Question

Difficulty: MediumRational and Radical Expressions and Equations

For all real numbers xx such that x0x \neq 0 and x3x \neq 3, what is the real solution to the equation xx3+2x=3x3\frac{x}{x-3} + \frac{2}{x} = \frac{3}{x-3}?

Answer: -2

Answer

The correct answer is -2.
Subtracting xx3\frac{x}{x-3} from both sides yields 2x=3xx3\frac{2}{x} = \frac{3-x}{x-3}. Since 3x=(x3)3-x = -(x-3), the right side simplifies to 1-1 for all x3x \neq 3. The equation becomes 2x=1\frac{2}{x} = -1, which gives x=2x = -2. Since 2-2 does not violate the domain constraints, it is the correct solution.

Step-by-Step Solution

1
Subtract xx3\frac{x}{x-3} from both sides of the equation.
2x=3xx3\frac{2}{x} = \frac{3-x}{x-3}
To group terms with common denominators on the same side.
2
Simplify the fraction on the right side of the equation.
2x=1\frac{2}{x} = -1
The numerator 3x3-x is the negative of the denominator x3x-3, so their quotient is 1-1 for all x3x \neq 3.
3
Solve the simplified equation 2x=1\frac{2}{x} = -1 for xx.
x=2x = -2
Multiplying both sides by xx yields 2=x2 = -x, and dividing by 1-1 gives x=2x = -2.
4
Check the solution against the domain restrictions x0x \neq 0 and x3x \neq 3.
The solution x=2x = -2 is valid.
Since 2-2 is neither 00 nor 33, it does not cause any denominator in the original equation to equal zero.

Key Concept

Solving rational equations by isolating terms with common denominators and checking for extraneous solutions.
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