Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
83
Figure 5-9: The graph of y = 2 is a horizontal line two units above the x-axis.
Graphing with a Table of Values
Plug in some x’s, plot some points, call it a day
Note: Problems 5.9–5.11 refer to the linear equation 2x – y = 4.
5.9 How many coordinates are required to draw the graph of the equation?
Geometric principles dictate that two points on the same plane are required
to draw the line that contains those points. If, however, you use coordinates to
draw a linear graph, identifying and plotting at least one additional point is
advised. The third point serves as a quick way to check your work. If it lies on
the same line as the other two points, you’re far less likely to have made an error
in your calculations.
Note: Problems 5.9–5.11 refer to the linear equation 2x – y = 4.
5.10 Use a table of values to identify three points on the graph of the line.
A table of values is a brute force arithmetic technique that generates lists of
coordinate pairs satisfying the given equation. After a sufficient number of
points has been identified and plotted, all that remains is to “connect the dots”
on the coordinate plane to graph the equation. The number of points necessary
to create an accurate graph varies based on the complexity of the equation.
I call the
third point a
“check point.” If
you make a mistake
nding either of the
other two points, the
check point won’t be
on the graph. That’s
a red ag to go
back and check
your arithmetic
for all three
coordinates.
Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
84
Because 2x – y = 4 is a linear equation in two variables, a minimum of two points
is required to graph it. However, as explained in Problem 5.9, three coordinates
should be plotted to better ensure your calculations are correct.
Begin by solving the equation for y.
Construct a table with columns labeled, from left to right, “x,” “y = 2x – 4,” and
“y.” The outside columns, x and y, will eventually house the coordinates you will
plot on the graph. The inner column contains the equation just solved for y.
Choose three values of x to plug into the equation y = 2x – 4. Select x-values that
do not produce large or unnecessarily complicated results. Write the three x-
values you chose in the left column of the table.
Substitute each x-value into the equation to determine the corresponding values
of y and record those in the right column.
Combine the x- and y-values in each row to conclude that the graph contains
the points (–1,–6), (0,–4), and (1,–2).
The ab-
solute value
graphs at the end
of the chapter are
a little trickier. You
could get away with
using only two points
to make the graph,
but you’d be showing
off. Better to use
more points as the
equations get more
complicated.
See Problems 4.37–
4.43 if you need help
with this step.
Whenever
possible, choose
simple, small numbers
like x = –1, x = 0,
and x = 1.
Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
85
Note: Problems 5.9–5.11 refer to the linear equation 2x – y = 4.
5.11 Graph the linear equation using the table of values generated by Problem 5.10.
Plot the points identified in Problem 5.10 and connect them to create the linear
graph illustrated by Figure 5-10.
–4
Figure 5-9: The graph of 2x – y = 4 passes through the points (–1,–6), (0,–4), and
(1,–2).
Note: Problems 5.12–5.13 refer to the linear equation x + 3y = –2.
5.12 Use a table of values to identify three points on the graph of the equation.
Solve the equation for y.
Create a table of values based on three simple values of x, as demonstrated in
Problem 5.10.
This time
the table of
values does not
contain x = –1 and
x = 0. If you plug in x =
1, however, the fraction
turns into an integer.
That’s why the other
two values, x = –2 and
4, are used as well—
they eliminate the
fraction.
Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
86
According to the table of values, points (–2,0), (1,–1) and (4,–2) belong to the
graph of x + 3y = –2.
Note: Problems 5.12–5.13 refer to the linear equation x + 3y = –2.
5.13 Graph the linear equation using the points generated by Problem 5.12.
Plot the points (–2,0), (1,–1) and (4,–2) on the coordinate plane and connect
them to graph the linear equation.
Figure 5-11: The graph of x + 3y = –2 passes through the points (–2,0), (1,–1) and
(4,–2).
–(–2) is a
double sign and
should be rewritten
as 2, according to
Problem 1.11. You can
also interpret –(–2) as a
multiplication problem;
imagine that the
outer “–” equals –1:
–(–2) = (–1)(–2) = 2.
Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
87
5.14 Use a table of values to graph the linear equation 2x – (y – 4x) = y + 2.
Begin by solving the equation for y.
Substitute simple values of x into the equation to generate coordinates and
graph the linear equation, as illustrated by Figure 5-12.
Figure 5-12: A table of values containing x = –1, x = 0, and x = 1 produces three
coordinates: (–1,–4), (0,–1), and (1,2). Those three points are sufficient to
graph 2x – (y – 4x) = y + 2.
5.15 Through which of the following points, if any, does the graph of
pass?
A = (–12,1); B = (3,–5); C = (24,9)
Design a table of values using the x-coordinates of each point: x = –12, x = 3, and
x = 24. Substitute each into the equation to determine whether the resulting
y-values satisfy the equation.
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