For all x > 3, which of the following is equivalent to the expression \frac{2x^2 - 5x - 3}… — SAT

For all $x > 3$ , which of the following is equivalent to the expression $\frac{2x^2 - 5x - 3}{x^2 - 9} - \frac{x}{x+3}$ ?

\frac{x + 1}{x + 3}

To simplify the expression, we factor the first term:

\frac{2x^2 - 5x - 3}{x^2 - 9} = \frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)}

. Canceling the common factor of

x - 3

leaves

\frac{2x + 1}{x + 3}

. Subtracting the second term gives

\frac{2x + 1}{x + 3} - \frac{x}{x + 3} = \frac{2x + 1 - x}{x + 3} = \frac{x + 1}{x + 3}

. This matches the correct expression.

Factor the numerator and denominator of the first rational expression.

The numerator

2x^2 - 5x - 3

factors into

(2x + 1)(x - 3)

. The denominator

x^2 - 9

is a difference of squares and factors into

(x - 3)(x + 3)

Factoring allows for the simplification of the rational expression by identifying common factors in the numerator and denominator.

Simplify the first rational expression by canceling the common factor.

\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}

for all

x > 3

Since

x > 3

, the term

x - 3

is non-zero, allowing us to divide both the numerator and denominator by

x - 3

Subtract the second expression from the simplified first expression.

\frac{2x + 1}{x + 3} - \frac{x}{x + 3} = \frac{2x + 1 - x}{x + 3} = \frac{x + 1}{x + 3}

Since both fractions now have the same denominator,

x + 3

, we subtract their numerators.

Simplifying rational expressions by factoring polynomials and performing operations on fractions with common denominators.

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