1) &:&3\left( 1,800\right) =3\left( 3.6x-3.6y\right) \...

...+10.8y \\

&& \\

&& \\

(3) &:&11,880=21.6x \\

&& \\

&&

\end{eqnarray*}


Step 2:


Solve for x.

\begin{eqnarray*}&& \\

11,880 &=&21.6x \\

&& \\

x &=&550 \\

&& \\

&&

\end{eqnarray*}


Step 2:


Substitute x=550 in equation (1) and solve for y.

\begin{eqnarray*}&& \\

(1) &:&1,800=\left( x-y\right) 3.6 \\

&& \\

1,800 &...

...6 \\

&& \\

1,800 &=&1,980-3.6y \\

&& \\

y &=&50 \\

&&

\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...

... \\

& & & & \vert & & \\

3 & & 3 & & \vert & & 1,800

\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. Multiply row 1 by $\frac{1}{3.6}$ to form a new row 1.


$

\begin{array}{r}

(1) \\

\\

(2)

\end{array}

\left[

\begin{array}{rrrr...

... \\

& & & & \vert & & \\

3 & & 3 & & \vert & & 1,800

\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 -3 times row 1 to row 2 to form a new row 2.

\begin{eqnarray*}&& \\

-3\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...

...00 \\

& & & & \vert & & \\

0 & & 6 & & \vert & & 300

\end{array}

\right] $




Step 3:


Manipulate the matrix so that the cell 22 is 1. Do this by multiplying row 2 by 1/6.



$

\begin{array}{r}

(1) \\

\\

(2)

\end{array}

\left[

\begin{array}{rrrr...

...500 \\

& & & & \vert & & \\

0 & & 1 & & \vert & & 50

\end{array}

\right] $




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

\begin{eqnarray*}&& \\

\lbrack Row\ 2]+[Row\ 1] &=&[New\ Row\ 1] \\

&&

\end{eqnarray*}



$

\begin{array}{r}

(1) \\

\\

(2)

\end{array}

\left[

\begin{array}{rrrr...

...550 \\

& & & & \vert & & \\

0 & & 1 & & \vert & & 50

\end{array}

\right] $




You can read the answers off the matrix as x=550, and y=50.






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 review the solution to the next problem, 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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