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一元一次方程
1)兩個數之差為10,若小數的2倍與大數的4倍之和是70,求小數。 2)三個連續奇數之和是57，求該三個數
最佳解答:
1)兩個數之差為10,若小數的2倍與大數的4倍之和是70,求小數。 設小數是 x 大數是 x + 10 2x + 4(x+10) = 70 2x + 4x + 40 = 70 6x = 30 x = 5 2)三個連續奇數之和是57，求該三個數 設中位數為 x，則這三數是 x – 2，x 及 x + 2 則 x – 2 + x + x + 2 = 57 3x = 57 x = 19 所以這三數為 17，19及21
其他解答:
Let x be the smaller number [設小數為x] Let y be the larger number [設大數為y] y – x = 10 --------------------------1 2x + 4y = 70 -----------------------2 From (1), [從 (1),] y = x + 10 --------------------------3 Sub (3) into (2), [代(3)入(2)] 2x + 4 (x + 10) = 70 2x + 4x + 40 = 70 6x + 40 = 70 6x = 30 x = 5 ========================================================= Let the smallest odd number is x [設最小奇數為x] Then the other two odd numbers must be (x+2) and (x+4) [那麼另外兩個奇數便是(x+2)和(x+4)] x + (x+2) + (x+4) = 57 3x + 6 = 57 3x = 51 x = 17 From x = 17, [從x=17] 17+2=19 17+4=21 Therefore, these continuous odd numbers are,17,19 and 21. [所以,三個連續奇數是17,19和21]|||||1. Let the small no. be x then the large no. be (x+10) 2x + 4 (x+10) = 70 2x + 4x + 40 = 70 6x = 30 x = 5 2. Let the smallest no. is x the next no. is x+2 the largest no. is x+4 x + (x+2) + (x+4) = 57 3x + 6 = 57 3x = 51 x = 17 Hence, the smallest no. = 17 the next no. = 17+2 = 19 the largest no. = 17+2+2 = 21|||||1) 4(b+10) +2b =70 4b+40+2b=70 6b+40=70 6b=70-40 6b=30 b=30/6 b=5 2) 17+19+21=57|||||1. 兩個數之差為10,若小數的2倍與大數的4倍之和是70,求小數。 設小數為s。 2s+4(s+10)=70 2s+4s+40=70 6s=30 s=5 所以小數是5，大數是15。 2. 三個連續奇數之和是57，求該三個數 設最小的數是m。 m+(m+2)+(m+4)=57 m+m+2+m+4=57 3m+6=57 3m=51 m=17 所以這三個數分別是17﹑19及21，|||||(1)X-Y=10 2Y+4X=70 2Y+40+4Y=70 Y=5 X=15 ANS:::::::::15,,,5 (2)X+X+2+X+4=57 3X=51 X=17 AN S ::::17 19 21|||||1) 假設小數為X 2X + 4(10+X) = 70 2X + 40 + 4X = 70 6X = 30 X = 5 小數為5 2) 假設三個數為X,X+2,x+4 X + (X+2) + (X+4) = 57 3X + 6 = 57 3X = 51 X = 17 三個數為17,19,21
一元一次方程
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發問:1)兩個數之差為10,若小數的2倍與大數的4倍之和是70,求小數。 2)三個連續奇數之和是57，求該三個數
最佳解答:
1)兩個數之差為10,若小數的2倍與大數的4倍之和是70,求小數。 設小數是 x 大數是 x + 10 2x + 4(x+10) = 70 2x + 4x + 40 = 70 6x = 30 x = 5 2)三個連續奇數之和是57，求該三個數 設中位數為 x，則這三數是 x – 2，x 及 x + 2 則 x – 2 + x + x + 2 = 57 3x = 57 x = 19 所以這三數為 17，19及21
其他解答:
Let x be the smaller number [設小數為x] Let y be the larger number [設大數為y] y – x = 10 --------------------------1 2x + 4y = 70 -----------------------2 From (1), [從 (1),] y = x + 10 --------------------------3 Sub (3) into (2), [代(3)入(2)] 2x + 4 (x + 10) = 70 2x + 4x + 40 = 70 6x + 40 = 70 6x = 30 x = 5 ========================================================= Let the smallest odd number is x [設最小奇數為x] Then the other two odd numbers must be (x+2) and (x+4) [那麼另外兩個奇數便是(x+2)和(x+4)] x + (x+2) + (x+4) = 57 3x + 6 = 57 3x = 51 x = 17 From x = 17, [從x=17] 17+2=19 17+4=21 Therefore, these continuous odd numbers are,17,19 and 21. [所以,三個連續奇數是17,19和21]|||||1. Let the small no. be x then the large no. be (x+10) 2x + 4 (x+10) = 70 2x + 4x + 40 = 70 6x = 30 x = 5 2. Let the smallest no. is x the next no. is x+2 the largest no. is x+4 x + (x+2) + (x+4) = 57 3x + 6 = 57 3x = 51 x = 17 Hence, the smallest no. = 17 the next no. = 17+2 = 19 the largest no. = 17+2+2 = 21|||||1) 4(b+10) +2b =70 4b+40+2b=70 6b+40=70 6b=70-40 6b=30 b=30/6 b=5 2) 17+19+21=57|||||1. 兩個數之差為10,若小數的2倍與大數的4倍之和是70,求小數。 設小數為s。 2s+4(s+10)=70 2s+4s+40=70 6s=30 s=5 所以小數是5，大數是15。 2. 三個連續奇數之和是57，求該三個數 設最小的數是m。 m+(m+2)+(m+4)=57 m+m+2+m+4=57 3m+6=57 3m=51 m=17 所以這三個數分別是17﹑19及21，|||||(1)X-Y=10 2Y+4X=70 2Y+40+4Y=70 Y=5 X=15 ANS:::::::::15,,,5 (2)X+X+2+X+4=57 3X=51 X=17 AN S ::::17 19 21|||||1) 假設小數為X 2X + 4(10+X) = 70 2X + 40 + 4X = 70 6X = 30 X = 5 小數為5 2) 假設三個數為X,X+2,x+4 X + (X+2) + (X+4) = 57 3X + 6 = 57 3X = 51 X = 17 三個數為17,19,21
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