Which of the following expressions is equivalent to \frac{2x^2 - 3x - 9}{x^2 - 9} - \frac{… — SAT

Which of the following expressions is equivalent to $\frac{2x^2 - 3x - 9}{x^2 - 9} - \frac{x}{x+3}$ for all $x > 3$ ?

$1$ Answer
B
$\frac{x^2 + 6x - 9}{x^2 - 9}$
C
$\frac{x^2 - 6x - 9}{x^2 - 9}$
D
$\frac{3x+3}{x+3}$

Answer

1

The correct answer is

1

. Factoring the first term gives

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

. Since

x > 3

, we cancel the common non-zero factor of

x-3

to obtain

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

. Subtracting

\frac{x}{x+3}

from this result yields

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

Step-by-Step Solution

Factor the numerator and the denominator of the first rational expression.

The numerator

2x^2 - 3x - 9

factors as

(2x+3)(x-3)

and the denominator

x^2 - 9

factors as

(x+3)(x-3)

Factoring allows us to identify and divide out common factors to simplify the expression.

Simplify the first rational expression by cancelling the common factor.

For all

x > 3

, the factor

x-3

is non-zero, so the expression

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

simplifies to

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

This simplifies the subtraction by reducing the first fraction.

Subtract the second expression from the simplified first expression.

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

Since the denominators are identical, the numerators can be combined directly.

Key Concept

Simplifying rational expressions by factoring and performing algebraic operations.

Alternative Method

Instead of simplifying the first fraction first, find a common denominator immediately by multiplying the numerator and denominator of the second fraction by

x-3

. This yields:

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

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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