Question

Solve \(\dfrac{a}{ax-1}\dfrac{b}{bx-1}\)=a+b by factorization.
please solve it step by step by explaining the detail

Answer 1

do u know the ans???

Answer 2

is der + sign between a/ax-1 nd b/bx-1

Answer 3

@ganeshie8 @adilalvi @dan815 @iambatman @abhisar @mathsolver @phi

Answer 4

@Sheraz12345 @paki @thomaster @chmvijay can you expain this

Answer 5

What symbol between a/ax-1and b/bx-1

Answer 6

\(\dfrac{a}{ax-1}+\dfrac{b}{bx-1}\)=a+b i rpeat the question again
sorry @cj49 i didn't understand what u mean

Answer 7

@ash2326 @ParthKohli @sidsiddhartha can you help me

Answer 8

@bradely what happen

Answer 9

a/(ax-1) + b/(bx-1) = -(a+b)

Multiply by (ax-1)*(bx-1)

a(bx -1) + b(ax-1) = -(a+b)(ax-1)*(bx-1)
2abx - (a + b) = -(a+b)(abx^2 - (a+b)x + 1)
2abx = -(a + b)abx^2 + x(a+b)^2
(a + b)abx^2 - (a^2 + b^2) x = 0

This is a quadratic and can be solved using the standard form for the solution:

X = [-B +/- SQRT(B^2 - 4AC)] / 2A

x = [ (a^2 + b^2) +/- SQRT((a^2 + b^2)^2) ] / [ 2(a + b)ab]
x = [ (a^2 + b^2) +/- (a^2 + b^2 ] / [ 2(a + b)ab]

x = 0
x = (a^2 + b^2) / [ (a + b)ab]

Check: x= 0 yes.
Check: x = (a^2 + b^2) / [ (a + b)ab]

a/(ax - 1) = a/[(a^2 + b^2) / [ (a + b)b] - 1]
a/(ax - 1) = a/[(a^2 + b^2 - (a + b)b) / [ (a + b)b] ]
a/(ax - 1) = a/[(a^2 - ab) / [ (a + b)b] ]
a/(ax - 1) = b(a + b)/[(a - b) ]

b/(bx - 1) = b/[(a^2 + b^2) / [ (a + b)a] - 1]
b/(bx - 1) = a(a + b) / [b - a]

b(a + b)/[(a - b) ] + a(a + b) / [b - a]
(a - b)(a + b)/(b - a) = -(a + b)
-(a + b) = -(a + b)

Answer 10

http://www.wolframalpha.com/input/?i=solve++%28a%2F%28ax-1%29%29%2B%28b%2F%28bx-1%29%29%3Da%2Bb

Answer 11

in 3rd line -(a+b)(abx^2-(a+b)x+1 i don't understand @cj49

Answer 12

\((-a+b)(abx^2-(a+b)x+1\)) how inn mid a+b has come

Answer 13

from your 3rd line or equation i can't understand @cj49

Answer 14

(ax-1)(bx-1)=abx2-ax-bx+1
taking out x as a common factor in -ax-bx
abx2-(a+b)x+1

Answer 15

thank you @cj49