x + 4) + (x + 6) = 2[x + (x + 2)] 17
B.
(x + 2) + (x + 3) = 2[x + (x + 1)] 17
C.
(x + 3) + (x + 4) = 2[(x + 1) + (x + 2)] 17
D.
none of these
2006-11-02 06:13:42 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(x+2) + (x+3) + 17 = 2[x + (x+1)]
D. none of these
Also, the end of the question is imprecise - sum of the two (greatest or smallest)?
2006-11-02 06:20:57 · answer #1 · answered by a_blue_grey_mist 7 · 0⤊ 0⤋
None of these. If x = the smallest of the 4 consecutive intergers, then the equation is:
(x + 2) + (x + 3) + 17 = 2(x + (x + 1))
2x + 22 = 4x + 2
20 = 2x
x = 10,
so the 4 intergers are 10, 11, 12, and 13
2006-11-02 14:25:45 · answer #2 · answered by Gary H 6 · 0⤊ 0⤋
(x + 2) + (x + 3) + 17 = 2 ((x + 2) + (x + 3))
2006-11-02 14:16:13 · answer #3 · answered by DanE 7 · 0⤊ 0⤋
Let the integers be 'x' , 'x+1' , 'x+2' and 'x+3'
x + 2 + x+ 3 = 2(x+x+1) - 17
2x + 5 = 2(2x + 1) - 17
2x = 4x + 2 -17 -5
2x = 4x -20
2x - 4x = -20
-2x = -20
2x = 20
x = 20/2
x = 10
The integers are 10,11,12 and 13
2006-11-02 20:54:14 · answer #4 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
D. answer is "D". The four possibilities you've given all lack a plus sign, a +, before the 17 in the right half of the equation.
2006-11-02 14:18:11 · answer #5 · answered by Anonymous · 0⤊ 0⤋