Question

Solve the following pairs of equations by cross multiplication method:
1) $\begin{array}{c}\frac{x}{a}+\frac{y}{b}=a+b\\ \frac{x}{a^2}+\frac{y}{b^2}=2\end{array}$
2)$\begin{array}{c}ax-by=\frac{a-b}{2}\\ x+3y=2\end{array}$

Answer 1

(1)

$\frac{x}{a}+\frac{y}{b}=a+b$

$=>bx+ay=ab(a+b)$

$=>ay=ab(a+b)-bx$

$=>y=\frac{ab(a+b)-bx}{a}\frac{x}{a^2}+\frac{y}{b^2}=2$

$=>\frac{x}{a^2}+\frac{ab(a+b)-bx}{a}\times \frac{1}{b^2}=2$

$=>\frac{x}{a^2}+\frac{a^{2}b+a{b}^2-bx}{a{b}^2}=2$

$=>\frac{x{b}^2+a^{3}b+a^2{b}^2-abx}{a^2{b}^2}=2=>x{b}^2-abx+a^{3}b+a^2{b}^2=2a^2{b}^2$

$=>xb(b-a)=a^2{b}^2-a^{3}b$

$=>bx=\frac{a^{2}b(b-a)}{(b-a)}$

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

$=>y=\frac{a^{2}b+a{b}^2-a^{2}b}{a}$

$=>y=\frac{a{b}^2}{a}=b^2$

(2)

$ax-by=\frac{a-b}{2}$

$=>ax=by+\frac{a-b}{2}$

$=>ax=\frac{2by+a-b}{2}$

$=>x=\frac{2by+a-b}{2a}x+3y=2$

$=>\frac{2by+a-b}{2a}+3y=2$

$=>2by+a-b+6ay=4a$

$=>2by+6ay=3a+b$

$=>2y(b+3a)=3a+b$

$=>2y=\frac{3a+b}{3a+b}$

$=>2y=1$

$=>y=\frac{1}{2}$