the complete question:
Find three consecutive integers so that twice the first added to five times the second exceeds three times the third by 51.
2007-02-24 23:22:20 · 5 answers · asked by rolaine 1 in Science & Mathematics ➔ Mathematics
the three numbers are ((x-1),(x),(x+1))
so 2(x-1) + 5x = 3(x+1) + 51
then 2x - 2 + 5x = 3x + 54
then 4x = 56
x= 14
then the three numbers are 13,14,15
2007-02-24 23:44:37 · answer #1 · answered by A New Life 3 · 0⤊ 0⤋
There are multiple answers: any 3 consecutive integers GREATER than 13. e.g.
14,15,16
15,16,17,
...
Solve using inequalities:
2n + 5(n+1) > 3(n+2) + 51 ===> 7n +5 > 3n +57,
4n > 52 ==> n>13.
So the lst integer is any over 13.
2007-02-25 07:58:48 · answer #2 · answered by kyq 2 · 0⤊ 0⤋
13, 14, 15
2007-02-25 07:50:13 · answer #3 · answered by the_innman 1 · 0⤊ 0⤋
2x+5(x+1)=3(x+2)+51
7x+5=3x+6+51
4x=52
x=52/4=13
so the integers are 13,14,15
2007-02-25 07:45:16 · answer #4 · answered by raj 7 · 0⤊ 0⤋
The equality answers are correct.
The inequality answer is wrong.
The problem said "exceeds ... by", not "exceeds ... by AT LEAST"
2007-02-25 17:21:19 · answer #5 · answered by Curt Monash 7 · 0⤊ 0⤋