Question

Here's a tricky Question! Solve this Problem Please.

If $\sf y=f(x)=\dfrac{ax-b}{bx-a}$
, then show that the f(y) = x

No Spam. Please Please Solve this. I'll mark you as Brainlest.​

Answer 1

AnswEr :

$\bigstar\:\sf \underline{Given} : y =f(x)=\dfrac{ax-b}{bx-a}$

$\rule{100}{2}$

$\textsf{Let f(y) = \sf\dfrac{ay-b}{by-a} ; As we have to find} \\ \textsf{ \quad f(y) that's why we replaced x with y.}$

$\rule{250}{1}$

• Let's Head to the Question Now :

$:\implies\sf f(y) = \dfrac{ay - b}{by - a}\\\\\qquad\scriptsize{\bf{ \dag}\: \tt{But ;\:y = \dfrac{ax - b}{bx - a}}} \\\\:\implies \sf f(y) = \dfrac{a\bigg(\dfrac{ax - b}{bx - a}\bigg) - b}{b\bigg(\dfrac{ax - b}{bx - a}\bigg) - a}\\\\\\:\implies\sf f(y) = \dfrac{\dfrac{a(ax - b)}{bx - a} - b}{\dfrac{b(ax - b)}{bx - a} - a}\\\\\\:\implies\sf f(y) = \dfrac{\dfrac{a(ax - b) - b(bx - a)}{\cancel{bx - a}}}{\dfrac{b(ax - b) - a(bx - a)}{\cancel{bx - a}}}\\\\\\:\implies\sf f(y) = \dfrac{a(ax - b) - b(bx - a)}{b(ax - b) - a(bx - a)}\\\\\\:\implies\sf f(y) = \dfrac{a^2x - \cancel{ab} - b^2x + \cancel{ab}}{\cancel{abx} - b^2 - \cancel{abx} + a^2}\\\\\\:\implies\sf f(y) = \dfrac{a^2x - b^2x}{a^2 - b^2}\\\\\\:\implies\sf f(y) = \dfrac{x\cancel{(a^2 - b^2)}}{\cancel{a^2 - b^2}}\\\\\\:\implies\large\boxed{\sf f(y) = x}$

Answer 2

Question :--- if y = f(x) = (ax-b)/(bx-a) , than show that f(y) = x ?

if we prove x = (ay-b)/(by-a) than we will get our Required answer .

Lets try to solve it with basic Method First .

______________________________

✪ Solution (1) ✪

it is given that y = (ax-b)/(bx-a) ,

To prove :--- x = (ay-b)/(by-a)

Putting value of y From Given , in Question we get :--

→ (ay - b) / (by -a)

→ [a{(ax-b)/(bx-a)} - b ] / [ b{(ax-b)/(bx-a)} - a ]

Taking LCM in both Numerator and Denominator now ,

→ [ {a(ax-b) - b(bx-a)} / (bx-a) ] / [{b(ax-b) - a(bx-a)} / (bx-a) ]

Now, (bx-a) will be cancel From both Denominator as its in Divide ,

we get now ,,

→ [ {a(ax-b) - b(bx-a)} ] / [{b(ax-b) - a(bx-a)} ]

→ [ a²x - ab - b²x + ab ] / [ axb - b² - axb + a² ]

→ [ ax² - b²x ] / [ a² - b² ]

Taking x common From Numerator now,

→ x [ a² - b² ] / [ a² - b²]

Now, (a² - b²) will be cancel From Num. and Deno.

→ x

So,, F(y) = x

✪✪ Hence Proved ✪✪

______________________________

Lets Try To solve it in a Easy way now,

✪ Solution (2) ✪

it is given that f(x) = y = (ax - b)/(cx - a)

☛ y = (ax - b)/(cx - a)

Cross - Multiplying This we get :-

☛ y*(cx - a) = ax - b

☛ cxy - ay = ax - b

☛ cxy- ax = ay - b

☛ x*(cy - a) = ay - b

☛ x = (ay - b)/(cy - a)

So, f(y) = x = (ay - b)/(cy - a)

✪✪ Hence Proved ✪✪