Question

Solve the following Quadratic Equation [Using Factorisation method]
$\orange{\bold{\large\tt\dfrac{a}{ \: ax - 1 \: } + \dfrac{b}{ \: bx - 1 \: } = a + b}} \\ \green{ \bold{ \tt a + b \neq0 } }\\ \green{ \bold{\tt \: ab \neq0}}$
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Answer 1

Answer:

Step-by-step explanation:

$\sf{\dfrac{a}{ax-1} + \dfrac{b}{bx-1} = a + b}$

Now, bringing the terms which are going to make a difference are :- a and b.

Let's see the magic brought out by them:-

$\sf{\dfrac{a}{ax-1} - b= a - \dfrac{b}{bx-1}}$

Why I did this, to cancel out the terms that will be formed in future.

$\sf{\dfrac{a - bax + b}{ax-1}= \dfrac{abx - a - b}{bx-1}}$

$\sf{(a - abx + b)(bx-1) = (ax-1)(abx - a - b)}$

Method 1 :- Cancelling the terms.

We will taken (-) common in 2nd phase :-

$\sf{(a - abx + b)(bx-1) = - (ax-1)(abx - a - b)}$

$\sf{(a - abx + b)(bx-1) = (ax-1)(- abx + a + b)}$

$\sf{(bx-1) = (ax-1)}$

$\sf{bx + ax -2 = 0}$

$\sf{x(a+b)= 2}$

$\sf{x= \dfrac{2}{a+b}}$ Answer (1)

Wait, the answer is still not complete yet!

Method 2:- Bringing common the 2 terms :-

$\sf{(a - abx + b)(bx-1) = - (ax-1)(abx - a - b)}$

$\sf{(a - abx + b)(bx-1) +(ax-1)(abx - a - b) = 0}$

[Minus turns plus on either side]

Taking a-abx+b common,

$\sf{(a - abx + b)(bx-1 + ax -1) = 0 }$

$\sf{(a - abx + b)(bx- 2 + ax ) = 0 }$

Look carefully! The second equation is similar to answer 1, so we have already found it.

We need to simplify 1st one.

1st one = 0 , 2nd one = 0

$\sf{a - abx + b = 0 }$

$\sf{a + b = abx }$ [Minus turns plus]

$\sf{\dfrac{a + b }{ab}= x }$ Answer (2)

$\sf{x= \dfrac{2}{a+b}}$ , $\sf{\dfrac{a + b }{ab}= x }$ are the answers by factorisation.

Answer 2

Topic :-

Quadratic Equation

Given :-

$\mathtt{\dfrac{a}{ax-1}+\dfrac{b}{bx-1}=a+b}$

To Find :-

Value of x.

Solution :-

$\mathtt{\dfrac{a}{ax-1}+\dfrac{b}{bx-1}=a+b}$

Transpose (a + b) in LHS,

$\mathtt{\dfrac{a}{ax-1}+\dfrac{b}{bx-1}-a-b=0}$

Rearrange it,

$\mathtt{\left( \dfrac{a}{ax-1}-b \right )+\left ( \dfrac{b}{bx-1}-a \right )=0}$

Take LCM,

$\mathtt{\left( \dfrac{a-b(ax-1)}{ax-1} \right )+\left ( \dfrac{b-a(bx-1)}{bx-1}\right )=0}$

Solve brackets,

$\mathtt{\left( \dfrac{a-abx+b}{ax-1} \right )+\left ( \dfrac{b-abx+a}{bx-1}\right )=0}$

Take (a + b -abx ) common,

$\mathtt{(a+b-abx)\left( \dfrac{1}{ax-1}+ \dfrac{1}{bx-1}\right )=0}$

Take LCM again,

$\mathtt{(a+b-abx)\left( \dfrac{bx-1+ax-1}{(ax-1)(bx-1)}\right )=0}$

$\mathtt{(a+b-abx)\left( \dfrac{bx+ax-2}{(ax-1)(bx-1)}\right )=0}$

$\mathtt{ \dfrac{(a+b-abx)(ax+bx-2)}{(ax-1)(bx-1)}=0}$

Cross Multiply,

$\mathtt{(a+b-abx)(ax+bx-2)=0(ax-1)(bx-1)}$

$\mathtt{(a+b-abx)(ax+bx-2)=0}$

So,

$\mathtt{Either\:(a+b-abx) = 0\:or\:(ax+bx-2) = 0.}$

Case 1,

$\mathtt{a+b-abx=0}$

$\mathtt{abx=a+b}$

$\mathtt{x=\dfrac{a+b}{ab}}$

Case 2,

$\mathtt{ax+bx-2=0}$

$\mathtt{(a+b)x-2=0}$

$\mathtt{(a+b)x=2}$

$\mathtt{x=\dfrac{2}{a+b}}$

Answer :-

So,

$\mathtt{x=\dfrac{a+b}{ab}\;or}$

$\mathtt{x=\dfrac{2}{a+b}}$