Question 117496:
the sum of the recipricals of two consecutive integers is 7/12. find the two intergers Found 2 solutions by edjones, checkley71:Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! Let the integers be x and x+1
1/x + 1/(x+1) = 7/12
12(x+1)+12x=7x(x+1) Multiply each side by 12x(x+1) to eliminate fractions.
12x+12+12x=7x^2+7x
7x^2-17x-12=0
7*-12=-84 What factors of -84 when added equal -17? Ans.: -21, 4
7x^2-21x+4x-12
7x(x-3)+4(x-3) Factor by grouping.
(7x+4)(x-3)=0
x=3 only answer that is an integer.
3, 4 Ans.
.
Check:
1/3 + 1/4
=4/12 + 3/12
=7/12
.
Ed
You can put this solution on YOUR website! let the integers be x & (x+1).
(1/x)+[1/(x+1)]=7/12 find a common denominator and combine these fractions on the left. x(x+1)
(x+1+x)/x(x+1)=7/12
(2x+1)/(x^2+x)=7/12 now cross multiply
7(x^2+x)=12(2x+1)
7x^2+7x=24x+12
7x^2+7x-24x-12=0
7x^2-17x-12=0
(7x+4)(x-3)=0
x-3=0
x=3 answer
so the integers are 3 & 4
proof
1/3+1/4=7/12
(4+3)/12=7/12
7/12=7/12
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