Jose had a herd of cattle. Then he bought more cattle so that his total number was 4 times the original number. He was given 6 more cattle. He then sold half of the herd and finally lost 4 cattle. At this time Jose had 39 cattle left. How many cattle did he have at the start?
2006-12-03 10:11:32 · 4 answers · asked by lynda c 1 in Science & Mathematics ➔ Mathematics
The equation use to solve this problem is:
(4x+6)-((4x+6)/2)-4=39
Once you work it out, Jose originally has 20 cattle at the begining of the problem.
2006-12-03 10:17:07 · answer #1 · answered by Dom 2 · 1⤊ 0⤋
let x be the number of the original cattle
we have
(4x+6)/2 - 4 = 39 => x = 20
2006-12-03 10:29:34 · answer #2 · answered by James Chan 4 · 0⤊ 0⤋
he started with 20
2006-12-03 10:15:54 · answer #3 · answered by mastermind 4 · 1⤊ 0⤋
(4x+6)-((4x+6)/2)-4=39
(4x+6)-(2x+3)-4=39
4x+6-2x-3-4=39
2x+3-4=39
2x-1=39
2x=39+1
2x=40
x=40/2
x=20
2006-12-03 10:35:59 · answer #4 · answered by Chez 4 · 0⤊ 1⤋