How to do this q ?

Let the number of $2 be m & the number of $5 be y

m+y=12 -------------(1)
2(m)+ 5(y) = 39 ----(2)

by (1), m=12-y

sub into (2),
2(12-y) +5(y)=39
24-2y+5y=39
24+3y=39
3y=15
y=5

Then, m=12-5 =3
Ans. The numbers of $2 and $5 are 3 and 5 respectively

2007-11-20 21:44:00 補充：
Sorry~ Amendment is as follows:Then, m=12-5 =7Ans. The numbers of $2 and $5 are 7 and 5 respectively

2007-11-28 00:13:21 補充：
Tiffany_bb82
Many thanks for your appreciation

Let $2 coin = y, and $5 coin = z
y + z = 12.......equation 1 ---> y = 12-z
2y + 5z = 39.....equation 2
Put equation 1 into equation 2
2(12-z) + 5z = 39
24-2z+5z=39
24+3z=39
3z = 15
z=5 ---> if z = 8, y = 12-5 so y = 7
Eric has 7 $2 coin and 5 $5 coin

Let Eric have x $2 coins and (12 - x) $5 coins.
2(x) + 5(12 - x) = 39
2x + 60 - 5x = 39
-3x = -21
x = 7

Therefore, Eric has 7 $2 coins and 5 $5 coins.

If there is a mistake, please inform me.