Originally Posted by

**ticbol** Umm, it's been a "long" time since I last practice on Math. I want to resume from easier topics first.

Linear modelling questions are about straight line graphs or models, I understand. So we want to see such lines representing the given facts in the problem. Straight lines are either vertical, horizontal, inclined upwards to the right (positive slope), or inclined downwards to the left (negative slope).

If in your lesson/class you are are already dealing with the "y = mx +b" form of the equation ot the straight line, then follow the solution shown you by TKHunny above. If not yet, then here one of many more solutions to your problem.

In your problem there are two companies involved. Let us study first the Jessica Comms.

*For 10 messages the cost is $8. For 40 messages the cost is $14*. We draw the line at once on the x,y rectangular axes. Since the cost depends on the number of calls, we use x for the number of calls and y for the cost.

*For 10 messages the cost is $8* ....x=10, y=8.......or point (10,8).

*For 40 messages the cost is $14*..........................point (40,14).

A minimum of two points determines a straight line. So we can now plot or sketch the model, or graph, of the Jessica Comms on the x,y axes.

From that sketch we can solve for the linear model of the graphical model. One way is by getting the slope of the line from the two given points, and then by using the point-slope form of the equation of the line.

slope, m = (y2 -y1)/(x2 -x1) = (14 -8)/(40 -10) = 6/30 = 1/5

Point-slope form of the line,

(y-y1) = m(x -x1)

Using point(10,8) as point (x1,y1),

y -8 = (1/5)(x -10)

y = (1/5)x -2 +8

y = (1/5)x +6 ------------the line or linear model for Jessica's.

Check that against the second point (40,14) so that you'd be sure your line is correct.

---------------------

As for the Tarren Telephones, following our solution on Jessica Comms above, here is a brief one:

*For 20 messages, they charge $8. For 60 messages they charge $24.*

Point (20,8) an point (60,24) ..........plot those on the same x,y axes as Jessica's. And connect these two new points with a straight line. That is your graphical model for Tarren's.

Slope, m2 = (24 -8)/(60 -20) = 16/40 = 2/5

Using (20,8) as (x1,y1) in the point-slope form of the line,

(y -8) = (2/5)(x -20)

y = (2/5)x -8 +8

y = (2/5)x -----------------the linear model for Tarren's.

Check that against the other point (60,24),

24 =? (2/5)(60)

24 =? 24

Yes, so, Ok.

=====================

I tend to talk too much because I like explaining. I am re-learning while I am explaining. I am out of practice yet.