The Solution of the equations p/x+q/y = m, q/x + p/y = n (a) x = q^2-p^2/mp-nq, y= p^2-q^2/np-mq

Question

The solution of the equations \(\frac{p}{X}+\frac{q}{y}\) = m, \(\frac{q}{X}+\frac{p}{y}\) = n is

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

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

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

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

Answer 1

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

Let \(\frac{1}{X}\) = a and \(\frac{1}{y}\) = b. Then, the given equations reduce to

pa + qb = m  .............(i)

qa + pb = n  ..............(ii)

Multiplying eqn (i) by q and eqn (ii) by p, we get

pqa + q²b = mq …......(iii)

qpa + p²b = np .......…(iv)

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

\((p^2-q^2)b\) = np-mq

\(\Rightarrow\)b = \(\frac{np-mq}{p^2-q^2}\)  \(\Rightarrow\) y = \(\frac{1}{b}\) = \(\frac{p^2-q^2}{np-mq}\)

Substituting this value of b in (i), we have

pa + \(\frac{q(np-mq)}{p^2-q^2}\) = m

\(\Rightarrow\)pa = m - \(\frac{(pqn-mq^2)}{p^2-q^2}\)

\(\Rightarrow\)pa = 

\(\Rightarrow\) pa = \(\frac{p(mp-qn)}{p^2-q^2}\)

\(\Rightarrow\) pa = \(\frac{mp-qn}{p^2-q^2}\)

\(\Rightarrow\) x = \(\frac{1}{a}\) = \(\frac{p^2-q^2}{mp-qn}\)