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~~~~~~use 方程式~~~~ 2) Sam had a number of $1 coins and $5 coins . It was knomn that half the number of $1coins was grater than the number of $5coins by 2. If the total value of the coins was $32.How many $1 coins and how many $5 coins did Sam has??

最佳解答:

Let the number of $1 coins is x, the number of $5 coins is y. half the number of $1coins was grater than the number of $5coins by 2 Therefore, (1/2)x = y + 2 -----(1) the total value of the coins was $32, Therefore, x + 5y = 32 ----(2) From (1), x = 2y + 4 ----(3) sub (3) into (2), 2y + 4 + 5y = 32 . 7y = 28 . y = 4 ----(4) sub (4) into (3), x = 2(4) + 4 . x = 12 Sam had 12 $1 coins and 4 $5 coins.

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其他解答:

let X=$1coins (X/2)-2=$5coins 1X+5[(X/2)-2]=32 X+(5X/2)-10=32 7X/2=42 7X=84 x=12 $1coins has12 $5coins has (12/2)-2 =4|||||let X=$1coins (X/2)-2=$5coins 1X+5[(X/2)-2]=32 X+(5X/2)-10=32 7X/2=42 7X=84 x=12 $1coins has12 $5coins has (12/2)-2 =4|||||let X=$1coins (X/2)-2=$5coins 1X+5[(X/2)-2]=32 X+(5X/2)-10=32 7X/2=42 7X=84 x=12 $1coins has12 $5coins has (12/2)-2 =4
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