Question : If $a x+b y=1$ and $b x+a y=\frac{2 a b}{a^2+b^2}$, then the value of $x$ (in terms of $a$ and $b$) is:
Option 1: $\frac{2 b}{a^2+b^2}$
Option 2: $\frac{a}{a^2+b^2}$
Option 3: $\frac{b}{a^2+b^2}$
Option 4: $\frac{2 a}{a^2+b^2}$
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
Correct Answer: $\frac{a}{a^2+b^2}$
Solution :
Given, $ax+by=1$ ....(1)
And, $bx+ay=\frac{2ab}{a^2+b^2}$ ......(2)
From $ax+by=1$, we get,
⇒ $y=\frac{1-ax}{b}$
Putting in (2), we get,
$bx+a\times \frac{1-ax}{b}=\frac{2ab}{a^2+b^2}$
⇒ $b^2x+a(1-ax)=\frac{2ab^2}{a^2+b^2}$
⇒ $b^2x+a-a^2x=\frac{2ab^2}{a^2+b^2}$
⇒ $(b^2-a^2)x+a=\frac{2ab^2}{a^2+b^2}$
⇒ $(b^2-a^2)x=\frac{2ab^2}{a^2+b^2}-a$
⇒ $(b^2-a^2)x=\frac{2ab^2-a(a^2+b^2)}{a^2+b^2}$
⇒ $x=\frac{2ab^2-a^3-ab^2}{(a^2+b^2)(b^2-a^2)}$
⇒ $x=\frac{ab^2-a^3}{(a^2+b^2)(b^2-a^2)}$
⇒ $x=\frac{a(b^2-a^2)}{(a^2+b^2)(b^2-a^2)}$
$\therefore x=\frac{a}{a^2+b^2}$
Hence, the correct answer is $\frac{a}{a^2+b^2}$.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Answer Key | Eligibility | Application | Admit Card | Preparation Tips | Result | Cutoff
Get Updates BrochureYour Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.




