Answer :
To solve the equation [tex]\(\frac{a}{(x-b)} + \frac{b}{(x-a)} = 2 \)[/tex], let's go through the process step-by-step.
### Step 1: Initial Equation
We start with the given equation:
[tex]\[ \frac{a}{x-b} + \frac{b}{x-a} = 2 \][/tex]
### Step 2: Eliminate Denominators
To eliminate the denominators, we'll multiply both sides of the equation by [tex]\((x-b)(x-a)\)[/tex]:
[tex]\[ a(x-a) + b(x-b) = 2(x-a)(x-b) \][/tex]
### Step 3: Simplify Both Sides
Expand and simplify both sides of the equation.
On the left-hand side:
[tex]\[ a(x-a) + b(x-b) = ax - a^2 + bx - b^2 = ax + bx - a^2 - b^2 \][/tex]
On the right-hand side, expand [tex]\((x-a)(x-b)\)[/tex]:
[tex]\[ 2(x-a)(x-b) = 2(x^2 - (a+b)x + ab) = 2x^2 - 2(a+b)x + 2ab \][/tex]
### Step 4: Equate the Simplified Expressions
Equate the simplified expressions from both sides:
[tex]\[ ax + bx - a^2 - b^2 = 2x^2 - 2(a+b)x + 2ab \][/tex]
### Step 5: Gather Like Terms
Combine like terms on both sides:
[tex]\[ 0 = 2x^2 - 2(a+b)x + 2ab - ax - bx + a^2 + b^2 \][/tex]
Rearrange the terms for clarity:
[tex]\[ 2x^2 - (2a + 2b)x + 2ab - ax - bx + a^2 + b^2 = 0 \][/tex]
### Step 6: Simplification
Simplify the expression by collecting terms:
[tex]\[ 2x^2 - (2a+2b)x + a^2 + b^2 - ax - bx = 2x^2 - (2a+2b)x + 2ab - ax - bx + a^2 + b^2 = 0 \][/tex]
### Step 7: Find Coefficients and Verify Identity
We need to check if this equation holds true for all [tex]\(x\)[/tex]. This means the coefficients of [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant must each individually equal zero for the equation to hold true as an identity.
### Step 8: Setting Coefficients to Zero
- For [tex]\(x^2\)[/tex]:
[tex]\[ 2 = 0 \][/tex]
This term does not contribute any additional insight as it is already balanced.
- For [tex]\(x\)[/tex]:
[tex]\[ -2(a+b) - a - b = 0 \][/tex]
Simplifying gives:
[tex]\[ -3a - 3b = 0 \][/tex]
Thus:
[tex]\[ a + b = 0 \][/tex]
- For the constant term:
[tex]\[ a^2 + b^2 + 2ab = 0 \][/tex]
But since [tex]\( a + b = 0 \)[/tex], substitute [tex]\( b = -a \)[/tex]:
[tex]\[ a^2 + (-a)^2 + 2a(-a) = 0 \][/tex]
[tex]\[ a^2 + a^2 - 2a^2 = 0 \][/tex]
[tex]\[ 0 = 0 \][/tex]
### Step 9: Summary of Solution
From the above, we found that in order for the equation [tex]\(\frac{a}{(x-b)} + \frac{b}{(x-a)} = 2 \)[/tex] to hold true for any [tex]\(x\)[/tex], the conditions [tex]\( a + b = 0 \)[/tex] and [tex]\(b = -a\)[/tex] must be satisfied.
Therefore, the given equation is true under the condition [tex]\( a + b = 0 \)[/tex].
### Step 1: Initial Equation
We start with the given equation:
[tex]\[ \frac{a}{x-b} + \frac{b}{x-a} = 2 \][/tex]
### Step 2: Eliminate Denominators
To eliminate the denominators, we'll multiply both sides of the equation by [tex]\((x-b)(x-a)\)[/tex]:
[tex]\[ a(x-a) + b(x-b) = 2(x-a)(x-b) \][/tex]
### Step 3: Simplify Both Sides
Expand and simplify both sides of the equation.
On the left-hand side:
[tex]\[ a(x-a) + b(x-b) = ax - a^2 + bx - b^2 = ax + bx - a^2 - b^2 \][/tex]
On the right-hand side, expand [tex]\((x-a)(x-b)\)[/tex]:
[tex]\[ 2(x-a)(x-b) = 2(x^2 - (a+b)x + ab) = 2x^2 - 2(a+b)x + 2ab \][/tex]
### Step 4: Equate the Simplified Expressions
Equate the simplified expressions from both sides:
[tex]\[ ax + bx - a^2 - b^2 = 2x^2 - 2(a+b)x + 2ab \][/tex]
### Step 5: Gather Like Terms
Combine like terms on both sides:
[tex]\[ 0 = 2x^2 - 2(a+b)x + 2ab - ax - bx + a^2 + b^2 \][/tex]
Rearrange the terms for clarity:
[tex]\[ 2x^2 - (2a + 2b)x + 2ab - ax - bx + a^2 + b^2 = 0 \][/tex]
### Step 6: Simplification
Simplify the expression by collecting terms:
[tex]\[ 2x^2 - (2a+2b)x + a^2 + b^2 - ax - bx = 2x^2 - (2a+2b)x + 2ab - ax - bx + a^2 + b^2 = 0 \][/tex]
### Step 7: Find Coefficients and Verify Identity
We need to check if this equation holds true for all [tex]\(x\)[/tex]. This means the coefficients of [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant must each individually equal zero for the equation to hold true as an identity.
### Step 8: Setting Coefficients to Zero
- For [tex]\(x^2\)[/tex]:
[tex]\[ 2 = 0 \][/tex]
This term does not contribute any additional insight as it is already balanced.
- For [tex]\(x\)[/tex]:
[tex]\[ -2(a+b) - a - b = 0 \][/tex]
Simplifying gives:
[tex]\[ -3a - 3b = 0 \][/tex]
Thus:
[tex]\[ a + b = 0 \][/tex]
- For the constant term:
[tex]\[ a^2 + b^2 + 2ab = 0 \][/tex]
But since [tex]\( a + b = 0 \)[/tex], substitute [tex]\( b = -a \)[/tex]:
[tex]\[ a^2 + (-a)^2 + 2a(-a) = 0 \][/tex]
[tex]\[ a^2 + a^2 - 2a^2 = 0 \][/tex]
[tex]\[ 0 = 0 \][/tex]
### Step 9: Summary of Solution
From the above, we found that in order for the equation [tex]\(\frac{a}{(x-b)} + \frac{b}{(x-a)} = 2 \)[/tex] to hold true for any [tex]\(x\)[/tex], the conditions [tex]\( a + b = 0 \)[/tex] and [tex]\(b = -a\)[/tex] must be satisfied.
Therefore, the given equation is true under the condition [tex]\( a + b = 0 \)[/tex].
