Certainly! Let's solve the given equation step-by-step:
Given:
[tex]\[
\frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}
\][/tex]
To solve for [tex]\( x \)[/tex], let's start by finding a common denominator for the fractions on the right-hand side.
On the right-hand side, we have:
[tex]\[
\frac{1}{a} + \frac{1}{b} + \frac{1}{x}
\][/tex]
The common denominator for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( x \)[/tex] is [tex]\( abx \)[/tex]. Rewriting each fraction with this common denominator, we get:
[tex]\[
\frac{bx}{abx} + \frac{ax}{abx} + \frac{ab}{abx}
\][/tex]
This simplifies to:
[tex]\[
\frac{bx + ax + ab}{abx}
\][/tex]
So, our original equation can now be written as:
[tex]\[
\frac{1}{a + b + x} = \frac{bx + ax + ab}{abx}
\][/tex]
Next, we need to cross-multiply to solve for [tex]\( x \)[/tex]. Cross-multiplying gives us:
[tex]\[
abx = (a + b + x)(bx + ax + ab)
\][/tex]
Expanding the right-hand side:
[tex]\[
abx = a \cdot bx + a \cdot ax + a \cdot ab + b \cdot bx + b \cdot ax + b \cdot ab + x \cdot bx + x \cdot ax + x \cdot ab
\][/tex]
Simplifying each term:
[tex]\[
abx = abx + a^2x + a^2b + abx + b^2x + b^2a + bx^2 + ax^2 + abx
\][/tex]
Combining like terms:
[tex]\[
abx = 3abx + a^2x + abx + b^2x + bx^2 + ax^2
\][/tex]
To simplify, we can group the terms involving [tex]\( x \)[/tex] and the constant terms:
[tex]\[
abx = 3abx + a^2x + b^2x + (a + b) x^2 + a^2 b + b^2 a
\][/tex]
Let's isolate the [tex]\( x \)[/tex]-terms and the constant terms:
[tex]\[
abx - 3abx = a^2x + b^2x + (a + b) x^2 + a^2 b + b^2 a
\][/tex]
Simplifying further, we get:
[tex]\[
-2abx = a^2x + b^2x + (a + b) x^2
\][/tex]
Rearranging to make [tex]\( x \)[/tex] the subject, we get:
[tex]\[
-2abx - a^2x - b^2x - (a + b) x^2 = 0
\][/tex]
Factoring [tex]\( x \)[/tex] out:
[tex]\[
x(-2ab - a^2 - b^2 - (a + b) x) = 0
\][/tex]
This implies that:
[tex]\[
-2ab - a^2 - b^2 - (a + b) x = 0
\][/tex]
Solving for [tex]\( x \)[/tex], we find the roots [tex]\( x = -a \)[/tex] or [tex]\( x = -b \)[/tex].
Therefore, the solutions to the equation are:
[tex]\[
x = -a \quad \text{or} \quad x = -b
\][/tex]
