Subtracting 15 from each side gives us
− 10 r+ 5 s=− 15
When we add 10r to each side, we get
5 s= 10 r− 15
Finally, we can divide through by 5, obtaining
s= 2 r− 3
That’s identical to the first original equation! There are infinitely many solutions here; an
infinite number of ordered pairs (r,s) will make the equation true. When a linear system
consists of two equations that look different but can both be morphed to get a single
equation, the system is said to be redundant. Some texts use the word dependent instead.
A redundant linear system always has infinitely many solutions.
Chapter 17
- In Fig. 17-9, line L passes through the points (−3, 0) and (0, 4). The y-intercept is equal
to 4. We can travel along L from (−3, 0) to (0, 4) if we move to the right by Δx= 3
units and upward by Δy= 4 units. The slope is therefore
m=Δy/Δx
= 4/3
Now that we know the slope and the y-intercept for line L, we can write its SI equation as
y= (4/3)x+ 4
- In Fig. 17-9, line M passes through (−3, 0) and (0, −2). The y-intercept is −2. We can
travel from (−3, 0) to (0, −2) if we move along the line to the right by Δx= 3 units and
upward by Δy=−2 units (the equivalent of going downward by 2 units). The slope is
therefore
m=Δy/Δx
=−2/3
We now have the slope and the y-intercept for M, so we can write its SI equation as
y= (−2/3)x− 2
- We have the equations for lines L and M from Fig. 17-9 in SI form. Together, they
constitute a two-by-two linear system:
y= (4/3)x+ 4
Chapter 17 645