Solve the systems of equations :
$a x + b y = c; b x - a y = c$

\frac{c (a - b)}{a^{2} + b^{2}}, - \frac{c (a + b)}{a^{2} + b^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{c (a - b)}{a^{2} + b^{2}}, \frac{c (a + b)}{a^{2} + b^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{c (a - b)}{a^{2} + b^{2}}, \frac{c (a + b)}{a^{2} - b^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{c (a + b)}{a^{2} + b^{2}}, - \frac{c (a - b)}{a^{2} + b^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{c (a + b)}{a^{2} + b^{2}}, - \frac{c (a - b)}{a^{2} + b^{2}}$
$a x + b y = c, b x - a y = c$
Using the cross-multiplication method,
$\Rightarrow \frac{x}{- a c - b c} = \frac{y}{a c - b c} = \frac{1}{- a^{2} - b^{2}}$
$\Rightarrow x = \frac{- a c - b c}{- a^{2} - b^{2}} = \frac{- c (a + b)}{- (a^{2} + b^{2})} = \frac{c (a + b)}{a^{2} + b^{2}}$
and $y = \frac{a c - b c}{- a^{2} - b^{2}} = \frac{c (a - b)}{- (a^{2} + b^{2})} = - \frac{c (a - b)}{a^{2} + b^{2}}$
Therefore, $x = \frac{c (a + b)}{a^{2} + b^{2}}, y = - \frac{c (a - b)}{a^{2} + b^{2}}$

Suggest Corrections

Similar questions

a^{2} + b^{2} + c^{2} - a b - b c - c a = \frac{1}{2} [{(b - c)}^{2} + {(c - a)}^{2} + {(a - b)}^{2}]

(a^{2} - 2 b c - b^{2} - c^{2}) \div (a - b - c) (a + b + c)

In a triangle ABC, $2 c a s i n \frac{A - B + C}{2}$ is equal to

a^{2} (b + c)^{2} + b^{2} (c + a)^{2} + c^{2} (a + b) + a b c (a + b + c) + (a^{2} + b^{2} + c^{2}) (b c + c a + a b)

Join BYJU'S Learning Program