Question

Solve the following pair of linear equations:
$\left(i\right)px+qy=p-q;qx-py=p+q$

$\left(ii\right)ax+by=c;bx+ay=1+c$

Answer 1

(i) $px+qy=p-q\dots \left(i\right)$

$qx-py=p+q\dots \left(ii\right)$

Multiplying $\left(i\right)$ by p and $\left(ii\right)$ by $q$ we get $\left(iii\right)$ and $\left(iv\right)$ respectively as

$p^{2}x+pqy=p^2-pq\dots \left(iii\right)$

$q^{2}x-pqy=pq+q^2\dots \left(iv\right)$

Adding $\left(iii\right)$ and $\left(iv\right)$ we get,

$(p^2+q^2)x=(p^2+q^2)$

$\Rightarrow x=1$

Substituting value of $x$ in $\left(i\right)$, we get

$p+qy=p-q$

$qy=-q$

$y=-1$

Hence, $x=1,y=-1$

(ii) $ax+by=c\dots \left(i\right)$

$bx+ay=1+c\dots \left(ii\right)$

Multiplying $\left(i\right)$ by $b$ and $\left(ii\right)$ by a we get $\left(iii\right)$ and $\left(iv\right)$ respectively as

$abx+b^{2}y=bc\dots \left(iii\right)$

$abx+a^{2}y=a+ac\dots \left(iv\right)$

Subtracting $\left(iv\right)$ from $\left(iii\right)$, we get

$(b^2-a^2)y=(b-a)c-a$

$y=\frac{(b-a)c-a}{(b^2-a^2)}$

Substituting value of $y$ in $\left(i\right)$, we get

$ax+b\times \frac{(b-a)c-a}{(b^2-a^2)}=c$

$ax=c-\frac{(b-a)bc-ab}{(b^2-a^2)}$

$ax=\frac{c(b^2-a^2)-b^{2}c+abc+ab}{(b^2-a^2)}$

$x=\frac{c{b}^2-c{a}^2-b^{2}c+abc+ab}{a(b^2-a^2)}$

$x=\frac{ab+abc-c{a}^2}{a(b^2-a^2)}$

$x=\frac{b-bc-ca}{(b^2-a^2)}$

$x=\frac{b+c(b-a)}{(b^2-a^2)}$