Question

Hi I just signed up.. pulling my hair out (since yesterday morning - sad i know) trying to figure out this linear equation;

An electricity supplier has two tariff rates for charging, and a consumer is free to choose whichever rate is desired;
RATE A; 10c/unit for lighting and 2c/unit for power
RATE B: a fixed charge of $10 per quarter, plus 3c/unit for either lighting or power.

ok, question is...
If a consumer uses 1000 units for power each quarter find the least number of units he/she would need to use for lighting in order to make RATE B the cheaper rate?
ANSWER: 286

Please

Answer 1

what does 10 c mean here

Answer 2

Hi ! sorry it means cents. charged 10 cents per unit

Answer 3

ok :)

Answer 4

Let x = # light units used
y = # power units used

Cost Rate A = .10x + .02 y
Cost Rate B = 10 + .03x + .03y

Answer 5

0.10L + .02(1000) < 10 + .03(1000)

I must be setting it up wrong....because this does not equal 256..it equals 200

Answer 6

oops..not 256...I meant does not equal 286

Answer 7

Rate B says .03 for EITHER lighting or power......so I just subbed in power

Answer 8

go with what perl said....that equals 286 when rounded

Answer 9

Thank you !! I'm still going through it ..

Answer 10

only I don't think I would put the equal sign in there because you want rate B less then...not less then or equal

Answer 11

right :)
woops

Answer 12

Let x = # light units used
y = # power units used

Cost with Rate A = .10x + .02 y
Cost with Rate B = 10 + .03x + .03y

We are given y = 1000

Cost Rate A = .10x + .02*1000
Cost Rate B = 10 + .03x + .03*(1000)


Cost Rate A = .10x + 20
Cost Rate B = .03x + 40


Solve , when is
cost rate B <= cost rate A

Solve:
.03x + 40 < .10x + 20

the solution is x > 286 , rounding up

Answer 13

You can see in the graph that rate A starts out as the cheaper rate, but 286 units of light is the tipping point. For 286 units of light or greater, rate B is cheater.

Answer 14

actually 285.714 is the tipping point

Answer 15

we can solve that inequality algebraically

Answer 16

Wow thank you so much!! I can see clearly how you have achieved this result. Thank you very much for responding. I really appreciate it, thank you! I am having trouble with all these types of question ie linear equation word problems. Every time i really struggle to form the equation. Solving the equation itself is the easy part. I spent all day yesterday and so far today dedicated to practicing similar problems online. Thats how i found this site (through 'Pauls Online Maths' webpage - doing his linear equation stuff). Anyway THANK YOU SO SO MUCH :D

Answer 17

your welcome :)

Answer 18

thats why perl is a qualified helper......very good at explaining stuff

Answer 19

I wish i could come up with the equation in the first place!

Answer 20

thats what I kept messing up on...lol. To me, thats the hardest part

Answer 21

Do you believe its a practice thing? Because if Im honest I dont think I'm getting any better at it ...
Some things are like that, I just want to know an honest answer from someone who knows

Answer 22

its practice and paying special attention to the wording of the problem

Answer 23

because some word problems are tricky and they throw info in the problem that is not even needed to solve the problem

Answer 24

heh - i hope so. Its like translating another entire language or something. i just cant see how to make what they're asking in maths terms

Answer 25

make sure you understand each step in the solution, thats how i learn for the next problem.
each problem you solve gives you tools to work on other problems.

Answer 26

thats about right...take it step by step

Answer 27

surely it is the amount of practice and attention to pay to learn them
i usually have difficulties with wording problems and that's because of lack of practiced problems i take sometime to understand what i have to do and then attack the problem
sometimes they are tricky so once you learn that trick from a problem it gets to next one
and so on :)
just put a lot of effort

Answer 28

Perl - that is golden advice, i will try to apply it the next time I dont know where to start heh. (which will no doubt be in about 3minutes - ie after I've written outy the next one to practice on)

Answer 29

believe you will find yourself masted it! just good amount of practice :)

Answer 30

I hope so xapproachesinfinity, thanks... they do my head in but hopefully soon it will start to turn around :)

Answer 31

also try to pick out 'forms' or ideas that recur in problems.
like in this problem we have, to get cost,
cost per unit * # units

cost / unit * # units , notice that the the units cancel

Answer 32

yeah right. hmmmmmm :D

Answer 33

Anything else ??? Hahaha :)

Answer 34

and the fixed charge of 10, well we just add to add that to B

Answer 35

Cost of plan A = (cost per unit light) * (# units of light ) + (cost per unit power) * (# units power )

Answer 36

yes I see now that too is a hint, to add it.
And the next line you're helping me with re the cost i can see too here that you are really breaking it all right down. Thank you for going through it with me

Answer 37

Cost of plan B = (cost per unit light) * (# units of light ) + (cost per unit power) * (# units power ) + fixed charge

Answer 38

$$ \Large{
\frac {cost }{unit } \cdot \#~ units = \frac {cost }{\cancel{unit} } \cdot \cancel{ \# ~units}
\\~\\

}
$$

Answer 39

Im re writing the question out in my workbook and dissecting it/adding all your comments and notes to try and expose the method behind extracting the ACTUAL question and FORMING THE EQUATION

Answer 40

there you go :)
then next time you will be better prepared to translate the world problem into math