we move to the right, and ramps gradually upward as we move to the left. Let’s plug in v=− 12
and calculate:
w= (−1/2)× (−12)− 5
= 6 − 5
= 1
This is in the field of view, so we can plot the point (−12, 1) as a small open circle, and draw
the line for the second equation through it and (0, −5).
Are you confused?
What happens when one of the lines in a graph has an undefined slope? How can you solve and graph a
linear system such as this, for example?
y= 2 x− 2and
x= 3You can’t reduce the second equation to SI form. Its graph is a vertical line passing through (3, 0). The
first equation has a slope of 2. Because this sloping line extends forever both ways, upward and to the right
as well as downward and to the left, it must cross the vertical line x= 3, which extends forever straight
upward and straight downward, at some point. This is a perfectly decent two-by-two linear system. For
extra credit, you can solve it and graph it.
Here’s a trick!
Look again at the system we graphed in Fig. 17-5. The original two equations are:
− 7 v+w+ 10 = 0and
4 v+ 8 w=− 40Suppose we want to draw a graph of this system with w as the independent variable and v as the dependent
variable. Instead of deriving the SI versions all over again with the variables reversed, we can do a trick with
the coordinate system and the lines drawn on it.
For a minute, imagine Fig. 17-5 as an unbreakable unit with the axes, points, and lines “all glued
together.” The positive v axis goes toward the right and the positive w axis goes upward. Now think: How
can we morph this figure so the positive w axis goes toward the right and the positive v axis goes upward?
If we can do that, we’ll end up with a graph of the two-by-two linear system with w as the independent
variable and v as the dependent variable.
Actually, this process takes only three steps. We rotate the drawing in Fig. 17-5 by 90° (one-quarter
of a turn) counterclockwise. Then we mirror the drawing left-to-right. Finally, we relabel the points by
We Renamed, We Replaced, We Can Graph 273