To begin solving for x, we first rearrange the equation by moving the term (1)/(x) from the right side to the left side. This groups terms involving x together.

We expand the right side of the equation: x(a+b+x) = x² + (a+b)x. Moving -ab to the right side, we get x² + (a+b)x + ab = 0. This is a quadratic equation in terms of x.

We can factor the quadratic equation x² + (a+b)x + ab = 0 into (x+a)(x+b) = 0. This is because a × b = ab and a + b = a+b.

Solve for x in terms of a and b: $\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$ (Where a, b, x ≠ 0 and a + b + x ≠ 0)

Answer: $x = -a \quad \text{or} \quad x = -b$

Step-by-step solution

Step 1: Rearrange the equation

To begin solving for $x$ , we first rearrange the equation by moving the term $\frac{1}{x}$ from the right side to the left side. This groups terms involving $x$ together.

Step 2: Combine fractions on both sides

Next, we combine the fractions on both sides of the equation by finding a common denominator for each side. On the left, the common denominator is x(a+b+x), and on the right, it is $ab$ .

Step 3: Simplify the equation

We simplify the numerator on the left side: $x - (a+b+x) = x - a - b - x = -a - b$ . This gives us $\frac{-(a+b)}{x(a+b+x)} = \frac{a+b}{ab}$ .

Step 4: Cross-multiply and solve for x

Since $a+b \neq 0$ (otherwise the original equation would be undefined), we can divide both sides by (a+b). This simplifies the equation to $\frac{-1}{x(a+b+x)} = \frac{1}{ab}$ . Then, we cross-multiply to get $-ab = x(a+b+x)$ .

Step 5: Form a quadratic equation

We expand the right side of the equation: $x(a+b+x) = x^2 + (a+b)x$ . Moving $-ab$ to the right side, we get $x^2 + (a+b)x + ab = 0$ . This is a quadratic equation in terms of $x$ .

Step 6: Factor the quadratic equation

We can factor the quadratic equation $x^2 + (a+b)x + ab = 0$ into $(x+a)(x+b) = 0$ . This is because $a \times b = ab$ and $a + b = a+b$ .

Step 7: Solve for x

From the factored form $(x+a)(x+b) = 0$ , we set each factor equal to zero to find the possible values for $x$ . This gives us $x+a=0 \implies x=-a$ or $x+b=0 \implies x=-b$ .

More from this chapter

All Quadratic Equations solutions →

Solve your own maths question free →

Solve for x in terms of a and b: 1a+b+x=1a+1b+1x\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}a+b+x1​=a1​+b1​+x1​(Where a, b, x ≠ 0 and a + b + x ≠ 0)