$$
begin{align*}
text{a);;} dfrac{1}{3}+dfrac{5}{4}&=dfrac{1(4)+5(3)}{3(4)}\\&=dfrac{4+15}{12}\\&=dfrac{19}{12}end{align*}
$$

Use general the rule:

$$
dfrac{a}{b}+dfrac{c}{d}=dfrac{ad+bc}{bd}
$$

$$
begin{align*} {text{b)};;;} dfrac{2x}{5}+dfrac{6x}{2}&=dfrac{(2x)(2)+(6x)(5)}{5(2)}\\&=dfrac{4x+30x}{10}\\&=dfrac{34x}{10}\\&=dfrac{2(17x)}{2(5)}\\&=dfrac{17x}{5}end{align*}
$$

Use the general the rule:

$$
dfrac{a}{b}+dfrac{c}{d}=dfrac{ad+bc}{bd}
$$

c) $dfrac{5}{4x^2}+dfrac{1}{7x^3}$

$=dfrac{5(7x)}{(4x^2)(7x)}+dfrac{1(4)}{(7x^3)(4)}$

$=dfrac{5(7x)}{28x^3}+dfrac{4}{28x^3}$

$=dfrac{35x+4}{28x^3}$ ; $xneq 0$

Make the denominator the same by multiplying both numerator and denominator with a certain factor, then perform the operation in the numerator.

$$
dfrac{a}{c}pm dfrac{b}{c}= dfrac{apm b}{c}
$$

d) $dfrac{2}{x}+dfrac{6}{x^2}$

$=dfrac{2(x)}{x(x)}+dfrac{6}{x^2}$

$=dfrac{2x}{x^2}+dfrac{6}{x^2}$

$=dfrac{2x+6}{x^2}$ ; $xneq 0$

Make the denominator the same by multiplying both numerator and denominator with a certain factor, then perform the operation in the numerator.

$$
dfrac{a}{c}pm dfrac{b}{c}= dfrac{apm b}{c}
$$

a) $dfrac{19}{12}$

b) $dfrac{17x}{5}$

c) $dfrac{35x+4}{28x^3}$ ; $xneq 0$

d) $dfrac{2x+6}{x^2}$ ; $xneq 0$

a) $dfrac{5}{9}-dfrac{2}{3}$

$=dfrac{5}{9}-dfrac{2(3)}{3(3)}$

$=dfrac{5}{9}-dfrac{6}{9}$

$=dfrac{5-6}{9}$

$$
=-dfrac{1}{9}
$$

Make the denominator the same by multiplying both numerator and denominator with a certain factor, then perform the operation in the numerator.

$$
dfrac{a}{c}pm dfrac{b}{c}= dfrac{apm b}{c}
$$

b) $dfrac{5y}{3}-dfrac{y}{2}$

$=dfrac{(5y)(2)}{3(2)}-dfrac{y(3)}{2(3)}$

$=dfrac{(5y)(2)}{6}-dfrac{3y}{6}$

$=dfrac{5y-3y}{6}$

$=dfrac{2y}{6}$

$=dfrac{2y}{2(3)}$

$=dfrac{y}{3}$

Make the denominator the same by multiplying both numerator and denominator with a certain factor, then perform the operation in the numerator.

$$
dfrac{a}{c}pm dfrac{b}{c}= dfrac{apm b}{c}
$$

c) $dfrac{5}{3x^2}-dfrac{7}{5}$

$=dfrac{5(5)-7(3x^2)}{(3x^2)(5)}$

$=dfrac{25-21x^2}{15x^2}$

Use the general rule for rational expressions:

$$
dfrac{a}{b}-dfrac{c}{d}=dfrac{ad-bc}{bd}
$$

d) $dfrac{6}{3xy}-dfrac{5}{y^2}$

$=dfrac{6(y)}{3xy(y)}-dfrac{5(3x)}{y^2(3x)}$

$=dfrac{6y}{3xy^2}-dfrac{5(3x)}{3xy^2}$

$$
=dfrac{6y-15x}{3xy^2}
$$

Make the denominator the same by multiplying both numerator and denominator with a certain factor, then perform the operation in the numerator.

$$
dfrac{a}{c}pm dfrac{b}{c}= dfrac{apm b}{c}
$$

a) $-dfrac{-1}{9}$

b) $dfrac{7y}{6}$

c) $dfrac{25-21x^2}{15x^2}$ ; $xneq 0$

d) $dfrac{6y-15x}{3xy^2}$ , $yneq 0$

a) Using the $LCD=
(x-3)(5x-1)
,$ the given expression, $dfrac{3}{x-3}-dfrac{7}{5x-1}
,$ is equivalent to

$$
begin{align*}
&
dfrac{3}{x-3}cdotdfrac{5x-1}{5x-1}-dfrac{7}{5x-1}cdotdfrac{x-3}{x-3}
\\&=
dfrac{3(5x-1)}{(x-3)(5x-1)}-dfrac{7(x-3)}{(5x-1)(x-3)}
\\&=
dfrac{15x-3-7x+21}{(x-3)(5x-1)}
\\&=
dfrac{8x+18}{(x-3)(5x-1)}
\\&=
dfrac{2(4x+9)}{(x-3)(5x-1)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{2(4x+9)}{(x-3)(5x-1)}
text{, }xneleft{ dfrac{1}{5},3 right}
.$

b) The factored form of the given expression, $dfrac{2}{x+3}+dfrac{7}{x^2-9}
,$ is

$$
begin{align*}
dfrac{2}{x+3}+dfrac{7}{(x+3)(x-3)}
.end{align*}
$$

Using the $LCD=
(x+3)(x-3)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{2}{x+3}cdotdfrac{x-3}{x-3}+dfrac{7}{(x+3)(x-3)}
\\&=
dfrac{2(x-3)+7}{(x+3)(x-3)}
\\&=
dfrac{2x-6+7}{(x+3)(x-3)}
\\&=
dfrac{2x+1}{(x+3)(x-3)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{2x+1}{(x+3)(x-3)}
text{, }xneleft{ -3,3 right}
.$

c) The factored form of the given expression, $dfrac{5}{x^2-4x+3}-dfrac{9}{x^2-2x+1}
,$ is

$$
begin{align*}
dfrac{5}{(x-3)(x-1)}-dfrac{9}{(x-1)(x-1)}
.end{align*}
$$

Using the $LCD=
(x-1)(x-1)(x-3)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{5}{(x-3)(x-1)}cdotdfrac{x-1}{x-1}-dfrac{9}{(x-1)(x-1)}cdotdfrac{x-3}{x-3}
\\&=
dfrac{5(x-1)-9(x-3)}{(x-3)(x-1)(x-1)}
\\&=
dfrac{5x-5-9x+27}{(x-3)(x-1)(x-1)}
\\&=
dfrac{-4x+22}{(x-3)(x-1)(x-1)}
\\&=
dfrac{-2(2x-11)}{(x-3)(x-1)(x-1)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{-2(2x-11)}{(x-3)(x-1)(x-1)}
text{, }xneleft{ 1,3 right}
.$

a) $dfrac{2(4x+9)}{(x-3)(5x-1)}
text{, }xneleft{ dfrac{1}{5},3 right}$

b) $dfrac{2x+1}{(x+3)(x-3)}
text{, }xneleft{ -3,3 right}$

c) $dfrac{-2(2x-11)}{(x-3)(x-1)(x-1)}
text{, }xneleft{ 1,3 right}$

a) Substituting $x=5$ in the given expression, $dfrac{2}{(x^2-9)}+dfrac{3}{(x-3)}
,$ results to

$$
begin{align*}
&
dfrac{2}{(5^2-9)}+dfrac{3}{(5-3)}
\\&=
dfrac{2}{25-9}+dfrac{3}{2}
\\&=
dfrac{2}{16}+dfrac{3}{2}
\\&=
dfrac{1}{8}+dfrac{3}{2}
.end{align*}
$$

Using the $LCD=
8
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{1}{8}+dfrac{3}{2}cdotdfrac{4}{4}
\\&=
dfrac{1}{8}+dfrac{12}{8}
\\&=
dfrac{13}{8}
.end{align*}
$$

b) The factored form of the given expression, $dfrac{2}{(x^2-9)}+dfrac{3}{(x-3)}
,$ is

$$
begin{align*}
&
dfrac{2}{(x+3)(x-3)}+dfrac{3}{(x-3)}
.end{align*}
$$

Using the $LCD=
(x+3)(x-3)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{2}{(x+3)(x-3)}+dfrac{3}{(x-3)}cdotdfrac{x+3}{x=3}
\\&=
dfrac{2+3(x+3)}{(x+3)(x-3)}
\\&=
dfrac{2+3x+9}{(x+3)(x-3)}
\\&=
dfrac{3x+11}{(x+3)(x-3)}
.end{align*}
$$

c) Substituting $x=5$ in the expression of Item b, $dfrac{3x+11}{(x+3)(x-3)}
,$ results to

$$
begin{align*}
&
dfrac{3(5)+11}{(5+3)(5-3)}
\\&=
dfrac{15+11}{(8)(2)}
\\&=
dfrac{26}{16}
\\&=
dfrac{cancel2(13)}{cancel2(8)}
\\&=
dfrac{13}{8}
.end{align*}
$$

The value $dfrac{13}{8}$ is the same value derived in Item a.

a) $dfrac{13}{8}$

b) $dfrac{3x+11}{(x+3)(x-3)}$

c) same value

a) Using the $LCD=
12
,$ the given expression, $dfrac{2x}{3}+dfrac{3x}{4}-dfrac{x}{6}
,$ is equivalent to

$$
begin{align*}
&
dfrac{2x}{3}cdotdfrac{4}{4}+dfrac{3x}{4}cdotdfrac{3}{3}-dfrac{x}{6}cdotdfrac{2}{2}
\\&=
dfrac{8x}{12}+dfrac{9x}{12}-dfrac{2x}{12}
\\&=
dfrac{15x}{12}
\\&=
dfrac{cancel3cdot5x}{cancel3cdot4}
\\&=
dfrac{5x}{4}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{5x}{4}
text{, no restriction}
.$

b) Using the $LCD=
10t^4
,$ the given expression, $dfrac{3}{t^4}+dfrac{1}{2t^2}-dfrac{3}{5t}
,$ is equivalent to

$$
begin{align*}
&
dfrac{3}{t^4}cdotdfrac{10}{10}+dfrac{1}{2t^2}cdotdfrac{5t^2}{5t^2}-dfrac{3}{5t}cdotdfrac{2t^3}{2t^3}
\\&=
dfrac{30}{10t^4}+dfrac{5t^2}{10t^4}-dfrac{6t^3}{10t^4}
\\&=
dfrac{-6t^3+5t^2+30}{10t^4}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{-6t^3+5t^2+30}{10t^4}
text{, }tne0
.$

c) Converting the terms to an equivalent fraction that uses the $LCD=
60y^4
,$ the given expression, $dfrac{2x}{3y}-dfrac{x^2}{4y^3}+dfrac{3}{5y^4}
,$ is equivalent to

$$
begin{align*}
&
dfrac{2x}{3y}cdotdfrac{20y^3}{20y^3}-dfrac{x^2}{4y^3}cdotdfrac{15y}{15y}+dfrac{3}{5y^4}cdotdfrac{12}{12}
\\&=
dfrac{40xy^3}{60y^4}-dfrac{15x^2y}{60y^4}+dfrac{36}{60y^4}
\\&=
dfrac{-15x^2y+40xy^3+36}{60y^4}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{-15x^2y+40xy^3+36}{60y^4}
text{, }yne0
.$

d) Converting the terms to an equivalent fraction that uses the $LCD=
mn
,$ the given expression, $dfrac{n}{m}+dfrac{m}{n}-m
,$ is equivalent to

$$
begin{align*}
&
dfrac{n}{m}cdotdfrac{n}{n}+dfrac{m}{n}cdotdfrac{m}{m}-mcdotdfrac{mn}{mn}
\\&=
dfrac{n^2}{mn}+dfrac{m^2}{mn}-dfrac{m^2n}{mn}
\\&=
dfrac{n^2+m^2-m^2n}{mn}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{n^2+m^2-m^2n}{mn}
text{, }mne0text{, }nne0
.$

a) $dfrac{5x}{4}
text{, no restriction}$

b) $dfrac{-6t^3+5t^2+30}{10t^4}
text{, }tne0$

c) $dfrac{-15x^2y+40xy^3+36}{60y^4}
text{, }yne0$

d) $dfrac{n^2+m^2-m^2n}{mn}
text{, }mne0text{, }nne0$

a) Converting the terms to an equivalent fraction that uses the $LCD=
a(a-4)
,$ the given expression, $dfrac{7}{a-4}+dfrac{2}{a}
,$ is equivalent to

$$
begin{align*}
&
dfrac{7}{a-4}cdotdfrac{a}{a}+dfrac{2}{a}cdotdfrac{a-4}{a-4}
\\&=
dfrac{7a}{a(a-4)}+dfrac{2a-8}{a(a-4)}
\\&=
dfrac{7a+2a-8}{a(a-4)}
\\&=
dfrac{9a-8}{a(a-4)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{9a-8}{a(a-4)}
text{, }aneleft{ 0,4 right}
.$

b) Converting the terms to an equivalent fraction that uses the $LCD=
3x-2
,$ the given expression, $dfrac{4}{3x-2}+6
,$ is equivalent to

$$
begin{align*}
&
dfrac{4}{3x-2}+6cdotdfrac{3x-2}{3x-2}
\\&=
dfrac{4}{3x-2}+dfrac{18x-12}{3x-2}
\\&=
dfrac{4+18x-12}{3x-2}
\\&=
dfrac{18x-8}{3x-2}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{18x-8}{3x-2}
text{, }xnedfrac{2}{3}
.$

c) Converting the terms to an equivalent fraction that uses the $LCD=
(x+4)(x+3)
,$ the given expression, $dfrac{5}{x+4}+dfrac{7}{x+3}
,$ is equivalent to

$$
begin{align*}
&
dfrac{5}{x+4}cdotdfrac{x+3}{x+3}+dfrac{7}{x+3}cdotdfrac{x+4}{x+4}
\\&=
dfrac{5x+15}{(x+4)(x+3)}+dfrac{7x+28}{(x+4)(x+3)}
\\&=
dfrac{12x+43}{(x+4)(x+3)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{12x+43}{(x+4)(x+3)}
text{, }xneleft{ -4,-3 right}
.$

d) Converting the terms to an equivalent fraction that uses the $LCD=
(2n-3)(n-5)
,$ the given expression, $dfrac{6}{2n-3}-dfrac{4}{n-5}
,$ is equivalent to

$$
begin{align*}
&
dfrac{6}{2n-3}cdotdfrac{n-5}{n-5}-dfrac{4}{n-5}cdotdfrac{2n-3}{2n-3}
\\&=
dfrac{6n-30}{(2n-3)(n-5)}-dfrac{8n-12}{(2n-3)(n-5)}
\\&=
dfrac{6n-30-8n+12}{(2n-3)(n-5)}
\\&=
dfrac{-2n-18}{(2n-3)(n-5)}
\\&=
dfrac{-2(n+9)}{(2n-3)(n-5)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{-2(n+9)}{(2n-3)(n-5)}
text{, }nneleft{ dfrac{3}{2},5 right}
.$

e) Converting the terms to an equivalent fraction that uses the $LCD=
(x+4)(x-6)
,$ the given expression, $dfrac{7x}{x+4}+dfrac{3x}{x-6}
,$ is equivalent to

$$
begin{align*}
&
dfrac{7x}{x+4}cdotdfrac{x-6}{x-6}+dfrac{3x}{x-6}cdotdfrac{x+4}{x+4}
\\&=
dfrac{7x^2-42x}{(x+4)(x-6)}+dfrac{3x^2+12x}{(x+4)(x-6)}
\\&=
dfrac{10x^2-30x}{(x+4)(x-6)}
\\&=
dfrac{10x(x-3)}{(x+4)(x-6)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{10x(x-3)}{(x+4)(x-6)}
text{, }nneleft{ -4,6 right}
.$

f) The factored form of the given expression, $dfrac{7}{2x-6}+dfrac{4}{10x-15}
,$ is

$$
begin{align*}
&
dfrac{7}{2(x-3)}+dfrac{4}{5(2x-3)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
10(x-3)(2x-3)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{7}{2(x-3)}cdotdfrac{5(2x-3)}{5(2x-3)}+dfrac{4}{5(2x-3)}cdotdfrac{2(x-3)}{2(x-3)}
\\&=
dfrac{35(2x-3)}{10(x-3)(2x-3)}+dfrac{8(2x-3)}{10(x-3)(2x-3)}
\\&=
dfrac{70x-105}{10(x-3)(2x-3)}+dfrac{16x-24}{10(x-3)(2x-3)}
\\&=
dfrac{86x-129}{10(x-3)(2x-3)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{86x-129}{10(x-3)(2x-3)}
text{, }xneleft{ dfrac{3}{2},3 right}
.$

a) $dfrac{9a-8}{a(a-4)}
text{, }aneleft{ 0,4 right}$

b) $dfrac{18x-8}{3x-2}
text{, }xnedfrac{2}{3}$

c) $dfrac{12x+43}{(x+4)(x+3)}
text{, }xneleft{ -4,-3 right}$

d) $dfrac{-2(n+9)}{(2n-3)(n-5)}
text{, }nneleft{ dfrac{3}{2},5 right}$

e) $dfrac{10x(x-3)}{(x+4)(x-6)}
text{, }nneleft{ -4,6 right}$

f) $dfrac{86x-129}{10(x-3)(2x-3)}
text{, }xneleft{ dfrac{3}{2},3 right}$

a) The factored form of the given expression, $dfrac{3}{x+1}+dfrac{4}{x^2-3x-4}
,$ is

$$
begin{align*}
&
dfrac{3}{x+1}+dfrac{4}{(x-4)(x+1)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
(x+1)(x-4)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{3}{x+1}cdotdfrac{x-4}{x-4}+dfrac{4}{(x-4)(x+1)}
\\&=
dfrac{3x-12}{(x+1)(x-4)}+dfrac{4}{(x+1)(x-4)}
\\&=
dfrac{3x-8}{(x+1)(x-4)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{3x-8}{(x+1)(x-4)}
text{, }xneleft{ -1,4 right}
.$

b) The factored form of the given expression, $dfrac{2t}{t-4}-dfrac{5t}{t^2-16}
,$ is

$$
begin{align*}
&
dfrac{2t}{t-4}-dfrac{5t}{(t-4)(t+4)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
(t-4)(t+4)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{2t}{t-4}cdotdfrac{t+4}{t+4}-dfrac{5t}{(t-4)(t+4)}
\\&=
dfrac{2t^2+8t}{(t-4)(t+4)}-dfrac{5t}{(t-4)(t+4)}
\\&=
dfrac{2t^2+3t}{(t-4)(t+4)}
\\&=
dfrac{t(2t+3)}{(t-4)(t+4)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{t(2t+3)}{(t-4)(t+4)}
text{, }tneleft{ -4,4 right}
.$

c) The factored form of the given expression, $dfrac{3}{t^2+t-6}+dfrac{5}{(t+3)^2}
,$ is

$$
begin{align*}
&
dfrac{3}{(t+3)(t-2)}+dfrac{5}{(t+3)(t+3)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
(t+3)(t+3)(t-2)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{3}{(t+3)(t-2)}cdotdfrac{t+3}{t+3}+dfrac{5}{(t+3)(t+3)}cdotdfrac{t-2}{t-2}
\\&=
dfrac{3t+9}{(t+3)(t+3)(t-2)}+dfrac{5t-10}{(t+3)(t+3)(t-2)}
\\&=
dfrac{8t-1}{(t+3)(t+3)(t-2)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{8t-1}{(t+3)(t+3)(t-2)}
text{, }tneleft{ -3,2 right}
.$

d) The factored form of the given expression, $dfrac{4x}{x^2+6x+8}-dfrac{3x}{x^2-3x-10}
,$ is

$$
begin{align*}
&
dfrac{4x}{(x+4)(x+2)}-dfrac{3x}{(x-5)(x+2)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
(x+4)(x+2)(x-5)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{4x}{(x+4)(x+2)}cdotdfrac{x-5}{x-5}-dfrac{3x}{(x-5)(x+2)}cdotdfrac{x+4}{x+4}
\\&=
dfrac{4x^2-20x}{(x+4)(x+2)(x-5)}-dfrac{3x^2+12x}{(x+4)(x+2)(x-5)}
\\&=
dfrac{x^2-32x}{(x+4)(x+2)(x-5)}
\\&=
dfrac{x(x-32)}{(x+4)(x+2)(x-5)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{x(x-32)}{(x+4)(x+2)(x-5)}
text{, }xneleft{ -4,-2,5 right}
.$

e) The factored form of the given expression, $dfrac{x-1}{x^2-9}+dfrac{x+7}{x^2-5x+6}
,$ is

$$
begin{align*}
&
dfrac{x-1}{(x+3)(x-3)}+dfrac{x+7}{(x-3)(x-2)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
(x+3)(x-3)(x-2)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{x-1}{(x+3)(x-3)}cdotdfrac{x-2}{x-2}+dfrac{x+7}{(x-3)(x-2)}cdotdfrac{x+3}{x+3}
\\&=
dfrac{(x-1)(x-2)}{(x+3)(x-3)(x-2)}+dfrac{(x+7)(x+3)}{(x+3)(x-3)(x-2)}
\\&=
dfrac{(x^2-3x+2)+(x^2+10x+21)}{(x+3)(x-3)(x-2)}
\\&=
dfrac{2x^2+7x+23}{(x+3)(x-3)(x-2)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{2x^2+7x+23}{(x+3)(x-3)(x-2)}
text{, }xneleft{ -3,2,3 right}
.$

f) The factored form of the given expression, $dfrac{2t+1}{2t^2-14t+24}+dfrac{5t}{4t^2-8t-12}
,$ is

$$
begin{align*}
&
dfrac{2t+1}{2(t^2-7t+12)}+dfrac{5t}{4(t^2-2t-3)}
\\&=
dfrac{2t+1}{2(t-3)(t-4)}+dfrac{5t}{4(t-3)(t+1)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
4(t-3)(t-4)(t+1)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{2t+1}{2(t-3)(t-4)}cdotdfrac{2(t+1)}{2(t+1)}+dfrac{5t}{4(t-3)(t+1)}cdotdfrac{t-4}{t-4}
\\&=
dfrac{2(2t+1)(t+1)}{4(t-3)(t-4)(t+1)}+dfrac{5t(t-4)}{4(t-3)(t-4)(t+1)}
\\&=
dfrac{2(2t^2+3t+1)}{4(t-3)(t-4)(t+1)}+dfrac{5t^2-20t}{4(t-3)(t-4)(t+1)}
\\&=
dfrac{4t^2+6t+2}{4(t-3)(t-4)(t+1)}+dfrac{5t^2-20t}{4(t-3)(t-4)(t+1)}
\\&=
dfrac{9t^2-14t+2}{4(t-3)(t-4)(t+1)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{9t^2-14t+2}{4(t-3)(t-4)(t+1)}
text{, }tneleft{ -1,3,4 right}
.$

a) $dfrac{3x-8}{(x+1)(x-4)}
text{, }xneleft{ -1,4 right}$

b) $dfrac{t(2t+3)}{(t-4)(t+4)}
text{, }tneleft{ -4,4 right}$

c) $dfrac{8t-1}{(t+3)(t+3)(t-2)}
text{, }tneleft{ -3,2 right}$

d) $dfrac{x(x-32)}{(x+4)(x+2)(x-5)}
text{, }xneleft{ -4,-2,5 right}$

e) $dfrac{2x^2+7x+23}{(x+3)(x-3)(x-2)}
text{, }xneleft{ -3,2,3 right}$

f) $dfrac{9t^2-14t+2}{4(t-3)(t-4)(t+1)}
text{, }tneleft{ -1,3,4 right}$

a) The factored form of the given expression, $dfrac{3}{4x^2+7x+3}-dfrac{5}{16x^2+24x+9}
,$ is

$$
begin{align*}
&
dfrac{3}{(4x+3)(x+1)}-dfrac{5}{(4x+3)(4x+3)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
(4x+3)(4x+3)(x+1)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{3}{(4x+3)(x+1)}cdotdfrac{4x+3}{4x+3}-dfrac{5}{(4x+3)(4x+3)}cdotdfrac{x+1}{x+1}
\\&=
dfrac{12x+9}{(4x+3)(4x+3)(x+1)}-dfrac{5x+5}{(4x+3)(4x+3)(x+1)}
\\&=
dfrac{7x+4}{(4x+3)(4x+3)(x+1)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{7x+4}{(4x+3)(4x+3)(x+1)}
text{, }xneleft{ -1,-dfrac{3}{4} right}
.$

b) The factored form of the given expression, $dfrac{a-1}{a^2-8a+15}-dfrac{a-2}{2a^2-9a-5}
,$ is

$$
begin{align*}
&
dfrac{a-1}{(a-3)(a-5)}-dfrac{a-2}{(2a+1)(a-5)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
(a-3)(a-5)(2a+1)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{a-1}{(a-3)(a-5)}cdotdfrac{2a+1}{2a+1}-dfrac{a-2}{(2a+1)(a-5)}cdotdfrac{a-3}{a-3}
\\&=
dfrac{(a-1)(2a+1)}{(a-3)(a-5)(2a+1)}-dfrac{(a-2)(a-3)}{(a-3)(a-5)(2a+1)}
\\&=
dfrac{(2a^2-a-1)-(a^2-5a+6)}{(a-3)(a-5)(2a+1)}
\\&=
dfrac{2a^2-a-1-a^2+5a-6}{(a-3)(a-5)(2a+1)}
\\&=
dfrac{a^2+4a-7}{(a-3)(a-5)(2a+1)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{a^2+4a-7}{(a-3)(a-5)(2a+1)}
text{, }aneleft{ -dfrac{1}{2},3,5 right}
.$

c) The factored form of the given expression, $dfrac{3x+2}{4x^2-1}+dfrac{2x-5}{4x^2+4x+1}
,$ is

$$
begin{align*}
&
dfrac{3x+2}{(2x+1)(2x-1)}+dfrac{2x-5}{(2x+1)(2x+1)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
(2x+1)(2x+1)(2x-1)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{3x+2}{(2x+1)(2x-1)}cdotdfrac{2x+1}{2x+1}+dfrac{2x-5}{(2x+1)(2x+1)}cdotdfrac{2x-1}{2x-1}
\\&=
dfrac{(3x+2)(2x+1)}{(2x+1)(2x+1)(2x-1)}+dfrac{(2x-5)(2x-1)}{(2x+1)(2x+1)(2x-1)}
\\&=
dfrac{(6x^2+7x+2)+(4x^2-12x+5)}{(2x+1)(2x+1)(2x-1)}
\\&=
dfrac{10x^2-5x+7}{(2x+1)(2x+1)(2x-1)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{10x^2-5x+7}{(2x+1)(2x+1)(2x-1)}
text{, }xneleft{ -dfrac{1}{2},dfrac{1}{2} right}
.$

a) $dfrac{7x+4}{(4x+3)(4x+3)(x+1)}
text{, }xneleft{ -1,-dfrac{3}{4} right}$

b) $dfrac{a^2+4a-7}{(a-3)(a-5)(2a+1)}
text{, }aneleft{ -dfrac{1}{2},3,5 right}$

c) $dfrac{10x^2-5x+7}{(2x+1)(2x+1)(2x-1)}
text{, }xneleft{ -dfrac{1}{2},dfrac{1}{2} right}$

a) The given expression, $dfrac{2x^3}{3y^2}timesdfrac{9y}{10x}-dfrac{2y}{3x}
,$ is equivalent to

$$
begin{align*}
&
dfrac{2x^3(9y)}{3y^2(10x)}-dfrac{2y}{3x}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{cancel{2x}cdot x^2(cancel{3y}cdot3)}{cancel{3y}cdot y(cancel{2x}cdot5)}-dfrac{2y}{3x}
\\&=
dfrac{3x^2}{5y}-dfrac{2y}{3x}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
15xy
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{3x^2}{5y}cdotdfrac{3x}{3x}-dfrac{2y}{3x}cdotdfrac{5y}{5y}
\\&=
dfrac{9x^3}{15xy}-dfrac{10y^2}{15xy}
\\&=
dfrac{9x^3-10y^2}{15xy}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{9x^3-10y^2}{15xy}
text{, }xne0text{, }yne0
.$

b) The given expression, $dfrac{x+1}{2x-6}divdfrac{2(x+1)^2}{2-x}+dfrac{11}{x-2}
,$ is equivalent to

$$
begin{align*}
&
dfrac{x+1}{2x-6}cdotdfrac{2-x}{2(x+1)^2}+dfrac{11}{x-2}
\\&=
dfrac{x+1}{2(x-3)}cdotdfrac{-(x-2)}{2(x+1)(x+1)}+dfrac{11}{x-2}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{cancel{x+1}}{2(x-3)}cdotdfrac{-(x-2)}{2(cancel{x+1})(x+1)}+dfrac{11}{x-2}
\\&=
dfrac{-(x-2)}{4(x-3)(x+1)}+dfrac{11}{x-2}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
4(x-3)(x+1)(x-2)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{-(x-2)}{4(x-3)(x+1)}cdotdfrac{x-2}{x-2}+dfrac{11}{x-2}cdotdfrac{4(x-3)(x+1)}{4(x-3)(x+1)}
\\&=
dfrac{-(x^2-4x+4)}{4(x-3)(x+1)(x-2)}+dfrac{44(x^2-2x-3)}{4(x-3)(x+1)(x-2)}
\\&=
dfrac{-x^2+4x-4+44x^2-88x-132}{4(x-3)(x+1)(x-2)}
\\&=
dfrac{43x^2-84x-136}{4(x-3)(x+1)(x-2)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{43x^2-84x-136}{4(x-3)(x+1)(x-2)}
text{, }xne{-1,2,3}
.$

c) The given expression, $dfrac{p+1}{p^2+2p-35}+dfrac{p^2+p-12}{p^2-2p-24}timesdfrac{p^2-4p-12}{p^2+2p-15}
,$ is equivalent to

$$
begin{align*}
&
dfrac{p+1}{(p+7)(p-5)}+dfrac{(p+4)(p-3)}{(p-6)(p+4)}timesdfrac{(p-6)(p+2)}{(p+5)(p-3)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{p+1}{(p+7)(p-5)}+dfrac{(cancel{p+4})(cancel{p-3})}{(cancel{p-6})(cancel{p+4})}timesdfrac{(cancel{p-6})(p+2)}{(p+5)(cancel{p-3})}
\\&=
dfrac{p+1}{(p+7)(p-5)}+dfrac{p+2}{p+5}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
(p+7)(p-5)(p+5)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{p+1}{(p+7)(p-5)}cdotdfrac{p+5}{p+5}+dfrac{p+2}{p+5}cdotdfrac{(p+7)(p-5)}{(p+7)(p-5)}
\\&=
dfrac{p^2+6p+5}{(p+7)(p-5)(p+5)}+dfrac{p^3+4p^2-31p-70}{(p+7)(p-5)(p+5)}
\\&=
dfrac{p^3+5p^2-25p-65}{(p+7)(p-5)(p+5)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{p^3+5p^2-25p-65}{(p+7)(p-5)(p+5)}
text{, }pne{ -7,-5,5 }
.$

d) The given expression, $dfrac{5m-n}{2m+n}-dfrac{4m^2-4mn+n^2}{4m^2-n^2}divdfrac{6m^2-mn-n^2}{3m+15n}
,$ is equivalent to

$$
begin{align*}
&
dfrac{5m-n}{2m+n}-dfrac{4m^2-4mn+n^2}{4m^2-n^2}cdotdfrac{3m+15n}{6m^2-mn-n^2}
\\&=
dfrac{5m-n}{2m+n}-dfrac{(2m-n)(2m-n)}{(2m+n)(2m-n)}cdotdfrac{3(m+5n)}{(3m+n)(2m-n)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{5m-n}{2m+n}-dfrac{(cancel{2m-n})(cancel{2m-n})}{(2m+n)(cancel{2m-n})}cdotdfrac{3(m+5n)}{(3m+n)(cancel{2m-n})}
\\&=
dfrac{5m-n}{2m+n}-dfrac{3(m+5n)}{(2m+n)(3m+n)}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
(2m+n)(3m+n)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{5m-n}{2m+n}cdotdfrac{3m+n}{3m+n}-dfrac{3(m+5n)}{(2m+n)(3m+n)}
\\&=
dfrac{15m^2+2mn-n^2}{(2m+n)(3m+n)}-dfrac{3m+15n}{(2m+n)(3m+n)}
\\&=
dfrac{15m^2+2mn-n^2-3m-15n}{(2m+n)(3m+n)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{15m^2+2mn-n^2-3m-15n}{(2m+n)(3m+n)}
text{, }mneleft{ -dfrac{n}{2},-dfrac{n}{3} right}
.$

a) $dfrac{9x^3-10y^2}{15xy}
text{, }xne0text{, }yne0$

b) $dfrac{43x^2-84x-136}{4(x-3)(x+1)(x-2)}
text{, }xne{-1,2,3}$

c) $dfrac{p^3+5p^2-25p-65}{(p+7)(p-5)(p+5)}
text{, }pne{ -7,-5,5 }$

d) $dfrac{15m^2+2mn-n^2-3m-15n}{(2m+n)(3m+n)}
text{, }mneleft{ -dfrac{n}{2},-dfrac{n}{3} right}$

a) Converting the terms to an equivalent fraction that uses the $LCD=
10
,$ the given expression, $dfrac{3m+2}{2}+dfrac{4m+5}{5}
,$ is equivalent to

$$
begin{align*}
&
dfrac{3m+2}{2}cdotdfrac{5}{5}+dfrac{4m+5}{5}cdotdfrac{2}{2}
\\&=
dfrac{15m+10}{10}+dfrac{8m+10}{10}
\\&=
dfrac{23m+20}{10}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{23m+20}{10}
text{, no restriction}
.$

b) Converting the terms to an equivalent fraction that uses the $LCD=
4x^3
,$ the given expression, $dfrac{5}{x^2}-dfrac{3}{4x^3}
,$ is equivalent to

$$
begin{align*}
&
dfrac{5}{x^2}cdotdfrac{4x}{4x}-dfrac{3}{4x^3}
\\&=
dfrac{20x}{4x^3}-dfrac{3}{4x^3}
\\&=
dfrac{20x-3}{4x^3}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{20x-3}{4x^3}
text{, }xne0
.$

c) Converting the terms to an equivalent fraction that uses the $LCD=
(y+1)(y-2)
,$ the given expression, $dfrac{2}{y+1}-dfrac{3}{y-2}
,$ is equivalent to

$$
begin{align*}
&
dfrac{2}{y+1}cdotdfrac{y-2}{y-2}-dfrac{3}{y-2}cdotdfrac{y+1}{y+1}
\\&=
dfrac{2y-4}{(y+1)(y-2)}-dfrac{3y+3}{(y+1)(y-2)}
\\&=
dfrac{-y-7}{(y+1)(y-2)}
.end{align*}
$$

Hence, the simplified form of the given expression and its restrictions are $dfrac{-y-7}{(y+1)(y-2)}
text{, }yne{ -1,2 }
.$

d) The factored form of the given expression, $dfrac{2x}{x^2+x-6}+dfrac{5}{x^2+2x-8}
,$ is

$$
begin{align*}
&
dfrac{2x}{(x+3)(x-2)}+dfrac{5}{(x+4)(x-2)}
.end{align*}
$$

Using the $LCD=
(x+3)(x-2)(x+4)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{2x}{(x+3)(x-2)}cdotdfrac{x+4}{x+4}+dfrac{5}{(x+4)(x-2)}cdotdfrac{x+3}{x+3}
\\&=
dfrac{2x^2+8x}{(x+3)(x-2)(x+4)}+dfrac{5x+15}{(x+3)(x-2)(x+4)}
\\&=
dfrac{2x^2+13x+15}{(x+3)(x-2)(x+4)}
\\&=
dfrac{(2x+3)(x+5)}{(x+3)(x-2)(x+4)}
.end{align*}
$$

Since the denominator of the original equation cannot be zero, then the simplified form with its restriction is $dfrac{(2x+3)(x+5)}{(x+3)(x-2)(x+4)}
text{, }xneleft{ -4,-3,2 right}
.$

a) $dfrac{23m+20}{10}
text{, no restriction}$

b) $dfrac{20x-3}{4x^3}
text{, }xne0$

c) $dfrac{-y-7}{(y+1)(y-2)}
text{, }yne{ -1,2 }$

d) $dfrac{(2x+3)(x+5)}{(x+3)(x-2)(x+4)}
text{, }xneleft{ -4,-3,2 right}$

Solving for $R$ in the given equation, $dfrac{1}{R}=dfrac{1}{s}+dfrac{1}{t}
,$ results to

$$
begin{align*}
dfrac{1}{R}&=dfrac{1}{s}+dfrac{1}{t}
\\
Rstleft(dfrac{1}{R}right)&=left(dfrac{1}{s}+dfrac{1}{t}right)Rst
\\
st&=Rt+Rs
\\
st&=R(t+s)
\\
dfrac{st}{t+s}&=R
\\
R&=dfrac{st}{t+s}
.end{align*}
$$

With $s$ increased by $1$ unit and $t$ decreased by $1$ unit, the given equation becomes $dfrac{1}{R’}=dfrac{1}{s+1}+dfrac{1}{t-1}
.$ Solving for $R’$ results to

$$
begin{align*}
R'(s+1)(t-1)left( dfrac{1}{R’} right) &=left( dfrac{1}{s+1}+dfrac{1}{t-1} right) R'(s+1)(t-1)
\\
(s+1)(t-1)&=R'(t-1)+R'(s+1)
\\
st-s+t-1&=R'[(t-1)+(s+1)]
\\
st-s+t-1&=R'(t+s)
\\
dfrac{st-s+t-1}{t+s}&=R’
\\
R’&=dfrac{st-s+t-1}{t+s}
.end{align*}
$$

Subtracting $R’$ and $R,$ then the change in resistance is

$$
begin{align*}
&
dfrac{st-s+t-1}{t+s}-dfrac{st}{t+s}
\\&=
dfrac{st-s+t-1-st}{t+s}
\\&=
dfrac{-s+t-1}{t+s}
\\&=
dfrac{t-s-1}{t+s}
.end{align*}
$$

$$
dfrac{t-s-1}{t+s}
$$

a) Using $speed=dfrac{distance}{time},$ then

$$
begin{align*}
speed_1&=dfrac{2x}{3}
\\&text{and}\\
speed_2&=dfrac{x+100}{2}
.end{align*}
$$

The difference in the two speeds is

$$
begin{align*}
&
dfrac{2x}{3}-dfrac{x+100}{2}
.end{align*}
$$

Converting the terms to an equivalent fraction that uses the $LCD=
6
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{2x}{3}cdotdfrac{2}{2}-dfrac{x+100}{2}cdotdfrac{3}{3}
\\&=
dfrac{4x}{6}-dfrac{3x+300}{6}
\\&=
dfrac{x-300}{6}
.end{align*}
$$

b) With $speed_2>speed_1,$ then

$$
begin{align*}
dfrac{x+100}{2}&>dfrac{2x}{3}
.end{align*}
$$

Using the properties of inequality, the inequality above is equivalent to

$$
begin{align*}
6left( dfrac{x+100}{2} right) &>left( dfrac{2x}{3} right)6
\\
3(x+100)&>2x(2)
\
3x+300&>4x
\
3x-4x&>-300
\
-x&>-300
\
x&<(-300)(-1)
&text{ (reverse the inequality)}
\
x&<300
.end{align*}
$$

Hence, the speed for the second trip was greater than the speed for the first trip when $x<300
.$

a) $dfrac{x-300}{6}$

b) $x<300$

We are given with the formula for the sound intensity as a function of distance

$$
begin{equation*} I=frac{k}{d^2} end{equation*}
$$

When Matthew moves $x$ meters farther from the stage, the new intensity $I_2$ is

$$
begin{equation*} I_2=frac{k}{(d+x)^2}end{equation*}
$$

The decrease in intensity is $I-I_2$

$$
begin{equation*} I-I_2=frac{k}{(d)^2}-frac{k}{(d+x)^2}; ;; dneq 0, xneq dend{equation*}
$$

Use the general algebraic rule: $dfrac{a}{b}-dfrac{c}{d}=dfrac{ad-bc}{bd}$

$$
begin{align*} I_2-I&=frac{[k(d+x)^2]-k(d^2)}{d^2(d+x)^2}\
&=frac{[k(d^2+2dx+x^2)]-kd^2}{d^2(d+x)^2}\
&=frac{(kd^2+2kdx+kx^2)-kd^2}{d^2(d+x)^2}\
I_2-I&=frac{2kdx+kx^2}{d^2(d+x)^2} ; , ;dneq 0 , ; -x\
end{align*}
$$

$$
dfrac{2kdx+kx^2}{d^2(d+x)^2},; dneq 0,;-x
$$

a.i) An example of a pair of rational expressions where the lowest common denominator is one of the denominators is

$$
begin{align*}
dfrac{x}{6}
text{ and }
dfrac{2x}{3}
end{align*}
$$

since their $LCD$ is $6.$

a.ii) An example of a pair of rational expressions where the lowest common denominator is the product of the denominators is

$$
begin{align*}
dfrac{x}{2}
text{ and }
dfrac{2x}{3}
end{align*}
$$

since their $LCD$ is $2times3=6.$

a.iii) An example of a pair of rational expressions where the lowest common denominator is neither one of the denominators nor the product of the denominators is

$$
begin{align*}
dfrac{x}{4}
text{ and }
dfrac{x}{6}
end{align*}
$$

since their $LCD$ is $12
.$

b) To determine the $LCD$ of two simplified rational functions with different quadratic denominators, factor first the quadratic expressions. The $LCD$ is composed of the factors common to both quadratic equations (with highest exponent) and of factors not common to both denominators (with highest exponent.)

For example, the factored form of the rational expressions $dfrac{1}{x^2+2x+1}$ and $dfrac{1}{x^2+3x+2}$ is

$$
begin{align*}
dfrac{1}{(x+1)^2}
text{ and }
dfrac{1}{(x+1)(x+2)}
.end{align*}
$$

The $LCD$ is composed of the common factor (with the highest exponent), $(x+1)^2,$ and the factor not common to both denominators (with highest exponent), $(x+2).$ Hence, the $LCD$ is $(x+1)^2(x+2)
.$

a.i) $dfrac{x}{6}
text{ and }
dfrac{2x}{3}$

a.ii) $dfrac{x}{2}
text{ and }
dfrac{2x}{3}$

a.iii) $dfrac{x}{4}
text{ and }
dfrac{x}{6}$

b) see explanation and example

a) Combining the left side of the given equation, $dfrac{1}{n(n+1)}=dfrac{1}{n}-dfrac{1}{n+1}
,$ results to

$$
begin{align*}
dfrac{1}{n(n+1)}&=dfrac{1}{n}cdotdfrac{n+1}{n+1}-dfrac{1}{n+1}cdotdfrac{n}{n}
\\
dfrac{1}{n(n+1)}&=dfrac{n+1}{n(n+1)}-dfrac{n}{n+1}
\\
dfrac{1}{n(n+1)}&=dfrac{n+1-n}{n(n+1)}
\\
dfrac{1}{n(n+1)}&=dfrac{1}{n(n+1)}
text{ (TRUE)}
.end{align*}
$$

Since, the solution above ended with a TRUE statement, then the original equation is an identity. That is, the difference between reciprocals of consecutive positive integers is the reciprocal of their product.

b) Let $z=1,y=2,$ and $x=2.$ Then

$$
begin{align*}
dfrac{1}{2}&=dfrac{1}{1}-dfrac{1}{2}
\\
dfrac{1}{2}&=dfrac{2}{2}-dfrac{1}{2}
\\
dfrac{1}{2}&=dfrac{1}{2}
text{ (TRUE)}
.end{align*}
$$

Let $z=5,y=6,$ and $x=30.$ Then

$$
begin{align*}
dfrac{1}{30}&=dfrac{1}{5}-dfrac{1}{6}
\\
dfrac{1}{30}&=dfrac{1}{5}cdotdfrac{6}{6}-dfrac{1}{6}cdotdfrac{5}{5}
\\
dfrac{1}{30}&=dfrac{6}{30}-dfrac{5}{30}
\\
dfrac{1}{30}&=dfrac{1}{30}
text{ (TRUE)}
.end{align*}
$$

a) Let $a$ be the first integer, and $(a+2)$ be the next even/odd integer. Then

$$
begin{align*}
&
dfrac{1}{a}+dfrac{1}{a+2}
\\&=
dfrac{1}{a}cdotdfrac{a+2}{a+2}+dfrac{1}{a+2}cdotdfrac{a}{a}
\\&=
dfrac{a+2}{a(a+2)}+dfrac{a}{a(a+2)}
\\&=
dfrac{2a+2}{a(a+2)}
.end{align*}
$$

Showing that the sum of the squares of the numerator and the denominator is a perfect square results to

$$
begin{align*}
&
(2a+2)^2+(a(a+2))^2
\&=
(4a^2+8a+4)+(a^2(a^2+4a+4))
\&=
(4a^2+8a+4)+(a^4+4a^3+4a^2)
\&=
a^4+4a^3+8a^2+8a+4
\&=
a^4+8a^2+4+4a^3+8a
\&=
a^4+4a^2+4a^2+4+4a^3+8a
\&=
a^4+4a^2+4+4a^3+4a^2+8a
\&=
(a^2)^2+(2a)^2+(2)^2+2(a^2)(2a)+2(a^2)(2)+2(2a)(2)
\&=
(a^2+2a+2)^2
.end{align*}
$$

Hence, the sum of the squares of the numerator and the denominator is a perfect square.

The Pythagorean Triples are $2a+2,a(a+2),$ and $(a^2+2a+2)$. If $a=1,$ the triples are ${ 4,3,5 }$ since

$$
begin{align*}
2a+2&Rightarrow
2(1)+2Rightarrow4
\
a(a+2)&Rightarrow
1(1+2)Rightarrow3
\
a^2+2a+2&Rightarrow
1^2+2(1)+2Rightarrow5
.end{align*}
$$

a) $2x+2$, $x^2+2x$, and $x^2+2x+2$ are Pythagorean triples.

b) Example: when $x=1$, Pythagorean triple: 3, 4, 5