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How to do this q ?
發問:
How to do this q ? Please... Eric has some $2 coins and $5 coins . If the total number of coins is 12 and their total value is $39 , how many $2 coins and $5 coins does Eric have ? ( ) = 39 ? 更新: 唔該!!
最佳解答:
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.
How to do this q ?
發問:
How to do this q ? Please... Eric has some $2 coins and $5 coins . If the total number of coins is 12 and their total value is $39 , how many $2 coins and $5 coins does Eric have ? ( ) = 39 ? 更新: 唔該!!
最佳解答:
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
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其他解答: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.
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