Question

Solve:
ax+by=c
bx-ay=0

Answer 1

Answer:

Hence, on solving we obtain:

$x=\dfrac{ac}{a^2+b^2},y=\dfrac{bc}{a^2+b^2}$

Step-by-step explanation:

We have system of linear equations as:

        $ax+by=c---------(1)$

and   $bx-ay=0----------(2)$

from equation (2) we can write our expression as:

$bx=ay$

so we can write:

$x=\dfrac{a}{b}y-------(3)$ or  $y=\dfrac{b}{a}x$

Hence in either cases we can say that a≠0 and b≠0.

Now we put the value of x in term of y in equation (1) to obtain:

$a\times \dfrac{a}{b}y+by=c\\\\\dfrac{a^2}{b}y+by=c\\\\y\times (\dfrac{a^2}{b}+b})=c\\\\y\times (\dfrac{a^2+b^2}{b})=c\\\\y=\dfrac{bc}{a^2+b^2}$

Now on putting the value of y in equation (3) we have:

$x=\dfrac{a}{b}\times \dfrac{bc}{a^2+b^2}\\\\x=\dfrac{ac}{a^2+b^2}$

Hence, on solving we obtain:

$x=\dfrac{ac}{a^2+b^2},y=\dfrac{bc}{a^2+b^2}$

Answer 2

Hope It Helps...

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