SOLUTION: two consecutive odd integers are such, that one-third the smaller one is greater than one seventh of the larger one by 6. find the number

Question 700895:
two consecutive odd integers are such, that one-third the smaller one is greater than one seventh of the larger one by 6. find the number
Answer by checkley79(3341) (Show Source): You can put this solution on YOUR website!
LET X & X+2 BE THE 2 ODD INTEGERS.
(X/3)=[(X+2)/7]+6
X/3=(X+2+6*7)/7
X/3=(X+2+42)/7
X/3=(X+44)/7 CROSS MULTIPLY.
7X=3(X+44)
7X=3X+132
7X-3X=132
4X=132
X=132/4
X=33 ANS. FOR THE SMALLER ODD INTEGER.
33+2=35 ANS. FOR THE LARGER ODD INTEGER.
PROOF:
33/3=(35/7)+6
11=5+6
11=11
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