# Number and alegbra

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• May 5th 2009, 05:41 AM
econlover
Number and alegbra
Jane has 15 coins which include \$2-coins and \$5-coins only. If the total value of \$2-coins is less than that of \$5-coins by \$33, how many \$5-coins does Jane have?
• May 5th 2009, 05:47 AM
Rapha
Hello

Quote:

Originally Posted by econlover
Jane has 15 coins which include \$2-coins and \$5-coins only. If the total value of \$2-coins is less than that of \$5-coins by \$33, how many \$5-coins does Jane have?

let x be the amount of \$2-coins

y the amount of \$5-coins

Jane has 15 coins, therefor
x+y = 15

Suppose you have x=6 \$2-coins, then its worth is 6*2 =\$12, and you have y=9 coins (actually this is the solution) then you have 9*5 = \$45. You know the difference between the worth of all your \$2 coins and 5\$-coins is 33\$.
Hence
2x = 5y-33
(2 and 5 because of the worth of x and y)

You need to solve the equations
x+y = 15
2x = 5y-33

Regards,
Rapha
• May 5th 2009, 05:55 AM
econlover
Thanks for the above explanation Rapha. I do have the same equations with yours but I am stuck on how to take the next step through these equations
x+y=15
2x+33=5y
• May 5th 2009, 05:59 AM
Rapha
Quote:

Originally Posted by econlover
Thanks for the above explanation Rapha.

You're welcome

Quote:

Originally Posted by econlover
I do have the same equations with yours but I am stuck on how to take the next step through these equations
x+y=15
2x+33=5y

this is the first equation: x+y=15, solving for x => x = 15-y

now we know x, therefor:

2(15-y)+33 = 5y

and calculate this

30 - 2y +33 = 5y // +2y
63 = 7y // :7

63/7 = y

9 = y

Now our x depends on y, because we got x = 15-y. We know y exactly and thus

x = 15-9 = 6

Done.

Rapha
• May 5th 2009, 06:11 AM
econlover
Ah...thanks Rapha!(Clapping) I finally got it !