Answer
a. $\frac{x - 2}{x - 1}$
The restrictions are $x \ne 1, 2$.
b. $\frac{x^2 - x - 4}{x^2 + 6x + 5}$
The restrictions are $x \ne -5, -1$.
Work Step by Step
a. Factor all expressions to their simplest forms:
$\frac{x + 3}{x - 2} - \frac{6x - 7}{(x - 2)(x - 1)}$
The least common denominator, or LCD, incorporates all factors in the denominators of the fractions. In this case, the LCD is $(x - 2)(x - 1)$.
Convert each fraction to an equivalent one by multiplying its numerator with whatever factor is missing between its denominator and the LCD:
$\frac{(x + 3)(x - 1)}{(x - 2)(x - 1)} - \frac{6x - 7}{(x - 2)(x - 1)}$
Use the FOIL method to distribute terms. With the FOIL method, we multiply the first terms first, then the outer terms, the inner terms, and finally, the last terms:
$\frac{(x)(x) + (x)(-1) + (3)(x) + (3)(-1)}{(x - 2)(x - 1)} - \frac{6x - 7}{(x - 2)(x - 1)}$
Multiply to simplify:
$\frac{x^2 - x + 3x - 3}{(x - 2)(x - 1)} - \frac{6x - 7}{(x - 2)(x - 1)}$
Combine like terms:
$\frac{x^2 + 2x - 3}{(x - 2)(x - 1)} - \frac{6x - 7}{(x - 2)(x - 1)}$
Subtract the fractions:
$\frac{x^2 + 2x - 3 - (6x - 7)}{(x - 2)(x - 1)}$
Simplify:
$\frac{x^2 - 4x + 4}{(x - 2)(x - 1)}$
Factor the numerator:
$\frac{(x - 2)(x - 2)}{(x - 2)(x - 1)}$
Cancel out common factors in the numerator and denominator:
$\frac{x - 2}{x - 1}$
Restrictions occur when the expression becomes undefined, meaning when the denominator is $0$.
To find the restrictions, set each expression in the denominators of the original rational expressions as well as in the reciprocal equal to zero and solve:
Restriction:
$x - 2 = 0$
Add $2$ to each side of the equation:
$x = 2$
Restriction:
$x - 1 = 0$
Add $1$ to each side of the equation:
$x = 1$
Therefore, the restrictions are $x \ne 1, 2$.
b. Factor all expressions to their simplest forms:
$\frac{x - 1}{x + 5} - \frac{x + 3}{(x + 5)(x + 1)}$
The least common denominator, or LCD, incorporates all factors in the denominators of the fractions. In this case, the LCD is $(x + 5)(x + 1)$.
Convert each fraction to an equivalent one by multiplying its numerator with whatever factor is missing between its denominator and the LCD:
$\frac{(x - 1)(x + 1)}{(x + 5)(x + 1)} - \frac{x + 3}{(x + 5)(x + 1)}$
Use the FOIL method to distribute terms. With the FOIL method, we multiply the first terms first, then the outer terms, the inner terms, and finally, the last terms:
$\frac{(x)(x) + (x)(1) + (-1)(x) + (-1)(1)}{(x + 5)(x + 1)} - \frac{x + 3}{(x + 5)(x + 1)}$
Multiply to simplify:
$\frac{x^2 + x - x - 1}{(x + 5)(x + 1)} - \frac{x + 3}{(x + 5)(x + 1)}$
Combine like terms:
$\frac{x^2 - 1}{(x + 5)(x + 1)} - \frac{x + 3}{(x + 5)(x + 1)}$
Subtract the fractions:
$\frac{x^2 - 1 - (x + 3)}{(x + 5)(x + 1)}$
Simplify:
$\frac{x^2 - 1 - x - 3}{(x + 5)(x + 1)}$
Combine like terms:
$\frac{x^2 - x - 4}{(x + 5)(x + 1)}$
Since the numerator cannot be factored, use the FOIL method to expand the binomials in the denominator:
$\frac{x^2 - x - 4}{x^2 + 6x + 5}$
Restrictions occur when the expression becomes undefined, meaning when the denominator is $0$.
To find the restrictions, set each expression in the denominators of the original rational expressions as well as in the reciprocal equal to zero and solve:
Restriction:
$x + 5 = 0$
Subtract $5$ from each side of the equation:
$x = -5$
Restriction:
$x + 1 = 0$
Subtract $1$ from each side of the equation:
$x = -1$
Therefore, the restrictions are $x \ne -5, -1$.
