For all real numbers x such that x \neq 0 and x \neq -1, which of the following is equival… — ACT

For all real numbers $x$ such that $x \neq 0$ and $x \neq -1$ , which of the following is equivalent to the expression $\frac{1}{x} + \frac{1}{x+1}$ ?

A
$\frac{2}{2x+1}$
B
$\frac{2}{x^2+x}$
C
$\frac{2x+1}{x^2+1}$
$\frac{2x+1}{x^2+x}$ Answer
E
$\frac{3}{x+1}$

Answer

The correct expression is

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

The correct expression is obtained by finding a common denominator of

x(x+1)

, rewriting the numerators as

x+1

and

x

, and then adding them to get

2x+1

over the expanded denominator

x^2+x

Step-by-Step Solution

Find a common denominator for the two rational terms.

The common denominator for

\frac{1}{x}

and

\frac{1}{x+1}

x(x+1)

, which expands to

x^2+x

To add fractions with different denominators, they must share a common denominator.

Rewrite each fraction with the common denominator by multiplying their numerators and denominators by the appropriate factor.

\frac{1}{x} = \frac{x+1}{x(x+1)}

and

\frac{1}{x+1} = \frac{x}{x(x+1)}

Multiplying the numerator and denominator of each fraction by the missing factor of the common denominator preserves the value of the fraction.

Add the numerators together over the common denominator and simplify.

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

Combining the numerators and expanding the denominator gives the simplified equivalent expression.

Key Concept

Adding rational expressions by finding a common denominator

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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