If Bobby and Garnetta thought og four consecutive odd integers such that 5 times the sum of the first and the third was 22 greater than the product of 8 and the sum of the second and the fourth. Find the numbers.
2007-07-23 09:41:52 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
the difference between one odd integer to the next odd integer is 2
so, let x be the first odd integer
x + 2 is the the second integer
x + 4 is the third integer
and x + 6 is the fourth integer
5 (x + x + 4) = 8(x + 2 + x + 6) + 22
5(2x + 4) = 8(2x + 8) + 22
10x + 20 = 16x + 64 + 22
10x + 20 = 16x + 86
-6x + 20 = 86
-6x = 66
x = -11
so, the four integers are: -11, -9, -7 and -5
2007-07-23 09:50:04 · answer #1 · answered by 7 · 3⤊ 0⤋
The first is 2x+1.
The other 4 are 2x+3, 2x+5, and 2x+7.
5 (2x+1 + 2x+5) - 22 = 8 * (2x+3+2x+7)
5 (4x+6) - 22 = 8 (4x + 10)
20x + 30 - 22 = 32x + 80
8 = 12x + 80.
12x = -72
x = -6
So the numbers would be -11, -9. -7, and -5.
2007-07-23 16:53:20 · answer #2 · answered by Anonymous · 0⤊ 0⤋
i, (1+2), (i +4), (i + 6) are the integers
5 * (i + (i + 4)) = 22 + (8 * (i + 2)+ (i+6))
10i + 20 = 22 + 8 * (2i + 8)
10i + 20 = 16i + 86
6i = -66
i = -11
The numbers are -11, -9, -7, -5
5 * (-11 - 7) = -90
22 + 8 * (-9 - 4) = -90
QED
2007-07-23 16:52:17 · answer #3 · answered by gebobs 6 · 1⤊ 0⤋
-3
-1
1
3
maybe
2007-07-23 16:51:42 · answer #4 · answered by Anonymous · 0⤊ 0⤋