Question

c. calculate each sum and difference. describe any restriction(s) for the value of x and simplify the answer when possible.

1. \\(\frac{3}{x} + \frac{1}{x + 1}\\)
2. \\(\frac{2}{x - 2} - \frac{5}{x + 3}\\)
3. \\(\frac{x}{2x - 1} + \frac{x + 2}{x}\\)
4. \\(\frac{1}{x + 3} - \frac{1}{x - 3}\\)
5. \\(\frac{1}{x^2 - 4} - \frac{1}{x - 2}\\)
6. \\(\frac{x + 3}{x - 1} + \frac{x - 4}{x + 2}\\)
7. \\(\frac{x + 1}{x^2 - 16} - \frac{x}{x^2 + 7x + 12}\\)
8. \\(\frac{1}{x - 4} - \frac{x}{x + 2} + \frac{x}{x - 1}\\)
9. \\(\frac{x + 1}{x^2 - 3x - 4} + \frac{x - 3}{x - 2}\\)
10. \\(\frac{x + 2}{2x - 2} - \frac{-2x - 1}{x^2 - 4x + 3}\\)

Explanation:

1. $\boldsymbol{\frac{3}{x} + \frac{1}{x+1}}$

Step1: Find common denominator

Common denominator is $x(x+1)$

Step2: Rewrite fractions

$\frac{3(x+1)}{x(x+1)} + \frac{x}{x(x+1)}$

Step3: Combine numerators

$\frac{3(x+1)+x}{x(x+1)} = \frac{3x+3+x}{x(x+1)} = \frac{4x+3}{x(x+1)}$

Step4: State restrictions

Denominators cannot be zero: $x
eq 0$, $x
eq -1$

---

2. $\boldsymbol{\frac{2}{x-2} - \frac{5}{x+3}}$

Step1: Find common denominator

Common denominator is $(x-2)(x+3)$

Step2: Rewrite fractions

$\frac{2(x+3)}{(x-2)(x+3)} - \frac{5(x-2)}{(x-2)(x+3)}$

Step3: Combine numerators

$\frac{2(x+3)-5(x-2)}{(x-2)(x+3)} = \frac{2x+6-5x+10}{(x-2)(x+3)} = \frac{-3x+16}{(x-2)(x+3)}$

Step4: State restrictions

Denominators cannot be zero: $x
eq 2$, $x
eq -3$

---

3. $\boldsymbol{\frac{x}{2x-1} + \frac{x+2}{x}}$

Step1: Find common denominator

Common denominator is $x(2x-1)$

Step2: Rewrite fractions

$\frac{x^2}{x(2x-1)} + \frac{(x+2)(2x-1)}{x(2x-1)}$

Step3: Expand and combine numerators

$\frac{x^2 + 2x^2 -x +4x -2}{x(2x-1)} = \frac{3x^2+3x-2}{x(2x-1)}$

Step4: State restrictions

Denominators cannot be zero: $x
eq 0$, $x
eq \frac{1}{2}$

---

4. $\boldsymbol{\frac{1}{x+3} - \frac{1}{x-3}}$

Step1: Find common denominator

Common denominator is $(x+3)(x-3)$

Step2: Rewrite fractions

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

Step3: Combine numerators

$\frac{(x-3)-(x+3)}{(x+3)(x-3)} = \frac{x-3-x-3}{x^2-9} = \frac{-6}{x^2-9}$

Step4: State restrictions

Denominators cannot be zero: $x
eq -3$, $x
eq 3$

---

5. $\boldsymbol{\frac{1}{x^2-4} - \frac{1}{x-2}}$

Step1: Factor denominator

$x^2-4=(x-2)(x+2)$, common denominator is $(x-2)(x+2)$

Step2: Rewrite fractions

$\frac{1}{(x-2)(x+2)} - \frac{x+2}{(x-2)(x+2)}$

Step3: Combine numerators

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

Step4: State restrictions

Denominators cannot be zero: $x
eq 2$, $x
eq -2$

---

6. $\boldsymbol{\frac{x+3}{x-1} + \frac{x-4}{x+2}}$

Step1: Find common denominator

Common denominator is $(x-1)(x+2)$

Step2: Rewrite fractions

$\frac{(x+3)(x+2)}{(x-1)(x+2)} + \frac{(x-4)(x-1)}{(x-1)(x+2)}$

Step3: Expand and combine numerators

$\frac{x^2+5x+6 + x^2-5x+4}{(x-1)(x+2)} = \frac{2x^2+10}{(x-1)(x+2)} = \frac{2(x^2+5)}{(x-1)(x+2)}$

Step4: State restrictions

Denominators cannot be zero: $x
eq 1$, $x
eq -2$

---

7. $\boldsymbol{\frac{x+1}{x^2-16} - \frac{x}{x^2+7x+12}}$

Step1: Factor denominators

$x^2-16=(x-4)(x+4)$, $x^2+7x+12=(x+3)(x+4)$; common denominator is $(x-4)(x+4)(x+3)$

Step2: Rewrite fractions

$\frac{(x+1)(x+3)}{(x-4)(x+4)(x+3)} - \frac{x(x-4)}{(x-4)(x+4)(x+3)}$

Step3: Expand and combine numerators

$\frac{x^2+4x+3 -x^2+4x}{(x-4)(x+4)(x+3)} = \frac{8x+3}{(x-4)(x+4)(x+3)}$

Step4: State restrictions

Denominators cannot be zero: $x
eq 4$, $x
eq -4$, $x
eq -3$

---

8. $\boldsymbol{\frac{1}{x-4} - \frac{1}{x+2} + \frac{1}{x-1}}$

Step1: Find common denominator

Common denominator is $(x-4)(x+2)(x-1)$

Step2: Rewrite fractions

$\frac{(x+2)(x-1)}{(x-4)(x+2)(x-1)} - \frac{(x-4)(x-1)}{(x-4)(x+2)(x-1)} + \frac{(x-4)(x+2)}{(x-4)(x+2)(x-1)}$

Step3: Expand and combine numerators

$\begin{align*}
&\frac{x^2+x-2 - (x^2-5x+4) + x^2-2x-8}{(x-4)(x+2)(x-1)}\\
&=\frac{x^2+x-2-x^2+5x-4+x^2-2x-8}{(x-4)(x+2)(x-1)}\\
&=\frac{x^2+4x-14}{(x-4)(x+2)(x-1)}
\end{align*}$

Step4: State restrictions

Denominators cannot be zero: $x
eq 4$, $x
eq -2$, $x
eq 1$

---

9. $\boldsymbol{\frac{x+1}{x^2-3x-4} + \frac{x-3}{x-2}}$

Step1: Factor denominator

$x^2-3x-4=(x-4)(x+1)$; common denominator is $(x-4)(x+1)(x-2)$

Step2: Rewrite fractions

$\frac{x+1}{(x-4)(x+1)} + \frac{(x-3)(x-4)(x+1)}{(x-4)(x+1)(x-2)} = \frac{1}{x-4} + \frac{(x-3)(x-4)(x+1)}{(x-4)(x+1)(x-2)}$

Step3: Rewrite with common denominator

$\frac{(x+1)(x-2)}{(x-4)(x+1)(x-2)} + \frac{(x-3)(x-4)(x+1)}{(x-4)(x+1)(x-2)}$

Step4: Expand and combine numerators

$\begin{align*}
&\frac{x^2-x-2 + (x-3)(x^2-3x-4)}{(x-4)(x+1)(x-2)}\\
&=\frac{x^2-x-2 + x^3-3x^2-4x-3x^2+9x+12}{(x-4)(x+1)(x-2)}\\
&=\frac{x^3-5x^2+4x+10}{(x-4)(x+1)(x-2)}
\end{align*}$

Step5: State restrictions

Denominators cannot be zero: $x
eq 4$, $x
eq -1$, $x
eq 2$

---

10. $\boldsymbol{\frac{x+2}{2x-2} - \frac{-2x-1}{x^2-4x+3}}$

Step1: Factor denominators

$2x-2=2(x-1)$, $x^2-4x+3=(x-1)(x-3)$; common denominator is $2(x-1)(x-3)$

Step2: Rewrite fractions

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

Step3: Expand and combine numerators

$\begin{align*}
&\frac{x^2-x-6 +4x+2}{2(x-1)(x-3)}\\
&=\frac{x^2+3x-4}{2(x-1)(x-3)} = \frac{(x+4)(x-1)}{2(x-1)(x-3)} = \frac{x+4}{2(x-3)}
\end{align*}$

Step4: State restrictions

Denominators cannot be zero: $x
eq 1$, $x
eq 3$

Answer:

1. $\frac{4x+3}{x(x+1)}$, $x

eq 0, -1$

1. $\frac{-3x+16}{(x-2)(x+3)}$, $x

eq 2, -3$

1. $\frac{3x^2+3x-2}{x(2x-1)}$, $x

eq 0, \frac{1}{2}$

1. $\frac{-6}{x^2-9}$, $x

eq -3, 3$

1. $-\frac{x+1}{(x-2)(x+2)}$, $x

eq 2, -2$

1. $\frac{2(x^2+5)}{(x-1)(x+2)}$, $x

eq 1, -2$

1. $\frac{8x+3}{(x-4)(x+4)(x+3)}$, $x

eq 4, -4, -3$

1. $\frac{x^2+4x-14}{(x-4)(x+2)(x-1)}$, $x

eq 4, -2, 1$

1. $\frac{x^3-5x^2+4x+10}{(x-4)(x+1)(x-2)}$, $x

eq 4, -1, 2$

1. $\frac{x+4}{2(x-3)}$, $x

eq 1, 3$