Let’s find another point on each line to make our line-drawing task easier. If we plug in
t= 3 to the SI version of the first equation, we get
u= (−2/5)t− 7/5
= (−2/5)× 3 − 7/5
=−6/5− 7/5
=−13/5
That gives us the point (3, −13/5), shown as a small open circle on the line for the first equation.
If we plug in t= 1 to the second equation, we get
u= 4 t− 3
= 4 × 1 − 3
= 4 − 3
= 1
That gives us the point (1, 1), shown as a small open circle on the line for the second equation.
Are you confused?
When we want to find well-spaced points to draw lines in situations like this, a little common sense is a
big help. We don’t want any of the points to land off the scale on either axis, but we have to get them far
enough away from the other points so we can easily draw the lines. The calculations are usually simple if
we use integers for the “plug-ins.”
Here’s a challenge!
Draw a graph of the above system with the variables transposed. That is, make u the independent variable
andt the dependent variable.
Solution
Let’s tackle this problem from scratch. To begin, we must get the two original equations into SI form with
t as the dependent variable. That will make it easy to find the t-intercepts. Then we’ll make up a coordi-
nate system with a horizontal u axis and a vertical t axis and plot the points. Finally, we’ll draw the lines
through the points.
Here are the original equations, each followed by a step-by-step manipulation to get it into SI form
witht as the dependent variable. By now, you can figure out the reasoning behind each step, so we don’t
have to drag ourselves through the justifications!
2 t+ 5 u=− 7
2 t=− 5 u− 7
t= (−5/2)u− 7/2
We Added, We Eliminated, We Can Graph 269