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Zorluk: OrtaRational and Radical Expressions and Equations

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

Cevap: -2

Cevap

The correct answer is -2.
Subtracting xx−3\frac{x}{x-3} from both sides yields 2x=3−xx−3\frac{2}{x} = \frac{3-x}{x-3}. Since 3−x=−(x−3)3-x = -(x-3), the right side simplifies to −1-1 for all x≠3x \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.

Adım Adım Çözüm

1
Subtract xx−3\frac{x}{x-3} from both sides of the equation.
2x=3−xx−3\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 3−x3-x is the negative of the denominator x−3x-3, so their quotient is −1-1 for all x≠3x \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 x≠0x \neq 0 and x≠3x \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.

Anahtar Kavram

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

Rational and Radical Expressions and Equations

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