<para>This module is from<link document="col10614">Elementary Algebra</link>by Denny Burzynski and Wade Ellis, Jr.</para>This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr.
In this chapter, the emphasis is on the mechanics of equation solving, which clearly explains how to isolate a variable. The goal is to help the student feel more comfortable with solving applied problems. Ample opportunity is provided for the student to practice translating words to symbols, which is an important part of the "FiveStep Method" of solving applied problems (discussed in modules (<link document="m21980"/>) and (<link document="m21979"/>)).
Objectives of this module: be able to identify the solution of a linear equation in two variables, know that solutions to linear equations in two variables can be written as ordered pairs.
Overview
 Solutions to Linear Equations in Two Variables
 Ordered Pairs as Solutions
Solutions to linear equations in two variables
Solution to an equation in two variables
We have discovered that an equation is a mathematical way of expressing the relationship of equality between quantities. If the relationship is between two quantities, the equation will contain two variables. We say that an equation in two variables has a solution if an ordered
pair of values can be found such that when these two values are substituted into the equation a true statement results. This is illustrated when we observe some solutions to the equation
$y=2x+5$ .

$x=4,\text{\hspace{0.17em}}y=13;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{since}\text{\hspace{0.17em}}13=2(4)+5\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{true}\text{}$ .

$x=1,\text{\hspace{0.17em}}y=7;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{since}\text{\hspace{0.17em}}7=2(1)+5\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{true}\text{}$ .

$x=0,\text{\hspace{0.17em}}y=5;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{since}\text{\hspace{0.17em}}5=2(0)+5\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{true}\text{}$ .

$x=6,\text{\hspace{0.17em}}y=7;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{since}\text{\hspace{0.17em}}7=2(6)+5\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{true}\text{}$ .
Ordered pairs as solutions
It is important to keep in mind that a solution to a linear equation in two variables is an ordered pair of values, one value for each variable. A solution is not completely known until the values of
both variables are specified.
Independent and dependent variables
Recall that, in an equation, any variable whose value can be freely assigned is said to be an
independent variable. Any variable whose value is determined once the other value or values have been assigned is said to be a
dependent variable. If, in a linear equation, the independent variable is
$x$ and the dependent variable is
$y$ , and a solution to the equation is
$x=a$ and
$y=b$ , the solution is written as the
ORDERED PAIR
$(a,\text{\hspace{0.17em}}b)$
Ordered pair
In an
ordered pair ,
$(a,\text{\hspace{0.17em}}b)$ , the first component,
$a$ , gives the value of the independent variable, and the second component,
$b$ , gives the value of the dependent variable.
We can use ordered pairs to show some solutions to the equation
$y=6x7$ .
$(0,7)$ .
If
$x=0$ and
$y=7$ , we get a true statement upon substitution and computataion.
$$\begin{array}{lllll}\hfill y& =\hfill & 6x7\hfill & \hfill & \hfill \\ 7\hfill & =\hfill & 6(0)7\hfill & \hfill & \text{Is}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{correct?}\hfill \\ 7\hfill & =\hfill & 7\hfill & \text{}\hfill & \text{Yes,}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{correct}\text{.}\hfill \end{array}$$
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$(8,\text{\hspace{0.17em}}41)$ .
If
$x=8$ and
$y=41$ , we get a true statement upon substitution and computataion.
$$\begin{array}{lllll}y\hfill & =\hfill & 6x7\hfill & \hfill & \hfill \\ 41\hfill & =\hfill & 6(8)7\hfill & \hfill & \text{Is}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{correct?}\hfill \\ 41\hfill & =\hfill & 487\hfill & \hfill & \text{Is}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{correct?}\hfill \\ 41\hfill & =\hfill & 41\hfill & \text{}\hfill & \text{Yes,}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{correct}\text{.}\hfill \end{array}$$
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$(4,\text{\hspace{0.17em}}31)$ .
If
$x=4$ and
$y=31$ , we get a true statement upon substitution and computataion.
$$\begin{array}{lllll}\hfill y& =\hfill & 6x7\hfill & \hfill & \hfill \\ 31\hfill & =\hfill & 6(4)7\hfill & \hfill & \text{Is}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{correct?}\hfill \\ 31\hfill & =\hfill & 247\hfill & \hfill & \text{Is}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{correct?}\hfill \\ 31\hfill & =\hfill & 31\hfill & \text{}\hfill & \text{Yes,}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{correct}\text{.}\hfill \end{array}$$
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