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in Linear Equations by (49.2k points)

Solve the following system of equations for x and y, x, y \(\ne\) 0.

\(\frac{a}{X} -\frac{b}{y}\) = 0, \(\frac{ab^2}{X}\) + \(\frac{a^2b}{y}\) = \(a^2+b^2\)

(a) x = b, y = a 

(b) x = a, y = b 

(c) x = –a, y = –b 

(d) x = –b, y = – a

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1 Answer

+1 vote
by (53.1k points)

(a) x = b, y = a

Let \(\frac{1}{X}\) = p, \(\frac{1}{y}\)= q. Then,

ap – bq = 0 ..........(i)

ab2p + a2bq = a2 + b2 \(\Rightarrow\) ab (bp + aq) = a2 + b2

\(\Rightarrow\) bp + aq = \(\frac{a^2+b^2}{ab}\)  ...........(ii)

Multiplying eqn (i) by b and eqn (ii) by a, we get 

abp – b2q = 0 ..........(iii)

abp + a2q = \(\frac{a^2+b^2}{b}\) .......(iv)

Now subtracting eqn (iii) from eqn (iv), we get

(a2 +b2)q = \(\frac{a^2+b^2}{b}\) \(\Rightarrow\) q = \(\frac{1}{b}\) \(\Rightarrow\) \(\frac{1}{X}\) = \(\frac{1}{b}\) \(\Rightarrow\) x = b

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