Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
148
Graphing Linear Systems
Graph two lines at once
8.1 Explain why the solution to a system of linear equations in two variables is the
coordinate at which the graphs of the equations intersect.
The graph of a linear equation in two variables is a line in the coordinate plane.
Each of the points (x,y) on that line, when substituted into the equation, results
in a true statement. The solution to a system of linear equations consists of the
coordinate pair (or coordinate pairs) that make all the equations in a system
true.
Graphing a system of equations in two variables on the same coordinate plane
produces multiple lines. If all the lines intersect at a single point (a,b), then x = a
and y = b satisfy each of the equations in the system and, therefore, (x,y) = (a,b)
is the solution to the system.
8.2 Solve the following system of equations graphically.
Graph each equation of the system on the same coordinate plane. The graph
of x = –4 is a vertical line four units left of the y-axis, and the graph of y = 1 is a
horizontal line one unit above the x-axis, as illustrated by Figure 8-1.
Figure 8-1: The graphs of x = –4 and y = 1 intersect at the point (–4,1).
Because the graphs in Figure 8-1 intersect at the point (–4,1), the solution to
the system is (x,y) = (–4,1).
For example,
the graph of
x + y = 4 passes
through the point (3,1),
because plugging
x = 3 and y = 1 into the
equation gives you the
true statement
3 + 1 = 4.
If all
the lines
intersect
at (a,b), then
every line passes
through (a,b), and
(x,y) = (a,b) is a
solution of each
equation. Hence,
it is also a
solution to the
system.
You could also
write the solution like
this: x = –4, y = 1.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
149
8.3 Solve the below system of equations graphically and verify the result.
Both equations of the system are in slope-intercept form. Graph them using the
technique described in Problems 6.21–6.26 and illustrated in Figure 8-2.
Figure 8-2: The graphs of y = x + 3 and y = –2x – 6, the equations of the system.
The graphs appear to intersect at the point (x,y) = (–3,0). Substitute x = –3 and
y = 0 into both equations to verify that the coordinate pair satisfies each and,
therefore, is the solution to the system.
Because the coordinate pair (x,y) = (–3,0) is a solution to both equations of the
system, it is the solution to the system of equations.
The y-intercept
of the rst equa-
tion is 3. The slope is
, so go up 1 and right 1
to reach another point on
the line: (1,4). The second
equation has y-intercept
–6. Its slope is
, so go
down 2 and right 1 (or up
2 and left 1) to get
to another point on
that line.
This is
what stinks
about solving
a system
graphically. It
LOOKS like the lines
intersect at
(–3,0), but looks
aren’t everything.
You have to be
SURE x = –3 and
y = 0 make both
equations into true
statements when
you substitute
them in.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
150
8.4 Solve the below system of equations graphically and verify the result.
Graph the equations of the system, as illustrated in Figure 8-3.
Figure 8-3: The graphs of y = 3x – 7 and , the equations of the system.
The lines in Figure 8-3 appear to intersect at (2,–1). Verify that (x,y) = (2,–1) is
the solution to the system by substituting x = 2 and y = –1 into the equations.
Because (x,y) = (2,–1) is a solution to both equations of the system, it is the
solution to the system of equations.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
151
8.5 Solve the below system of equations graphically and verify the result.
Graph the equations by plotting the x- and y-intercepts of each, as explained in
Problems 5.20–5.25. The equation x – 2y = 12 has x-intercept 12 and y-intercept
–6. The equation 5x + 3y = 8 has x-intercept and y-intercept
, as illustrated in Figure 8-4.
Figure 8-4: The graphs of x – 2y = 12 and 5x + 3y = 8, the equations of the system.
Verify that (x,y) = (4,–4) is the intersection point of the lines and the solution to
the system by substituting the coordinate pair into the equations of the system.
Plug x = 0
in to see what
the y-intercept
is and plug y = 0 in
to see what the x-
intercept is. You don’t
HAVE to graph this
way—you could always
solve for y and graph
using slope-intercept
form—but using
intercepts is
faster.
RULE
OF THUMB:
The intercepts
of 5x + 3y = 8
are ugly fractions,
which are hard to
graph accurately by
hand. Inaccurate
graphs lead to
inaccurate solutions,
so solving systems
by graphing works
best when fractions
aren’t involved in
the graph or the
solution.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
152
8.6 Describe the solution to the below system of equations.
Solve the equation 3x + 6y = –6 for y to express it in slope-intercept form.
Notice that this equation is exactly the same as the other equation in the
system. Two lines with the same slope and the same y-intercept also share the
same graph. Therefore, the graphs of and 3x + 6y = –6 overlap,
intersecting at every point along the shared line and result in an infinitely large
solution set that consists of all the points on the line.
Systems of equations that have infinitely many solutions are described as
“dependent.”
8.7 Describe the solution to the below system of equations.
Solve the equation x – 3y = –9 for y to express it in slope-intercept form.
Both of the equations in the system have slope , so the lines are parallel, as
illustrated by Figure 8-5.
The equations
in a dependent
system have the
same graph. Usually
that means you can
start with one of the
equations and turn it
into the other one. In
this case, solving the
second equation for y
turns it into the rst
equation.
Parallel
lines have
slopes that are
equal. Perpendicular
lines have slopes
that are opposite
reciprocals.
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