Question

Solve the given pair of linear equations.

(ax/b) - (by/a) = a + b

ax - by = 2ab​

Answer 1

SOLUTION:-

Given:

$\frac{ax}{b} - \frac{by}{a} = a + b ............(1) \\ \\ ax - by = 2ab...............(2)$

Multiply by ab to eq (1) & a to eq. (2), we get;

=) a²x - b²y = a²b + ab²..........(3)

=) a²x - aby = 2a²b................(4)

Therefore,

Subtracting equation (4) from equation (3), we get;

=)(a²x - a²x)+(-aby)-(-b²y)=(2a²b-a²b)-ab²

=) -aby + b²y = a²b -ab²

=) by(b-a) = ab(b-a)

=) y= b(b-a) / ab(a-b)

=) y = -a

Substitute y value in equation (2), we get;

=) ax -b(-a) = 2ab

=) ax + ab = 2ab

=) ax = ab

=) x= b

Hence,

The solution is x = b & y= -a.

Hope it helps ☺️

Answer 2

Solution:

Given Equations:

$\sf{\implies \dfrac{ax}{b} -\dfrac{by}{a} =a+b}$

$\sf{\implies ax-by=2ab}$

Divide 2nd equation by (a), we get

$\sf{\implies \dfrac{ax}{b} -\dfrac{by}{a} =a+b\;\;\;.......(1)}$

$\sf{\implies x-\dfrac{by}{a}=2b\;\;\;......(2)}$

Now, we subtracting equation (2) from equation (1), we get

$\sf{\implies \dfrac{ax}{b} -x=a+b-2b}$

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

$\sf{\implies \dfrac{x(a-b)}{b}=a-b}$

$\large{\boxed{\boxed{\sf{\implies x = b}}}}$

Put the value of x in equation (2), we get

$\sf{\implies x - \dfrac{by}{a}=2b}$

$\sf{\implies b -\dfrac{by}{a} =2b}$

$\sf{\implies -b = \dfrac{by}{a}}$

$\large{\boxed{\boxed{\sf{\implies y=-a}}}}$

So,

=> x = b

=> y = -a