The solution of the equations \( \Large \frac{p}{x}+\frac{q}{y} \  and  \  \frac{q}{x}+\frac{p}{y} \) n is


A) \( \Large x=\frac{q^{2}-p^{2}}{mp-nq}, y=\frac{p^{2}-q^{2}}{np-mq} \)

B) \( \Large x=\frac{p^{2}-q^{2}}{mp-nq}, y=\frac{q^{2}-p^{2}}{np-mq} \)

C) \( \Large x=\frac{p^{2}-q^{2}}{mp-nq}, y=\frac{p^{2}-q^{2}}{np-mq} \)

D) \( \Large x=\frac{q^{2}-p^{2}}{mp-nq}, y=\frac{q^{2}-p^{2}}{np-mq} \)

Correct Answer:
C) \( \Large x=\frac{p^{2}-q^{2}}{mp-nq}, y=\frac{p^{2}-q^{2}}{np-mq} \)

Description for Correct answer:

\( \Large \frac{p}{x}+\frac{q}{y}=m \) ...(i)

\( \Large \frac{q}{x}+\frac{p}{y}=n \) ...(ii)

On multiplying Eq. (i) by q and Eq. (ii) by p and subtracting, we get

\( \Large \frac{pq}{x}+\frac{q^{2}}{y}=mq \)

\( \Large \frac{pq}{x}+\frac{p^{2}}{y}=np \)

\( \Large \frac{q^{2}}{y}-\frac{p^{2}}{y}=mq-np \)

Therefore, \( \Large q^{2}-p^{2}=y \left(mq-np\right) \)

Therefore, \( \Large y = \frac{q^{2}-p^{2}}{mq-np}=\frac{p^{2}-q^{2}}{np-mq} \)

Again, on multiplying Eq. (i) by p and Eq. (ii) by q and subtracting, we get

\( \Large \frac{p^{2}}{x}+\frac{pq}{y}=mp \)

\( \Large \frac{q^{2}}{x}+\frac{pq}{y}=nq \)

\( \Large \frac{p^{2}}{x}-\frac{q^{2}}{x}=mp-nq \)

= \( \Large p^{2}-q^{2}=x \left(mp-nq\right) \)

= \( \Large x = \frac{p^{2}-q^{2}}{mp-nq} \)

Therefore, \( \Large x = \frac{p^{2}-q^{2}}{mp-nq} \)

and \( \Large y = \frac{p^{2}-q^{2}}{np-mq} \)


Part of solved Linear Equations questions and answers : >> Aptitude >> Linear Equations








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