Algebra Know-It-ALL

268 Two-by-Two Linear Graphs

When we subtract 2t from each side, we get

5 u=− 2 t− 7

We can divide through by 5 to obtain

u= (−2/5)t− 7/5

Theu-intercept for this line is −7/5, so the point (0, −7/5) is on it. Note that the ordered pairs

here are always of the form (t,u), because it’s customary to list the independent variable first and

the dependent variable after it. In Chap. 16, we found that the solution for the linear system

wast= 4/11 and u=−17/11. Therefore, the point (4/11, −17/11) is where the lines intersect.

Connect the points

We have the points we need to graph the lines. Our first line passes through (0, −3) and

(4/11,−17/11). The second line goes through (0, −7/5) and (4/11, −17/11). Unfortunately,

two of these points are so close together that it’s hard to draw the line through them precisely,

as can be seen by looking at the graph of this system (Fig. 17-3).

t

u

(0,–7/5)

(0,–3)

Solution =

2 t+ 5u= –7 (4/11,–17/11)

(3,–13/5)

u= 4t– 3

(1,1)

Each axis

increment

is 1/2 unit

Figure 17-3 Graphs of 2t+ 5 u=−7 and u= 4 t− 3

as a two-by-two linear system where

the independent variable is t and the

dependent variable is u. On both axes,

each increment represents 1/2 unit.