# Math Help - Linear Modelling People!!

1. ## Linear Modelling People!!

I am new *waves*

I badly need help with some linear modelling questions.

Here is the question:
Jessica Communications is a phone company that charges for SMS messages. For 10 messages the cost is $8. For 40 messages the cost is$14. Tarren Telephones also charge for SMS messages. For 20 messages, they charge $8. For 60 messages they charge$24.
(a) Find a linear model for each of the phone companies.
(b) Sketch both models on the same axes.

Thank you to anyone who replies!

Also I am not good at maths so if you could help by explaining it in a simpler way.

2. You must have learned that linear models have two features:

1) A constant term.
2) A variable term with only one variable.

On a graph, they are all lines.

y = m*x + b is a common form known as the Slope-Intercept Form

ALL Linear Models can be put in this form.

Are you saying you have not heard ANY of this?

"For 10 messages the cost is $8."$8 = m*(10) + b

For 40 messages the cost is $14.$14 = m*(40) + b

Use those two expressions and solve for 'm' and 'b'.

Have you any experience in solving two equations in two unknowns?

"Also I am not good at maths so if you could help by explaining it in a simpler way."

It is a very common misconception that there is a "simpler way" to do things. This rather assumes that there is a great conspiracy to make mathematics as difficult as possible. This simply is false. There is also a common tendency to decide that one is "not good at maths" based on very limited experience and generally very foolish conclusions.

1) Get over it. Forget whatever it is in your past that is telling you that you are no good. Just stop it. Right now. Never even think it again.

2) Part of the learning of mathematics is leaning to think logically. Not everyone takes to this easily. It may require work. This does not mean that you are bad at it.

3) There is no conspiracy. Some portions of mathematics are easier for some and harder for others. So what!? Name me one thing for which this is not the case? Stop being afraid. Fear is of no value, here.

3. Originally Posted by SuperMax
I am new *waves*

I badly need help with some linear modelling questions.

Here is the question:
Jessica Communications is a phone company that charges for SMS messages. For 10 messages the cost is $8. For 40 messages the cost is$14. Tarren Telephones also charge for SMS messages. For 20 messages, they charge $8. For 60 messages they charge$24.
(a) Find a linear model for each of the phone companies.
(b) Sketch both models on the same axes.

Thank you to anyone who replies!

Also I am not good at maths so if you could help by explaining it in a simpler way.
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.

4. 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.
There's been a lot of changes while you've been away, ticbol. You don't know me, for starters! And there's a cow that has made quite a splash. TPH is now a social worker and suffers fools gladly. Welcome back.

Actually, I was only joking about TPH. His axe is swinging faster and harder than ever Especially since Russia lost.

5. Originally Posted by mr fantastic
There's been a lot of changes while you've been away, ticbol. You don't know me, for starters! And there's a cow that has made quite a splash. TPH is now a social worker and suffers fools gladly. Welcome back.

Actually, I was only joking about TPH. His axe is swinging faster and harder than ever Especially since Russia lost.
Hello, mr fantastic! Thanks for the welcome.

I see from your countless green megadots that you are `1one of the many cows that have splashed wonderfully in this forum while I was pre-occupied somewhere (bird-watching, for example). Welcome to you and to the others too!

I talk a lot and I write mistakes a lot too----like, lines leaning downwards to the left (negative slope), lol-----but that's me. If you haven't scratched your head oftenly in confusion yet, you'd start doing that in your reading my solutions later on.

I may not have the chance to answer questions as often as I want because there are too many you wonderful people answering here already but I'd show my butt from time to time on the easier ones.

TPH, he-he. Him and Captain Black were my good friends before. We traded jokes back then. The Russians lost to Turkey? I'm sorry, I'm into NBA and NFL only.