Question

(x-a)/ b + (x- b) / a + (x-3a-3b)/a+b= 0
find x.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$x= \frac{ a^3+b^3+4ab(a+b)}{a^2+3ab+b^2}$

Step-by-step explanation:

Given:

$\frac{(x-a)}{b}+\frac{(x-b)}{a} +\frac{(x+3a-3b)}{a+b} =0$

Taking LCM,

$\frac{[(x-b)\times a\times (a+b)]+[(x-b)\times b \times (a+b)]+[ (x-3a-3b)\times a\times b]}{a\times b\times (a+b)}=0$

$\frac{[(x-a)(a^2+ab)]+[(x-b)(ab+b^2)]+[(x-3a-3b)(ab)]}{a^2b+ab^2}=0$

$\frac{xa^2+xab-a^3-a^2b+xab+xb^2-ab^2-b^3+xab-3a^2b-3ab^2}{a^2b+ab^2}=0$

$xa^2+3xab-a^3-4a^2b+xb^2-4ab^2-b^3 = (a^2b+ab^2)\times 0$

$xa^2+3xab-a^3-4a^2b+xb^2-4ab^2-b^3 = 0$

$xa^2+3xab+xb^2 = a^3+b^3+4ab^2+4a^2b$

$x(a^2+3ab+b^2} = a^3+b^3+4ab^2+4a^2b$

$x= \frac{ a^3+b^3+4ab^2+4a^2b}{a^2+3ab+b^2}$

$x= \frac{ a^3+b^3+4ab(a+b)}{a^2+3ab+b^2}$