We are tasked with solving the equation:

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

Step 1: Combine the fractions

To combine the two fractions, we need a common denominator. The common denominator will be $(x-b)(x-a)$. Thus, we rewrite the equation as:

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

Step 2: Simplify the equation

Now, combine the numerators on the left side:

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

Distribute $a$ and $b$ in the numerators:

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

Combine like terms:

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

Step 3: Eliminate the denominator

To eliminate the denominator, multiply both sides of the equation by $(x-b)(x-a)$:

$(a+b)x - (a^2 + b^2) = 2(x-b)(x-a)$

Step 4: Expand both sides

First, expand the right side:

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

Thus, the equation becomes:

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

Now expand the right side:

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

Step 5: Rearrange the equation

Move all terms to one side:

$0 = 2x^2 - 2(a+b)x + 2ab - (a+b)x + (a^2 + b^2)$

Combine like terms:

$0 = 2x^2 - (2a + 2b + a + b)x + 2ab + a^2 + b^2$

Simplify the coefficients:

$0 = 2x^2 - (3a + 3b)x + 2ab + a^2 + b^2$

Step 6: Solve using the quadratic formula

This is now a quadratic equation in the form $Ax^2 + Bx + C = 0$:

$2x^2 - (3a + 3b)x + (a^2 + b^2 + 2ab) = 0$

We can solve this using the quadratic formula:

$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

Where $A = 2$, $B = -(3a + 3b)$, and $C = a^2 + b^2 + 2ab$.

Substitute these values into the quadratic formula:

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

Simplify the discriminant:

$(3a + 3b)^2 = 9a^2 + 18ab + 9b^2$

Now calculate the discriminant:

$\Delta = 9a^2 + 18ab + 9b^2 - 8(a^2 + b^2 + 2ab)$

$\Delta = 9a^2 + 18ab + 9b^2 - 8a^2 - 8b^2 - 16ab$

$\Delta = a^2 + 2ab + b^2$

Thus, the discriminant simplifies to:

$\Delta = (a + b)^2$

So the quadratic formula becomes:

$x = \frac{3a + 3b \pm (a + b)}{4}$

Step 7: Solve for the two possible values of $x$

Now we have two cases to solve:

Case 1: $+$

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

Case 2: $-$

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

Final Solution:

The two possible solutions for $x$ are:

$x = a + b \quad \text{or} \quad x = \frac{a + b}{2}$

Would you like further clarification on any of the steps? Here are some related questions for exploration:

1. How do you handle the case when the discriminant is zero?
2. What happens if $a = b$ in this equation?
3. How would you solve the same equation using a different method, such as substitution?
4. What are some real-life scenarios where this type of equation might occur?
5. How does the quadratic formula help in finding roots of any quadratic equation?

Tip: When solving quadratic equations, always check if the discriminant ($\Delta$) is a perfect square, as this helps identify if the roots are real or complex.