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Mathematics 14 Online
OpenStudy (flvskidd):

HELP Rita's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Rita $5.70 per pound, and type B coffee costs $4.15 per pound. This month, Rita made 148 pounds of the blend, for a total cost of $711.85 . How many pounds of type B coffee did she use?

OpenStudy (flvskidd):

@SolomonZelman

OpenStudy (flvskidd):

@phi

OpenStudy (flvskidd):

@sweetburger

OpenStudy (photon336):

you may need to make a system of equations for this.

OpenStudy (photon336):

one of our functions may look like this: \[Cost = No.Pounds~Coffee*(cost~per~pound)\]

OpenStudy (photon336):

follow?

OpenStudy (flvskidd):

@photon336 yes one second

OpenStudy (flvskidd):

the question changed but its the same idea

OpenStudy (flvskidd):

Isabel's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Isabel $4.65 per pound, and type B coffee costs $5.70 per pound. This month's blend used twice as many pounds of type B coffee as type A, for a total cost of $433.35 . How many pounds of type A coffee were used?

OpenStudy (photon336):

alright well we know that the total cost is 433.35

OpenStudy (flvskidd):

yes

OpenStudy (photon336):

here's what I'm thinking: we find a way to represent the total pounds of coffee for A. if each pound costs 4.65 then we can represent it like this. \[Coffee~A = 4.65(A) = f(A) ~~ A = no.Pounds \]

OpenStudy (photon336):

this will give us the total cost for coffee A.

OpenStudy (flvskidd):

I fill 4.65 and 5.70 into the equation

OpenStudy (flvskidd):

and 433.35

OpenStudy (flvskidd):

5.7a + 4.65(2a) = 433.35

OpenStudy (flvskidd):

ohh right

OpenStudy (photon336):

@FLVSkidd I'm reading the wrong question it should be now: it says, twice as many pounds of coffee B than type A \[4.65(A)+5.70(B) = 433.35\]

OpenStudy (photon336):

Then we're told that the number of pounds of coffee b are twice the amount as a. B = 2A

OpenStudy (photon336):

\[E_{1} | $4.65(A)+$5.70(B) = 433.35\] \[E_{2}| B = 2A\]

OpenStudy (photon336):

now, I agree with you. the second equation tells us that B = 2A where we see B substitute 2A

OpenStudy (photon336):

\[$4.65(A)+$5.70(2A) = $433.35 \] Now you have an equation of one variable which you can easily solve. for A. this will give you the number of pounds of A. then you can find the number of pounds of B by just taking what you got by A and substituting it into \[E_{2}|~~~ B = 2A\]

OpenStudy (flvskidd):

27 lbs

OpenStudy (flvskidd):

for a

OpenStudy (photon336):

what is B then?

OpenStudy (photon336):

if your answer is right when you plug in both A and B into E1 written above you should get 433.35

OpenStudy (flvskidd):

Where do I plug it in?

OpenStudy (photon336):

\[4.65(A)+5.70(B) = $433.35\]

OpenStudy (flvskidd):

is it 60?

OpenStudy (photon336):

do this first you got A = 27 right so you can easily find B. B = 2A

OpenStudy (flvskidd):

54lbs

OpenStudy (photon336):

yep now

OpenStudy (photon336):

plug both your answers into \[4.65(A)+5.70(B) = 433.35 \]

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