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March 1, 2015

March 1, 2015

Posted by **Sarah Mae B. Oquindo** on Friday, February 12, 2010 at 10:15am.

- algebra -
**drwls**, Friday, February 12, 2010 at 10:41amI have already solved two problems like this for you. Set up the algebraic equations that describe what you know, and solve them. There will always be as many independent equations as there are unknowns.

- algebra -
**Reiny**, Friday, February 12, 2010 at 10:49amnumber of $20's ---- x

number of $5's ----- y

number of $1's ----- z

but z = x+y - 1

x+y+z = 39

x + y + x+y-1 = 39

2x + 2y = 40

x + y = 20 (#1)

also

20x + 5y + z = 194

20x + 5y + x+y-1 = 194

21x + 6y = 195

7x + 2y = 65 (#2)

double #1 and subtract fron #2

5x = 25

x = 5

then in x+y = 20 , ----> y = 15

and finally

z = x+y-1 = 19

check: is 5+15+19 = 39 ? YES

is 20(5) + 5(15) + 19 = 195 ? YEA!!

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