HELP WITH LINEAR EQUATIONS!! will fan and medal :)
You just purchased a cellular phone and are trying to determine to which cell phone company you will give your business. When you contacted the Talks-A-Lot Company, they were offering a monthly plan of $40 for 600 minutes and $0.35 for each minute exceeding the 600 minutes. In the Sunday paper you see an ad for the Chat-Away Company, which offers a monthly plan of $50 for 600 minutes and $0.10 for each minute exceeding the 600 minutes. How many minutes would you have to talk over and above the 600 minutes for the cost to be the same with both companies? What would be the equal cost? Set up a linear system consisting of two equations. Assume you will talk for a minimum of 600 minutes. The first equation would be for the Talks-A-Lot Company. The total cost, y, equals the base fee plus cost per minute times the number of minutes exceeding 600 minutes. The second equation would be set up just like the first, only you need to use the information for the Chat-Away Company.
I would recommend you try to get all your figures, and company names in one place to begin. Eg. Talk A-lot > $40, 600mins + $0.35/min exceeding 600 mins.
ok one sec
It's also worth noting, there is more than one way to solve this. @ganeshie8 feel free to correct me.
wait a quick question... why did you write a > sign?
I apologise, it's just a symbol I used to dictate ''associated info''
Not greater than, it might aswell have been ->
ok thanks so is this good? Talk A-lot = $40, 600mins + $0.35/min exceeding 600 min Chat away = $50, 600mins + $0.10/min exceeding 600 mins.
Yeah, that's good. We always try to take the ''wordiness'' out of the question! So, any idea where you'd go next?
" The total cost, y, equals the base fee plus cost per minute times the number of minutes exceeding 600 minutes. " this is what it says in the 2nd one, so y=40+0.10x for the first one?
and is the other one going to be y=50+0.10x?
Just a moment, OpenStudy is being slow for me :)
its ok!
My question was, what type of mathematical solution are you familiar with? Have you done matrics, substitution?
I don't know what matrics is... but I am learning the substitution method.
Ah! Ok, then I will try to guide you to the answer, rather than just give you the solution OK? :)
sounds good!
y=40+0.10x y=50+0.10x Were your answers?
yup. I think?
Well firstly, y=40+0.10x? Can you see the mistake with the x value? :)
oh, y=40+0.35x correct?
Yes, but we are also forgetting something, in the equation, can you guess what it is?
?
Our 600 minutes! :)
what do you do with the 600 min? I thought bc you're assuming its over the 600 and the 40 was covering the first 600 I think..
it says to assume its over 600 min when writing the equation
It says ''assume you will talk for a minimum of 600 minutes.'' Yeah?
so where would the 600 go in the equation if its like, in the past?
How about (x-600) ?
ok... so y=40+0.35(x-600) y=50+0.10(x-600)
I think that is the right way to go. Do you understand to this point?
yes. Thank you so much
Ok, so we have two equations, both linear as the Q states. Since both equations are Y, we can set them equal to each other, make sense?
no
wait like solve the system of equations?
Ok. So: y=40+0.35(x-600) y=50+0.10(x-600) We have two unknowns, X & Y. Since both equations are Y. We can set 40+0.35(x-600) = 50+0.10(x-600)
Now we've only got 1 unknown :)
oh, ok.
So, do you know what to do next?
solve for x.
Try it :)
40+0.35(x-600) = 50+0.10(x-600) +600 +600 _______________________________ 40+0.35x = 50+0.10x ??????????????????
40+0.35(x-600)=50+0.10(x-600) 40+0.35x-210=50+0.10x-60
what?
Multiply out the brackets first.
how?
40+0.35(x-600)=50+0.10(x-600) 0.35(x-600) = 0.35x - 210 :)
Eg.
where does the 210 come from.
0.35 x 600 Do you know how to multiply brackets?
no
sorry to say this, but I only have like 3 minutes before I have to leave. I think I'll just meet with my teacher or review the lesson. I've really apreciated your help @Asylum15
If you don't know how to multiply out brackets, I suggest you mend that. Let me know how you get on. You solve for X, and substitute the X value back into one of the equations, to solve for Y.
thanks again @Asylum15
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