If two consecutive integers such that 4 times the larger exceeds 3 times the smaller by 23.
Please show me algebreically and please show me how to do it!
2006-12-10 07:27:15 · 7 answers · asked by tennis_babe 1 in Science & Mathematics ➔ Mathematics
Consecutive means one is x and one is x+1. That's all you need to know to do these.
4 times the larger
4(x + 1)
exceeds 3 times the smaller
3x
by 23
4(x+1) = 3x + 23
4x + 4 = 3x + 23
x = 19
x + 1 = 20
2006-12-10 07:29:29 · answer #1 · answered by Jim Burnell 6 · 1⤊ 0⤋
well, since they're consecutive, the two numbers can be x and x+1.
then according to your thing, 4(x+1) = 3x + 23
4x + 4 = 3x + 23
x = 19
so the smaller number is 19, and the larger is 20.
2006-12-10 15:30:31 · answer #2 · answered by car of boat 4 · 0⤊ 0⤋
translate to math language! let x be larger
4x = 3y + 23 but y is 1 less than x (they're consecutive)
so
4x = 3(x-1) + 23
4x = 3x -3 + 23
4x - 3x = 20
x = 20, so the other number is 19
2006-12-10 15:30:56 · answer #3 · answered by alia_vahed 3 · 0⤊ 0⤋
let first integer = x
let second = x + 1
4(x+1) = 3x + 23
4x + 4 = 3x + 23
x = 19
19, 20
2006-12-10 15:29:53 · answer #4 · answered by Alex 2 · 0⤊ 0⤋
4(x+1)=23+3x
x=smaller int
x+1=next consecutive (larger)
4x+4=23+3x
x+4=23
x=19
2006-12-10 15:30:07 · answer #5 · answered by Anonymous · 0⤊ 0⤋
x & (x+1) represent two consecutive integers.
4(x+1) = 3x + 23
Give x & (x+1) as answers.
2006-12-10 15:34:21 · answer #6 · answered by S. B. 6 · 0⤊ 0⤋
(x+1)+x
4x+4=3x+23
x=19
2006-12-10 17:05:43 · answer #7 · answered by XXX 2 · 0⤊ 0⤋