Question

1/x - 1/x-b = 1/a - 1/a-b solve

Answer 1

$\frac{1}{x} - \frac{1}{x - b} = \frac{1}{a} - \frac{1}{a - b} \\ there \: are \: two \: possibilities \\ 1 \\ x = a \: \: \: \: x - b = a - b \\ x = a \\ 2 \\ x = - (a - b) \: \: \: \: \: \: - (x - b) = a \\ x = b - a \\ therefore \: x = a \: \: \: x = b - a$
hope it helps

Answer 2

$\underline\bold{\huge{ANSWER \: :}}$


<< FACTS AT THE BEGINNING >>

▶The general form of the quadratic equation is (ax²+bx+c) = 0, for any problems, but (a ≠ 0)

▶The term "b²-4ac" is called discriminant of the quadratic equation.

▶ When b²-4ac = 0, then the roots are real & equal.

▶ When b²-4ac > 0, then the roots are real and distinct.

▶ When b²-4ac < 0, then there will be no real roots.

______________________________

$\frac{1}{x} - \frac{1}{x - b} = \frac{1}{a} - \frac{1}{a - b}$

=> $\frac{x - b - x}{x(x - b)} = \frac{a - b - a}{a(a - b)}$

=> $\frac{ - b}{x(x - b)} = \frac{ - b}{a(a - b)}$

=> $\frac{1}{ {x}^{2} - \: bx } = \frac{1}{ {a}^{2} - \: ab }$

=> a² - ab = x² - bx

=> a² - x² - ab + bx = 0

=> ( a + x ) ( a - x ) - b ( a - x ) = 0

=> ( a - x ) ( a + x - b ) = 0

______________________________

$\bold{\huge{ULTIMATELY}}$

$\bold{EITHER}$

a + x - b = 0

=> x = (b - a)

______________________________

$\bold{OR}$

a - x = 0

=> a = x