Question

solve:- ax+by=5;bx+ay=3,where a & b are constant term.

Answer 1

Therefore ,

$x= \frac{(3b-5a)}{b^2-a^2}$      and    $y =\frac{5b-3a}{b^2-a^2}$

Step-by-step explanation:

Given,

ax+by = 5..........(1)

bx+ay=3...........(2)

Equation (1)×b -equation(2)×a we get

abx +b²y-abx-a²y=5b -3a

⇔(b²-a²)y=5b -3a

$\Leftrightarrow y =\frac{5b-3a}{b^2-a^2}$

Putting the value of y in equation (1)

$ax+b(\frac{5b-3a}{b^2-a^2})=5$

$\Leftrightarrow ax= 5-\frac{5b^2-3ab}{b^2-a^2}$

$\Leftrightarrow ax= \frac{5b^2-5a^2-5b^2+3ab}{b^2-a^2}$

$\Leftrightarrow ax= \frac{a(-5a+3b)}{b^2-a^2}$

$\Leftrightarrow x= \frac{(3b-5a)}{b^2-a^2}$

Therefore ,

$x= \frac{(3b-5a)}{b^2-a^2}$     and    $y =\frac{5b-3a}{b^2-a^2}$