Solving Linear Systems Graphically

This summary covers what the terms mean, how to solve a linear system by graphing, and how to tell how many solutions a system has without graphing.

Solving linear systems

What is a linear system? It is simply two equations of lines. For example:

$y=4x+3y=-2x+1$

What does it mean to solve a linear system of equations? It just means to find where the graphs of the equations cross. That is, to find the coordinates of the point where the two lines meet.

Example 1

Solve the given linear system by graphing.

$\begin{array}{rl}y& =x+1\\ 2x-y& =2\end{array}$

First, we graph each equation.

The first equation is in $y=mx+b$ form. So, graph by plotting the $y$-intercept, $b$. Then use the slope, $m$, and $\frac{\text{rise}}{\text{run}}$ to plot a few more points. Then draw the line.

The second equation has terms with both $x$ and $y$ on the left side. So, graph using the $x$- and $y$-intercepts. Use the “cover up” shortcut: complete the mini-table, plot the intercepts, and draw the line.

$x$	$y$
0	$-2$
$1$	0

The lines y = x + 1 and 2x - y = 2 cross at the marked point (3, 4)

Now, look at the graph and find where the lines meet.

The lines meet at $(3,4)$.

$\therefore$ the solution to the linear system is $(3,4)$.

Analyzing linear systems

There are three ways that two lines can meet, as shown below.

You can identify each case by analyzing the slope and $y$-intercept of each line.

Three small graphs: two lines crossing at one point, two parallel lines that never meet, and two coincident lines drawn on top of each other

	intersection	parallel	parallel, coincident
slope	different	same	same
$y$-intercept	different	different	same
solutions	one solution (lines cross at one point)	no solutions (lines do not cross)	infinitely many solutions (lines cross at infinitely many points)

Example 2

Without graphing, state whether each system of equations has one solution, no solutions, or infinitely many solutions.

a)

$\begin{array}{rl}y& =-2x+3\\ y& =-2x+1\end{array}$

The slopes are the same and the $y$-intercepts are different.

$\therefore$ no solutions

b)

$\begin{array}{rl}-x+2y& =-6\\ 2x+y& =-1\end{array}$

Solve each equation for $y$ so that you can analyze it.

$\begin{array}{rl}-x+2y& =-6\\ 2y& =x-6\\ y& =\frac{1}{2}x-\frac{6}{2}\\ y& =\frac{1}{2}x-3\end{array}\begin{array}{rl}2x+y& =-1\\ y& =-2x-1\end{array}$

The slopes are different. Now, stop! As soon as you see the slope is different, you know there is just one solution.

$\therefore$ one solution

Curriculum connection

B1.3

identify the relationship between the slopes of parallel and perpendicular lines, and use this relationship to solve related problems;

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B1.5

represent the equations of lines in different forms (e.g., $y=mx+b$, $Ax+By+C=0$, $Ax+By=D$) and translate between these forms, as appropriate for the context.

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