Algebra Know-It-ALL

270 Two-by-Two Linear Graphs

and

u= 4 t− 3

u+ 3 = 4 t

4 t=u+ 3

t= (1/4)u+ 3/4

Now we know that the t-intercepts are −7/2 and 3/4, so we can plot the points corresponding to the

ordered pairs (0,−7/2) and (0,3/4). The solution is still t= 4/11 and u=−17/11, but when we write

this as an ordered pair, it must be of the form (u,t), so we plot the intersection point as (−17/11, 4/11).

Figure 17-4 is a Cartesian graph of this situation. We’ve extended the negative (downward) t axis so we can

plot the t-intercept for the first line and have a little extra room to extend the line past the point.

t

u

Each axis

increment

is 1/2 unit

Solution =

(–17/11,4/11)

(0,–7/2)

(0,3/4)

u= 4t– 3

t= (1/4)u+ 3/4

2 t+ 5u= –7

t= (–5/2)u– 7/2

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

as a two-by-two linear system where

the independent variable is u and the

dependent variable is t. On both axes,

each increment represents 1/2 unit. The

SI forms of the equations are shown below

the originals.