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anonymous
 one year ago
Mike and his friends bought cheese wafers for $2 per packet and chocolate wafers for $1 per packet at a carnival. They spent a total of $25 to buy a total of 20 packets of wafers of the two varieties.
Part A: Write a system of equations that can be solved to find the number of packets of cheese wafers and the number of packets of chocolate wafers that Mike and his friends bought at the carnival. Define the variables used in the equations.
Part B: How many packets of chocolate wafers and cheese wafers did they buy?Explain how you got the answer and why you selected a particular me
anonymous
 one year ago
Mike and his friends bought cheese wafers for $2 per packet and chocolate wafers for $1 per packet at a carnival. They spent a total of $25 to buy a total of 20 packets of wafers of the two varieties. Part A: Write a system of equations that can be solved to find the number of packets of cheese wafers and the number of packets of chocolate wafers that Mike and his friends bought at the carnival. Define the variables used in the equations. Part B: How many packets of chocolate wafers and cheese wafers did they buy?Explain how you got the answer and why you selected a particular me

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright so system of eqations: Lets call the amount of chocolate wafer packets (c) and the amount of cheese wafer packets (h) They bought a total of 20 packets so: c + h = 20 They spent $25 dollars, and chocolate packets were $1 while cheese were $2, so: c(1) + h(2) = 25

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you get it so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright awesome :) next part, to solve for this, we just use substitution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think you should try this one out yourself before I give you any answers :P Hint: Solve for h first by making the first equation into c = bla bla bla

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you want to show me your steps and how you got there or are you in a bit of a rush to get something done?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, so show me step by step what you did and I will guide you in the right direction to the best of my abilities

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Remember, you're plugging the first equation into the second equation once you make the first one c = 20  h

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and because you have the variable c in the second equation, you just plug 20 h in as c, and therefore you only have to solve for one variable instead of two making things much easier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x+y=20 2x+5=25 then x=5 y=15

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So first thing, always use the given variables, don't change them or you have a lot bigger chance of becoming confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0try again with that in mind

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do i subtract 1 from 2 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry, I got distracted. Alright I will show you how to do it stepbystep

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0c + h = 20 c + 2h = 25 That is our system of equations, now we solve for one variable, then use that value to solve for the other variable: c + h = 20 > c = 20  h Using this, we plug it into the second equation: (20  h) + 2h = 25 > h = 5 h = 5 back into the first equation, c + 5 = 20, c = 15

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you need me to rewrite everything for you to answer the questions more easily tell me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No problem :) If you have more questions to answer just paste them in here quickly, and I will answer them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So whenever you're ready

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont have anymore, all done(:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am but im doing a "refresher course"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have to do my own final 4 tests for the second segment of Algebra II Honors however, anytime at all, just ask me and I will respond as fast as I can :)
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