Question

a/x-b +b/x-a = 2. Solve for x

Answer 1

Answer:

$x = a+b / 2\:\:\:OR\:\:\:x = a + b$

Step-by-step explanation:

$\frac{a}{x-b} +\frac{b}{x-a}=2 \:\:\:(*(x-a)(x-b))\\a(x-a) + b(x-b) =2 (x-a)(x-b)\\ax - a^2 + bx - b^2 = 2 (x-a) (x-b) \\ax - a^2+ bx - b^2 = (2x-2a) (x-b) \\ax - a^2 + bx - b^2 = 2x^2 - 2bx - 2ax + 2ab \\-2x^2 + 3bx + 3ax - a^2 - b^2 - 2ab = 0 \\2x^2 - 3bx - 3ax + a^2 + b^2 +2ab = 0 \\2x^2 - 3 (a+b) x + (a+b)^2 = 0 \\2x^2 - 2(a+b)x - 1(a+b)x + (a+b)^2 = 0 \\2x [x - (a+b)] - (a+b) [x - (a+b)] = 0 \\(2x - a - b ) (x - a - b) = 0 \\2x - a - b = 0\:\:\:\:OR\:\:\:\:x - a - b = 0$

Therefore,

$x = a+b / 2\:\:\:OR\:\:\:x = a + b$

Answer 2

Answer:

$x = \frac{a+b}{2}$ and x = a+b

Step-by-step explanation:

Give that:

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

then;

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

⇒$a(x-a)+b(x-b) = 2(x-a)(x-b)$

Using distributive property i.e, $a \cdot (b+c) = a\cdot b+ a\cdot c$

then;

$ax-a^2+bx-b^2 = 2(x^2-bx-ax+ab)$

⇒$ax-a^2+bx-b^2 = 2x^2-2bx-2ax+2ab$

⇒$3ax-a^2+3bx-b^2 = 2x^2+2ab$

⇒$3ax-a^2+3bx-b^2-2x^2-2ab=0$

⇒$2x^2-3bx-3ax+a^2+b^2+2ab=0$

Using identity rule: $(a+b)^2 = a^2+b^2+2ab$

⇒$2x^2-3x(a+b)+(a+b)^2=0$

⇒$2x^2-2x(a+b)-x(a+b)+(a+b)^2=0$

⇒$2x(x-(a+b))-(a+b)(x-(a+b))=0$

⇒$(2x-(a+b))((x-(a+b))=0$

By zero product property we have;

$2x = a+b$ or $x = a+b$

⇒$x = \frac{a+b}{2}$ or x = a+b

Therefore, the value of x are:

$x = \frac{a+b}{2}$ and x = a+b