Question

Solve the following equations :
(a) $\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}$ = 2$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
(b) $\frac{(x-a)(x-b)}{x-a-b}=\frac{(x-c)(x-d)}{x-c-d}$

Answer 1

(a) The given equation can be re written as
$\Sigma$ax-$\Sigma$ a${}^2$ = 2$\Sigma$bc
or x$\Sigma$ a = $\Sigma$ a${}^2$ + 2$\Sigma$bc
or x$\Sigma$a = ($\Sigma$ a)${}^2$ or $\Sigma$a (x - $\Sigma$a) = 0
$\therefore$ x=$\Sigma$a=a+b+c,
provided a + b + c $\ne$ 0
(b) L.H.S. = $\frac{x^2-ax-bx+ab}{x-a-b}=\frac{x(x-a-b)}{x-a-b}+\frac{ab}{x-a-b}=x+\frac{ab}{x-a-b}$
$\therefore$ $\frac{ab}{x-a-b}=\frac{cd}{x-c-d}$
or ab (x-c + d) = cd (x- a-b)
ac(b-a) + bd(a - c)
or x= $\frac{ac(b-d)+bd(a-c)}{ab-cd}$