Francisco E. answered • 08/08/14

Francisco; Civil Engineering, Math., Science, Spanish, Computers.

Morris H.

asked • 08/08/14I have the answer, just need to know how to work it.

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Francisco E. answered • 08/08/14

Tutor

5
(1)
Francisco; Civil Engineering, Math., Science, Spanish, Computers.

M1C1=M2C2+(M1-M2)C3

We know M1; C1; C2; C3; Then we can calculate M2 and M3

M3=M1-M2

3.6=0.28x+ 6.1- 0.61x

x=M2= 7.572 lbs and M3 = 2.428

M2 is the one with 28% and M3 is the one with 61%

Morris,

This is similar to the previous problem, except we now have 2 unknowns.

Let x be the quantity of silver at 28% and y the quantity at 61%

Then x*28% + y*61% = 10*36%

and x + y = 10

write x (or y) in terms of the other variable using the second equation, then substitute back into the 1st equation and solve.

Morris: for some reason it's not wanting me to save a comment to your comment so I'll ad the additional work here.

Let y = 10-x then

28*x + (10-x)*61 = 360

28*x + 610 - 61*x = 360

collecting like terms:

33*x = 250 and x = 250/33 = 7.58 lbs.

so y = 10 - x = 2.42 lbs.

check: 7.58*28 + 2.42*61 = 359.8, so our work is correct.

Stephen K.

tutor

Sure, Let y = 10-x

Then x*28 + (10-x)*61 = 360

28·x + 610 - 61·x = 360

Rearranging and collecting like terms:

33·x = 250

x = 250/33 =7.58 lbs

y = 10 - 7.58 = 2.42 lbs

Check:

7.58*28 + 2.42*61 = 212.2 + 147.6 = 359.8

Report

08/08/14

Stephen K.

tutor

I added additional work so that you can see how the numbers work out. Hope this helps.

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08/08/14

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Morris H.

_{I tried doing what you said, I think, and got answers that didn't work out. Would you mind explaining just a little more please?}08/08/14