Expert: Sherman D. - 10/13/2004

Question
Find four consecutive integers such that, twice the sum of the two larger integers, exceeds three times the first, by ninety-one.

Answer
1st = x
2nd = x + 1
3rd = x + 2
4th = x + 3

2((x + 3) + (x + 2)) + 91 = 3x

2(x + 3 + x + 2) + 91 = 3x

2(2x + 5) + 91 = 3x

4x + 10 + 91 = 3x

4x + 101 = 3x

101 = -x

x = -101

the numbers are

-101, -100, -99, -98

that would make -99 and -98 the 2 larger numbers, and if you are thinking how can -99 and -98 be larger than -101 and -100, its because the first away from 0 you get on the negative side, the lower the value of that number. Also exceeds means "in addition to", so when it said exceeds 3 times the first by ninety-one, thats saying exceeds 3x by 91, or y + 91 = 3x where as y is the sum of the 2 larger numbers times 2.