Menu Close

solve-for-x-a-ax-1-b-bx-1-a-b-

Tinku Tara 0 Comments
Pin




Question Number 698 by 9999 last updated on 01/Mar/15
solve for x (a/(ax−1))+(b/(bx−1))=a+b
solveforxaax−1+bbx−1=a+b
Commented by 123456 last updated on 28/Feb/15
a=0 (b/(bx−1))=b,b=0,∀x≠(1/b) bx−1=1 x=(2/b),b≠0 b=0 (a/(ax−1))=a,a=0,∀x≠(1/a) ax−1=1 x=(2/a),a≠0 a=b=0 0=0
a=0bbx−1=b,b=0,∀x≠1bbx−1=1x=2b,b≠0b=0aax−1=a,a=0,∀x≠1aax−1=1x=2a,a≠0a=b=00=0
Answered by prakash jain last updated on 01/Mar/15
(a/(ax−1))+(b/(bx−1))=a+b a(bx−1)+b(ax−1)=(a+b)(ax−1)(bx−1) 2abx−(a+b)=ab(a+b)x^2 −(a+b)^2 x+(a+b) ab(a+b)x^2 −2abx−(a+b)^2 x+2(a+b)=0 abx[(a+b)x−2]−(a+b)[(a+b)x−2]=0 [(a+b)x−2][abx−(a+b)]=0 x=(2/(a+b)) or x=((a+b)/(ab))
aax−1+bbx−1=a+ba(bx−1)+b(ax−1)=(a+b)(ax−1)(bx−1)2abx−(a+b)=ab(a+b)x2−(a+b)2x+(a+b)ab(a+b)x2−2abx−(a+b)2x+2(a+b)=0abx[(a+b)x−2]−(a+b)[(a+b)x−2]=0[(a+b)x−2][abx−(a+b)]=0x=2a+borx=a+bab

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *