# SOLUTION: the sum of the recipricals of two consecutive integers is 7/12. find the two intergers

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> SOLUTION: the sum of the recipricals of two consecutive integers is 7/12. find the two intergers      Log On

 Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Algebra: Polynomials, rational expressions and equations Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Polynomials-and-rational-expressions Question 117496: the sum of the recipricals of two consecutive integers is 7/12. find the two intergersFound 2 solutions by edjones, checkley71:Answer by edjones(7569)   (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 Answer by checkley71(8403)   (Show Source): 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