Question

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Answer 1

Step-by-step explanation:

★ Concept :-

Here the concept of Linear Equations in Two Variables has been used. We see that we are giveb two equations where we have two variables. We have to find these variable terms that is value of x and y. We no need to find the value of a, b and c since they are constant terms. Only the value of variables can be determined. So firstly we shall make relationships between two equations. Then we shall use any of the methods to solve both of them and find the values of each x and y seperately one - by - one.

Let's do it !!

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★ Solution :-

Given,

» ax + by = c

Let this be equation (i).

» bx - ay = 1 + c

Let this be equation (ii).

On multiplying equation (i) with a we get,

>> a(ax) + a(by) = a(c)

>> a²x + aby = ac

Let this be equation (iii).

On multiplying equation (ii) with b we get,

>> b(bx) - b(ay) = b(1 + c)

>> b²x - aby = b + bc

Let this be equation (iv).

Let this be equation (iv).On adding equation (iii) with equation (iv), we get

>>> a²x + aby + b²x - aby = ac + b + bc

Cancelling the terms with unlike sign but samw values, we get

>>> a²x + b²x = ac + b + bc

Taking x as common, we get

>>> x(a² + b²) = ac + b + bc

Now on transposing the term, we get

$\;\;\boxed{\bf{\mapsto\;\;x\;=\;\purple{\dfrac{ac\:+\;b\:+\;bc}{(a^{2}\:+\:b^{2})}}}}$

Now let's put this value of x which we got from previous equation, in the first equation.

Then,

$\;\;\tt{\rightarrow\;\;ax\;+\;by\;=\;c}$

$\;\;\tt{\rightarrow\;\;a\bigg(\dfrac{ac\:+\;b\:+\;bc}{(a^{2}\:+\:b^{2})}\bigg)\;+\;by\;=\;c}$

$\;\;\tt{\rightarrow\;\;c\;-\;a\bigg(\dfrac{ac\:+\;b\:+\;bc}{(a^{2}\:+\:b^{2})}\bigg)\;=\;by}$

$\;\;\tt{\rightarrow\;\;c\;-\;\dfrac{a^{2}c\:+\;ab\:+\;abc}{(a^{2}\:+\:b^{2})}\;=\;by}$

On taking L.C.M. at the L.H.S., we get

$\;\;\tt{\rightarrow\;\;\dfrac{a^{2}c\:+\:b^{2}c\:-\:a^{2}c\:-\;ab\:-\;abc}{(a^{2}\:+\:b^{2})}\;=\;by}$

$\;\;\tt{\rightarrow\;\;\dfrac{b^{2}c\:\:-\;ab\:-\;abc}{(a^{2}\:+\:b^{2})}\;=\;by}$

$\;\;\tt{\rightarrow\;\;\dfrac{b(bc\:\:-\;a\:-\;ac)}{(a^{2}\:+\:b^{2})}\;=\;by}$

$\;\;\tt{\rightarrow\;\;\dfrac{b(bc\:\:-\;a\:-\;ac)}{b(a^{2}\:+\:b^{2})}\;=\;y}$

On cancelling b from numerator and denominator, we get

$\;\;\boxed{\bf{\mapsto\;\;y\;=\;\red{\dfrac{(bc\:\:-\;a\:-\;ac)}{(a^{2}\:+\:b^{2})}}}}$