# Math Help - Number and alegbra

1. ## 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?

2. Hello

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

3. 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

4. Originally Posted by econlover
Thanks for the above explanation Rapha.
You're welcome

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

5. Ah...thanks Rapha! I finally got it !