This is “Linear Systems with Two Variables and Their Solutions”, section 3.1 from the book Advanced Algebra (v. 1.0).
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Realworld applications are often modeled using more than one variable and more than one equation. A system of equationsA set of two or more equations with the same variables. consists of a set of two or more equations with the same variables. In this section, we will study linear systemsA set of two or more linear equations with the same variables. consisting of two linear equations each with two variables. For example,
$$\{\begin{array}{c}2x3y=0\\ 4x+2y=8\end{array}$$
A solution to a linear systemGiven a linear system with two equations and two variables, a solution is an ordered pair that satisfies both equations and corresponds to a point of intersection., or simultaneous solutionUsed when referring to a solution of a system of equations., is an ordered pair (x, y) that solves both of the equations. In this case, (3, 2) is the only solution. To check that an ordered pair is a solution, substitute the corresponding x and yvalues into each equation and then simplify to see if you obtain a true statement for both equations.
Check: (3, 2) 


Equation 1: $2x3y=0$ 
Equation 2: $4x+2y=8$ 
$\begin{array}{ccc}\hfill 2\left({3}\right)3\left({2}\right)& =& 0\hfill \\ \hfill 66& =& 0\hfill \\ \hfill 0& =& 0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\u2713}\hfill \end{array}$ 
$\begin{array}{ccc}\hfill 4\left({3}\right)+2\left({2}\right)& =& 8\hfill \\ \hfill 12+4& =& 8\hfill \\ \hfill 8& =& 8\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\u2713}\hfill \end{array}$ 
Determine whether or not (1, 0) is a solution to the system $\{\begin{array}{ccc}xy& =& 1\\ 2x+3y& =& 5\end{array}$.
Solution:
Substitute the appropriate values into both equations.
Check: (1, 0) 


Equation 1: $xy=1$ 
Equation 2: $2x+3y=5$ 
$\begin{array}{ccc}\hfill \left({1}\right)\left({0}\right)& =& 1\hfill \\ \hfill 10& =& 1\hfill \\ \hfill 1& =& 1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\u2713}\hfill \end{array}$ 
$\begin{array}{ccc}\hfill 2\left({1}\right)+3\left({0}\right)& =& 5\hfill \\ \hfill 2+0& =& 5\hfill \\ \hfill 2& =& 5\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\u2717}\hfill \end{array}$ 
Answer: Since (1, 0) does not satisfy both equations, it is not a solution.
Try this! Is (−2, 4) a solution to the system $\{\begin{array}{ccc}\hspace{0.17em}xy& =& 6\\ 2x+3y& =& 16\end{array}$?
Answer: Yes
Geometrically, a linear system consists of two lines, where a solution is a point of intersection. To illustrate this, we will graph the following linear system with a solution of (3, 2):
$$\{\begin{array}{c}2x3y=0\\ 4x+2y=8\end{array}$$
First, rewrite the equations in slopeintercept form so that we may easily graph them.
$\begin{array}{ccc}\hfill 2x3y& =& 0\hfill \\ \hfill 2x3y{2x}& =& 0{2x}\hfill \\ \hfill 3y& =& 2x\hfill \\ \hfill \frac{3y}{{3}}& =& \frac{2x}{{3}}\hfill \\ \hfill y& =& \frac{2}{3}x\hfill \end{array}$ 
$\begin{array}{ccc}\hfill 4x+2y& =& 8\hfill \\ \hfill 4x+2y{\text{\hspace{0.17em}}+4x}& =& 8{\text{\hspace{0.17em}}+4x}\hfill \\ \hfill 2y& =& 4x8\hfill \\ \hfill \frac{2y}{{2}}& =& \frac{4x8}{{2}}\hfill \\ \hfill y& =& 2x4\hfill \end{array}$ 
Next, replace these forms of the original equations in the system to obtain what is called an equivalent systemA system consisting of equivalent equations that share the same solution set.. Equivalent systems share the same solution set.
$$\begin{array}{l}{Original\hspace{0.17em}system\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}Equivalent\hspace{0.17em}system}\\ \{\begin{array}{c}2x3y=0\\ 4x+2y=8\end{array}\stackrel{\hspace{1em}\hspace{1em}}{\Rightarrow}\{\begin{array}{c}y=\frac{2}{3}x\hfill \\ y=2x4\end{array}\end{array}$$
If we graph both of the lines on the same set of axes, then we can see that the point of intersection is indeed (3, 2), the solution to the system.
To summarize, linear systems described in this section consist of two linear equations each with two variables. A solution is an ordered pair that corresponds to a point where the two lines intersect in the rectangular coordinate plane. Therefore, one way to solve linear systems is by graphing both lines on the same set of axes and determining the point where they cross. This describes the graphing methodA means of solving a system by graphing the equations on the same set of axes and determining where they intersect. for solving linear systems.
When graphing the lines, take care to choose a good scale and use a straightedge to draw the line through the points; accuracy is very important here.
Solve by graphing: $\{\begin{array}{ccc}xy& =& 4\\ 2x+y& =& 1\end{array}$.
Solution:
Rewrite the linear equations in slopeintercept form.
$\begin{array}{ccc}\hfill xy& =& 4\hfill \\ \hfill y& =& x4\hfill \\ \hfill \frac{y}{{1}}& =& \frac{x4}{{1}}\hfill \\ \hfill y& =& x+4\hfill \end{array}$ 
$\begin{array}{ccc}\hfill 2x+y& =& 1\hfill \\ \hfill y& =& 2x+1\hfill \end{array}$ 
Write the equivalent system and graph the lines on the same set of axes.
$$\{\begin{array}{c}xy=4\\ 2x+y=1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\hspace{1em}\hspace{1em}}{\Rightarrow}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\{\begin{array}{l}y=x+4\\ y=2x+1\end{array}$$
$\begin{array}{c}{Line\hspace{0.17em}1:}\hspace{0.17em}y=x+4\\ yintercept:\left(0,4\right)\\ slope:m=1=\frac{1}{1}=\frac{rise}{run}\end{array}$ 
$\begin{array}{c}{Line\hspace{0.17em}2:}y=2x+1\\ yintercept:\left(0,1\right)\\ slope:m=2=\frac{2}{1}=\frac{rise}{run}\end{array}$ 
Use the graph to estimate the point where the lines intersect and check to see if it solves the original system. In the above graph, the point of intersection appears to be (−1, 3).
Check: (−1, 3) 


Line 1: $xy=4$ 
Line 2: $2x+y=1$ 
$\begin{array}{ccc}\hfill \left({1}\right)\left({3}\right)& =& 4\hfill \\ \hfill 13& =& 4\hfill \\ \hfill 4& =& 4\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\u2713}\hfill \end{array}$ 
$\begin{array}{ccc}\hfill 2\left({1}\right)+\left({3}\right)& =& 1\hfill \\ \hfill 2+3& =& 1\hfill \\ \hfill 1& =& 1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\u2713}\hfill \end{array}$ 
Answer: (−1, 3)
Solve by graphing: $\{\begin{array}{c}2x+y=2\\ 2x+3y=18\end{array}$.
Solution:
We first solve each equation for y to obtain an equivalent system where the lines are in slopeintercept form.
$$\{\begin{array}{c}2x+y=2\\ 2x+3y=18\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\hspace{1em}\hspace{1em}}{\Rightarrow}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\{\begin{array}{l}y=2x+2\\ y=\frac{2}{3}x6\end{array}$$
Graph the lines and determine the point of intersection.
Check: (3, −4) 


$\begin{array}{ccc}\hfill 2x+y& =& 2\hfill \\ \hfill 2\left({3}\right)+\left({4}\right)& =& 2\hfill \\ \hfill 64& =& 2\hfill \\ \hfill 2& =& 2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\u2713}\hfill \end{array}$ 
$\begin{array}{ccc}\hfill 2x+3y& =& 18\hfill \\ \hfill 2\left({3}\right)+3\left({4}\right)& =& 18\hfill \\ \hfill 612& =& 18\hfill \\ \hfill 18& =& 18\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\u2713}\hfill \end{array}$ 
Answer: (3, −4)
Solve by graphing: $\{\begin{array}{c}3x+y=6\\ y=3\end{array}$.
Solution:
$\{\begin{array}{c}3x+y=6\\ y=3\end{array}\stackrel{\hspace{1em}\hspace{1em}}{\Rightarrow}\{\begin{array}{c}y=3x+6\\ y=3\hfill \end{array}$
Check: (3, −3) 


$\begin{array}{ccc}\hfill 3x+y& =& 6\hfill \\ \hfill 3\left({3}\right)+\left({3}\right)& =& 6\hfill \\ \hfill 93& =& 6\hfill \\ \hfill 6& =& 6\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\u2713}\hfill \end{array}$ 
$\begin{array}{ccc}\hfill y& =& 3\hfill \\ \hfill \left({3}\right)& =& 3\hfill \\ \hfill 3& =& 3\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\u2713}\hfill \end{array}$ 
Answer: (3, −3)
The graphing method for solving linear systems is not ideal when a solution consists of coordinates that are not integers. There will be more accurate algebraic methods in sections to come, but for now, the goal is to understand the geometry involved when solving systems. It is important to remember that the solutions to a system correspond to the point, or points, where the graphs of the equations intersect.
Try this! Solve by graphing: $\{\begin{array}{c}x+y=6\hfill \\ \hspace{0.17em}5x+2y=2\end{array}$.
Answer: (−2, 4)
A system with at least one solution is called a consistent systemA system with at least one solution.. Up to this point, all of the examples have been of consistent systems with exactly one ordered pair solution. It turns out that this is not always the case. Sometimes systems consist of two linear equations that are equivalent. If this is the case, the two lines are the same and when graphed will coincide. Hence, the solution set consists of all the points on the line. This is a dependent systemA linear system with two variables that consists of equivalent equations. It has infinitely many ordered pair solutions, denoted by $\left(x,mx+b\right)$.. Given a consistent linear system with two variables, there are two possible results:
A solution to an independent systemA linear system with two variables that has exactly one ordered pair solution. is an ordered pair (x, y). The solution to a dependent system consists of infinitely many ordered pairs (x, y). Since any line can be written in slopeintercept form,$y=mx+b$, we can express these solutions, dependent on x, as follows:
$$\begin{array}{cc}\left\{\left(x,y\right)y=mx+b\right\}& {SetNotation}\hfill \\ \left(x,mx+b\right)& {Shortened\hspace{0.17em}Form}\hfill \end{array}$$
In this text we will express all the ordered pair solutions $(x,\hspace{0.17em}y)$ in the shortened form $(x,\hspace{0.17em}\text{\hspace{0.17em}}mx+b)$, where x is any real number.
Solve by graphing: $\{\begin{array}{c}2x+3y=9\\ \hspace{0.17em}\hspace{0.17em}4x6y=18\end{array}$.
Solution:
Determine slopeintercept form for each linear equation in the system.
$\begin{array}{ccc}\hfill 2x+3y& =& 9\hfill \\ \hfill 2x+3y& =& 9\hfill \\ \hfill 3y& =& 2x9\hfill \\ \hfill y& =& \frac{2x9}{3}\hfill \\ \hfill y& =& \frac{2}{3}x3\hfill \end{array}$ 
$\begin{array}{ccc}\hfill 4x6y& =& 18\hfill \\ \hfill 4x6y& =& 18\hfill \\ \hfill 6y& =& 4x+18\hfill \\ \hfill y& =& \frac{4x+18}{6}\hfill \\ \hfill y& =& \frac{2}{3}x3\hfill \end{array}$ 
$$\{\begin{array}{c}2x+3y=9\\ \hspace{0.17em}\hspace{0.17em}4x6y=18\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\hspace{1em}\hspace{1em}}{\Rightarrow}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\{\begin{array}{l}y=\frac{2}{3}x3\\ y=\frac{2}{3}x3\end{array}$$
In slopeintercept form, we can easily see that the system consists of two lines with the same slope and same yintercept. They are, in fact, the same line. And the system is dependent.
Answer: $(x,\hspace{0.17em}\text{\hspace{0.17em}}\frac{2}{3}x3\hspace{0.17em})$
In this example, it is important to notice that the two lines have the same slope and same yintercept. This tells us that the two equations are equivalent and that the simultaneous solutions are all the points on the line $y=\frac{2}{3}x3$. This is a dependent system, and the infinitely many solutions are expressed using the form $(x,\hspace{0.17em}\text{\hspace{0.17em}}mx+b)$. Other resources may express this set using set notation, {(x, y)  $y=\frac{2}{3}x3$}, which reads “the set of all ordered pairs (x, y) such that $y=\frac{2}{3}x3$.”
Sometimes the lines do not cross and there is no point of intersection. Such a system has no solution, Ø, and is called an inconsistent systemA system with no simultaneous solution..
Solve by graphing: $\{\begin{array}{ccc}2x+5y& =& 15\\ 4x+10y& =& 10\end{array}$.
Solution:
Determine slopeintercept form for each linear equation.
$\begin{array}{ccc}\hfill 2x+5y& =& 15\hfill \\ \hfill 2x+5y& =& 15\hfill \\ \hfill 5y& =& 2x15\hfill \\ \hfill y& =& \frac{2x15}{5}\hfill \\ \hfill y& =& \frac{2}{5}x3\hfill \end{array}$ 
$\begin{array}{ccc}\hfill 4x+10y& =& 10\hfill \\ \hfill 4x+10y& =& 10\hfill \\ \hfill 10y& =& 4x+10\hfill \\ \hfill y& =& \frac{4x+10}{10}\hfill \\ \hfill y& =& \frac{2}{5}x+1\hfill \end{array}$ 
$$\{\begin{array}{c}2x+5y=15\\ 4x+10y=10\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\hspace{1em}\hspace{1em}}{\Rightarrow}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\{\begin{array}{l}y=\frac{2}{5}x3\\ \hspace{0.17em}y=\frac{2}{5}x+1\end{array}$$
In slopeintercept form, we can easily see that the system consists of two lines with the same slope and different yintercepts. Therefore, the lines are parallel and will never intersect.
Answer: There is no simultaneous solution, Ø.
Try this! Solve by graphing: $\{\begin{array}{ccc}x+y& =& 1\\ 2x2y& =& 2\end{array}$.
Answer: $(x,x1)$
Determine whether or not the given ordered pair is a solution to the given system.
(3, −2);
$$\{\begin{array}{c}x+y=1\\ 2x2y=2\end{array}$$(−5, 0);
$$\{\begin{array}{c}x+y=1\\ 2x2y=2\end{array}$$(−2, −6);
$$\{\begin{array}{c}x+y=4\\ 3xy=12\end{array}$$(2, −7);
$$\{\begin{array}{c}3x+2y=8\\ 5x3y=11\end{array}$$(0, −3);
$$\{\begin{array}{c}5x5y=15\\ 13x+2y=6\end{array}$$$(\frac{1}{2},\text{\hspace{0.17em}}\frac{1}{4})$;
$$\{\begin{array}{c}x+y=\frac{1}{4}\\ 2x4y=0\end{array}$$$(\frac{3}{4},\text{\hspace{0.17em}}\frac{1}{4})$;
$$\{\begin{array}{c}xy=1\\ 4x8y=5\end{array}$$(−3, 4);
$$\{\begin{array}{c}\frac{1}{3}x+\frac{1}{2}y=1\\ \frac{2}{3}x\frac{3}{2}y=8\end{array}$$(−5, −3);
$$\{\begin{array}{c}y=3\\ 5x10y=5\end{array}$$(4, 2);
$$\{\begin{array}{c}\hfill x=4\\ \hfill 7x+4y=8\end{array}$$Given the graphs, determine the simultaneous solution.
Solve by graphing.
Assuming m is nonzero solve the system: $\{\begin{array}{l}y=mx+b\\ y=mx+b\end{array}$
Assuming b is nonzero solve the system: $\{\begin{array}{l}y=mx+b\\ y=mxb\end{array}$
Find the equation of the line perpendicular to $y=2x+4$ and passing through $\left(3,3\right)$. Graph this line and the given line on the same set of axes and determine where they intersect.
Find the equation of the line perpendicular to $yx=2$ and passing through $\left(5,1\right)$. Graph this line and the given line on the same set of axes and determine where they intersect.
Find the equation of the line perpendicular to $y=5$ and passing through $\left(2,5\right)$. Graph both lines on the same set of axes.
Find the equation of the line perpendicular to the yaxis and passing through the origin.
Use the graph of $y=\frac{2}{3}x+3$ to determine the xvalue where $y=3$. Verify your answer using algebra.
Use the graph of $y=\frac{4}{5}x3$ to determine the xvalue where $y=5$. Verify your answer using algebra.
Discuss the weaknesses of the graphing method for solving systems.
Explain why the solution set to a dependent linear system is denoted by $\left(x,mx+b\right)$.
Draw a picture of a dependent linear system as well as a picture of an inconsistent linear system. What would you need to determine the equations of the lines that you have drawn?
No
No
Yes
No
Yes
(5, 0)
(6, −6)
(0, 0)
$(x,2x+2)$
$\xd8$
(−2, 3)
(3, −1)
(5, 3)
(−3, −2)
(10, 0)
$\xd8$
(3, 4)
(−3, −5)
(6, 2)
$\xd8$
$(x,\text{\hspace{0.17em}}x)$
(2, −3)
(10, 5)
(−9, 6)
$(x,\text{\hspace{0.17em}}\frac{1}{3}x9)$
(−5, 4)
(0, 0)
$\xd8$
$\xd8$
$\left(0,b\right)$
$y=\frac{1}{2}x+\frac{3}{2}$; $\left(1,2\right)$
$x=2$
$x=9$
Answer may vary
Answer may vary