In the following exercises involving multiplication or division of rational numbers, remember the following algebraic rules:

$$
begin{equation*} dfrac{a}{b}times dfrac{c}{d}=dfrac{ac}{bd} end{equation*}
$$

$$
begin{equation*}dfrac{a}{b}div dfrac{c}{d}=dfrac{a}{b}times dfrac{d}{c}=dfrac{ad}{bc} end{equation*}
$$

$$
begin{equation*}dfrac{ab}{ac}=dfrac{b}{c}end{equation*}
$$

$$
begin{equation*}x^atimes x^b=x^{a+b}end{equation*}
$$

$$
begin{equation*} dfrac{x^a}{x^b}=x^{a-b} end{equation*}
$$

The values of variables that would make the denominator zero in the original expression are the restrictions on the variable of the simplified expression.

$$
begin{align*}
bold{a);;} dfrac{2}{3}times dfrac{5}{8} &=dfrac{(2)(5)}{(3)(8)}\\ &=dfrac{10}{24}\\ &=dfrac{5(2)}{12(2)}\\ &=dfrac{5}{12}end{align*}
$$

$$
begin{align*}
bold{b);;} dfrac{6x^2y}{5y^3}times dfrac{xy}{8}&=dfrac{(6x^2y)(xy)}{(5y^3)(8)}\\&=dfrac{6x^3y^2}{40y^3}\\&=dfrac{6x^3y^{2-3}}{40}\\&=dfrac{6x^3}{40y}\\&=dfrac{(2)(3)x^3}{2(20)y}\\&=dfrac{3x^3}{20y};; ;;; y neq 0
end{align*}
$$

$$
begin{align*}
bold{c);;} frac{(x+1)(x-5)}{(x+4)}times frac{(x+4)}{2(x-5)} &= dfrac{(x+1)(x-5)(x+4)}{(x+4)(2)(x-5)}\\
&=dfrac{(x+1)cancel{(x-5)}cancel{(x+4)}}{cancel{(x+4)}(2)cancel{(x-5)}}\\&=dfrac{x+1}{2}; ;;xneq -4,;5end{align*}
$$

$$
begin{align*}
bold{d);;} frac{x^2}{2x+1}timesfrac{6x+3}{5x}&=frac{(x^2(6x+3)}{(2x+1)(5x)} \\
&=frac{x^2(3)(2x+1)}{(2x+1)(5x)}\\&=frac{x^2(3)cancel{(2x+1)}}{cancel{(2x+1)}(5x)}\\&=frac{3x^2}{5x}\\&=frac{3x^{2-1}}{5}\\&=frac{3x}{5} ;; ; ;; xneq -frac{1}{2}, ;0 end{align*}
$$

a) $dfrac{5}{12}$

b) $dfrac{3x^3}{20y}$ ; $yneq 0$

c) $dfrac{x+1}{2}$ , $xneq -4,;5$

d) $dfrac{3x}{5}$ , $xneq -dfrac{1}{2},;0$

In the following exercises involving multiplication or division of rational numbers, remember the following algebraic rules:

$$
begin{equation*} dfrac{a}{b}times dfrac{c}{d}=dfrac{ac}{bd} end{equation*}
$$

$$
begin{equation*}dfrac{a}{b}div dfrac{c}{d}=dfrac{a}{b}times dfrac{d}{c}=dfrac{ad}{bc} end{equation*}
$$

$$
begin{equation*}dfrac{ab}{ac}=dfrac{b}{c}end{equation*}
$$

$$
begin{equation*}x^atimes x^b=x^{a+b}end{equation*}
$$

$$
begin{equation*} dfrac{x^a}{x^b}=x^{a-b} end{equation*}
$$

The values of variables that would make the denominator zero in the original expression are the restrictions on the variable of the simplified expression.

$$
begin{align*}
bold{a);;} dfrac{2x}{3}div dfrac{x^2}{5} &= dfrac{2x}{3}times dfrac{5}{x^2} \\
&= dfrac{(2x)(5)}{3(x^2)}\\
&=dfrac{10x}{3x^2}\\
&=dfrac{10x^{1-2}}{3}\\
&=dfrac{10x^{-1}}{3}\\
&=dfrac{10}{3x} ;; ; ;; xneq 0
end{align*}
$$

$$
begin{align*}
bold{b);;;} frac{x-7}{10}divdfrac{2x-14}{25} &= dfrac{x-7}{10}timesdfrac{25}{2x-14} \\
&=dfrac{(x-7)(25)}{10(2x-14)}\\ &=frac{(x-7)(25)}{10(2)(x-7)}\\ &=frac{cancel{(x-7)}(25)}{10(2)cancel{(x-7)}}\
&=frac{25}{20}\\ &=frac{5(5)}{5(4)} \\&=frac{5}{4} ;; ;;; xneq 7
end{align*}
$$

$$
begin{align*}
bold{c);;} dfrac{3x(x-6)}{(x+2)(x-7)}div dfrac{x-6}{x+2} &=frac{3x(x-6)}{(x+2)(x-7)}times frac{(x+2)}{(x-6)} \\
&=dfrac{3x(x-6)(x+2)}{(x+2)(x-7)(x-6)}\\&=dfrac{3xcancel{(x-6)}cancel{(x+2)}}{cancel{(x+2)}(x-7)cancel{(x-6)}}\\
&=dfrac{3x}{x-7} ;; ; ;; xneq -2,7,6end{align*}
$$

$$
begin{align*}
bold{d);;} dfrac{x^2-1}{x-1}div dfrac{x+1}{12-6x}&=dfrac{x^2-1}{x-2}times dfrac{12-6x}{x+1} \\
&=dfrac{(x^2-1)(12-6x)}{(x-2)(x+1)}\\ &=dfrac{(x+1)(x-1)(6)(2-x)}{(x-2)(x+1)}\\ &=dfrac{cancel{(x+1)}(x-1)(6)(2-x)}{(x-2)cancel{(x+1)}}\\&=dfrac{(x-1)(6)(2-x)}{x-2}\\&=dfrac{(x-1)(-6)(x-2)}{x-2}\\&=dfrac{(x-1)(-6)cancel{(x-2)}}{cancel{(x-2)}}\\&=(-6)(x-1)\\&=6(1-x) ;; ; ;; xneq 2,;-1 end{align*}
$$

a) $dfrac{10}{3x}$ , $xneq 0$

b) $dfrac{5}{4}$ , $xneq 7$

c) $dfrac{3x}{x-7}$ , $xneq -2, 6,7$

d) $6(1-x)$ ; $xneq 2,-1$

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

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

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

$$
begin{align*}
&dfrac{(cancel{x+1})(x+1)}{(x+3)(cancel{x-1})}cdotdfrac{(cancel{x-1})(x-1)}{(x+3)(cancel{x+1})}
\\&=
dfrac{(x+1)(x-1)}{(x+3)(x+3)}
\\&=
dfrac{x^2-1}{x^2+6x+9}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{x^2-1}{x^2+6x+9}
text{, }xne{-3,-1,1,3}
.$

b) The factored form of the given expression, $dfrac{2x+10}{x^2-4x+4}divdfrac{x^2-25}{x-2}
,$ is

$$
begin{align*}
&
dfrac{2x+10}{x^2-4x+4}cdotdfrac{x-2}{x^2-25}
\\&=
dfrac{2(x+5)}{(x-2)(x-2)}cdotdfrac{x-2}{(x-5)(x+5)}
.end{align*}
$$

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

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

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{2}{(x-2)(x-5)}
text{, }xne{ -5,2,5}
.$

a) $dfrac{x^2-1}{x^2+6x+9}
text{, }xne{-3,-1,1,3}$

b) $dfrac{2}{(x-2)(x-5)}
text{, }xne{ -5,2,5}$

a) The factored form of the given expression, $dfrac{2x^2}{7}timesdfrac{21}{x}
,$ is

$$
begin{align*}
&
dfrac{2xcdot x}{7}timesdfrac{3cdot7}{x}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{2xcdot cancel{x}}{cancel7}timesdfrac{3cdotcancel7}{cancel{x}}
\\&=
6x
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $6x
text{, }xne0
.$

b) The factored form of the given expression, $dfrac{7a}{3}divdfrac{14a^2}{5}
,$ is

$$
begin{align*}
&
dfrac{7a}{3}cdotdfrac{5}{14a^2}
\\&=
dfrac{7a}{3}cdotdfrac{5}{7acdot2a}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{7a}}{3}cdotdfrac{5}{cancel{7a}cdot2a}
\\&=
dfrac{5}{6a}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{5}{6a}
text{, }ane0
.$

c) The factored form of the given expression, $dfrac{2x^3y}{3xy^2}timesdfrac{9x}{4x^2y}
,$ is

$$
begin{align*}
&
dfrac{2x^2ycdot x}{3xcdot y^2}timesdfrac{3xcdot3}{2x^2ycdot2}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{2x^2y}cdot x}{cancel{3x}cdot y^2}timesdfrac{cancel{3x}cdot3}{cancel{2x^2y}cdot2}
\\&=
dfrac{3x}{2y^2}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{3x}{2y^2}
text{, }xne0text{, }yne0
.$

d) The factored form of the given expression, $dfrac{3a^2b^3}{2ab^2}divdfrac{9a^2b}{14a^2}
,$ is

$$
begin{align*}
&
dfrac{3a^2b^3}{2ab^2}cdotdfrac{14a^2}{9a^2b}
\\&=
dfrac{3a^2bcdot b^2}{2acdot b^2}cdotdfrac{2acdot7a}{3a^2bcdot3}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{3a^2b}cdot cancel{b^2}}{cancel{2a}cdot cancel{b^2}}cdotdfrac{cancel{2a}cdot7a}{cancel{3a^2b}cdot3}
\\&=
dfrac{7a}{3}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{7a}{3}
text{, }ane0text{, }bne0
.$

a) $6x
text{, }xne0$

b) $dfrac{5}{6a}
text{, }ane0$

c) $dfrac{3x}{2y^2}
text{, }xne0text{, }yne0$

d) $dfrac{7a}{3}
text{, }ane0text{, }bne0$

a) The factored form of the given expression, $dfrac{2(x+1)}{3}timesdfrac{x-1}{6(x+1)}
,$ is

$$
begin{align*}
&
dfrac{2(x+1)}{3}cdotdfrac{x-1}{2(x+1)cdot3}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{2(x+1)}}{3}cdotdfrac{x-1}{cancel{2(x+1)}cdot3}
\\&=
dfrac{x-1}{9}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{x-1}{9}
text{, }xne-1
.$

b) The factored form of the given expression, $dfrac{3a-6}{a+2}divdfrac{a-2}{a+2}
,$ is

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

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

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

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $3
text{, }ane{ -2,2 }
.$

c) The factored form of the given expression, $dfrac{2(x-2)}{9x^3}timesdfrac{12x^4}{2-x}
,$ is

$$
begin{align*}
&
dfrac{2(x-2)}{3x^3cdot3}cdotdfrac{3x^3cdot4x}{-1(x-2)}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{2(cancel{x-2})}{cancel{3x^3}cdot3}cdotdfrac{cancel{3x^3}cdot4x}{-1(cancel{x-2})}
\\&=
dfrac{8x}{-3}
\\&=
-dfrac{8x}{3}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $-dfrac{8x}{3}
text{, }xne{ 0,2 }
.$

d) The factored form of the given expression, $dfrac{3(m+4)^2}{2m+1}divdfrac{5(m+4)}{7m+14}
,$ is

$$
begin{align*}
&
dfrac{3(m+4)^2}{2m+1}cdotdfrac{7m+14}{5(m+4)}
\\&=
dfrac{3(m+4)(m+4)}{2m+1}cdotdfrac{7(m+2)}{5(m+4)}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{3(m+4)(cancel{m+4})}{2m+1}cdotdfrac{7(m+2)}{5(cancel{m+4})}
\\&=
dfrac{3(m+4)(7)(m+2)}{5(2m+1)}
\\&=
dfrac{21(m+4)(m+2)}{5(2m+1)}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{21(m+4)(m+2)}{5(2m+1)}
text{, }mneleft{ -4,-2,-dfrac{1}{2} right}
.$

a) $dfrac{x-1}{9}
text{, }xne-1$

b) $3
text{, }ane{ -2,2 }$

c) $-dfrac{8x}{3}
text{, }xne{ 0,2 }$

d) $dfrac{21(m+4)(m+2)}{5(2m+1)}
text{, }mneleft{ -4,-2,-dfrac{1}{2} right}$

a) Cancelling the common factors between the numerator and the denominator, the given expression, $dfrac{(x+1)(x-3)}{(x+2)^2}timesdfrac{2(x+2)}{(x-3)(x+3)}
,$ is equivalent to

$$
begin{align*}
&
dfrac{(x+1)(cancel{x-3})}{(x+2)^{cancel21}}timesdfrac{2(cancel{x+2})}{(cancel{x-3})(x+3)}
\\&=
dfrac{2(x+1)}{(x+2)(x+3)}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{2(x+1)}{(x+2)(x+3)}
text{, }xne{ -3,-2,3}
.$

b) The factored form of the given expression, $dfrac{2(n^2-7n+12)}{n^2-n-6}divdfrac{5(n-4)}{n^2-4}
,$ is

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

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

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

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{2(n-2)}{5}
text{, }nneleft{ -2,2,3,4 right}
.$

c) The factored form of the given expression, $dfrac{2x^2-x-1}{x^2-x-6}timesdfrac{6x^2-5x+1}{8x^2+14x+5}
,$ is

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

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

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

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{(x-1)(x-1)(6x+1)}{(x-3)(x+2)(4x+5)}
text{, }xneleft{ -dfrac{5}{4},-2,-dfrac{1}{2},3 right}
.$

d) The factored form of the given expression, $dfrac{9y^2-4}{4y-12}divdfrac{9y^2+12y+4}{18-6y}
,$ is

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

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

$$
begin{align*}
&
dfrac{(cancel{3y+2})(3y-2)}{cancel2cdot2(cancel{y-3})}cdotdfrac{cancel2cdot(-3)(cancel{y-3})}{(cancel{3y+2})(3y+2)}
\\&=
dfrac{-3(3y-2)}{2(3y+2)}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{-3(3y-2)}{2(3y+2)}
text{, }yneleft{ -dfrac{2}{3}, 3 right}
.$

a) $dfrac{2(x+1)}{(x+2)(x+3)}
text{, }xne{ -3,-2,3}$

b) $dfrac{2(n-2)}{5}
text{, }nneleft{ -2,2,3,4 right}$

c) $dfrac{(x-1)(x-1)(6x+1)}{(x-3)(x+2)(4x+5)}
text{, }xneleft{ -dfrac{5}{4},-2,-dfrac{1}{2},3 right}$

d) $dfrac{-3(3y-2)}{2(3y+2)}
text{, }yneleft{ -dfrac{2}{3}, 3 right}$

a) The factored form of the given expression, $dfrac{x^2-5xy+4y^2}{x^2+3xy-28y^2}timesdfrac{x^2+2xy+y^2}{x^2-y^2}
,$ is

$$
begin{align*}
&
dfrac{(x-4y)(x-y)}{(x+7y)(x-4y)}cdotdfrac{(x+y)(x+y)}{(x+y)(x-y)}
.end{align*}
$$

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

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

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{x+y}{x+7y}
text{, }xne-7ytext{, }xne4ytext{, }xne ytext{, }xne-y
.$

b) The factored form of the given expression, $dfrac{2a^2-12ab+18b^2}{a^2-7ab+10b^2}divdfrac{4a^2-12ab}{a^2-7ab+10b^2}
,$ is

$$
begin{align*}
&
dfrac{2a^2-12ab+18b^2}{a^2-7ab+10b^2}cdotdfrac{a^2-7ab+10b^2}{4a^2-12ab}
\\&=
dfrac{2(a^2-6ab+9b^2)}{(a-5b)(a-2b)}cdotdfrac{(a-2b)(a-5b)}{4a(a-3b)}
\\&=
dfrac{2(a-3b)(a-3b)}{(a-5b)(a-2b)}cdotdfrac{(a-2b)(a-5b)}{4a(a-3b)}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel2(cancel{a-3b})(a-3b)}{cancel{(a-5b)(a-2b)}}cdotdfrac{cancel{(a-5b)(a-2b)}}{cancel2cdot2a(cancel{a-3b})}
\\&=
dfrac{a-3b}{2a}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{a-3b}{2a}
text{, }ane0text{, }ane5btext{, }ane2btext{, }ane3b
.$

c) The factored form of the given expression, $dfrac{10x^2+3xy-y^2}{9x^2-y^2}divdfrac{6x^2+3xy}{12x+4y}
,$ is

$$
begin{align*}
&
dfrac{10x^2+3xy-y^2}{9x^2-y^2}cdotdfrac{12x+4y}{6x^2+3xy}
\\&=
dfrac{(5x-y)(2x+y)}{(3x+y)(3x-y)}cdotdfrac{4(3x+y)}{3x(2x+y)}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{(5x-y)(cancel{2x+y})}{(cancel{3x+y})(3x-y)}cdotdfrac{4(cancel{3x+y})}{3x(cancel{2x+y})}
\\&=
dfrac{4(5x-y)}{3x(3x-y)}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{4(5x-y)}{3x(3x-y)}
text{, }xneleft{ -dfrac{y}{2},-dfrac{y}{3}, 0,dfrac{y}{3} right}
.$

d) The factored form of the given expression, $dfrac{15m^2+mn-2n^2}{2n-14m}timesdfrac{7m^2-8mn+n^2}{5m^2+7mn+2n^2}
,$ is

$$
begin{align*}
&
dfrac{(3m-n)(5m+2n)}{2(n-7m)}timesdfrac{(7m-n)(m-n)}{(5m+2n)(m+n)}
\\&=
dfrac{(3m-n)(5m+2n)}{-2(7m-n)}timesdfrac{(7m-n)(m-n)}{(5m+2n)(m+n)}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{(3m-n)(cancel{5m+2n})}{-2(cancel{7m-n})}timesdfrac{(cancel{7m-n})(m-n)}{(cancel{5m+2n})(m+n)}
\\&=
dfrac{(3m-n)(m-n)}{-2(m+n)}
\\&=
-dfrac{(3m-n)(m-n)}{2(m+n)}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $-dfrac{(3m-n)(m-n)}{2(m+n)}
text{, }mneleft{ -dfrac{2n}{5},-n,dfrac{n}{7} right}
.$

a) $dfrac{x+y}{x+7y}
text{, }xne-7ytext{, }xne4ytext{, }xne ytext{, }xne-y$

b) $dfrac{a-3b}{2a}
text{, }ane0text{, }ane5btext{, }ane2btext{, }ane3b$

c) $dfrac{4(5x-y)}{3x(3x-y)}
text{, }xneleft{ -dfrac{y}{2},-dfrac{y}{3}, 0,dfrac{y}{3} right}$

d) $-dfrac{(3m-n)(m-n)}{2(m+n)}
text{, }mneleft{ -dfrac{2n}{5},-n,dfrac{n}{7} right}$

The factored form of the given expression, $dfrac{x^2+x-6}{(2x-1)^2}timesdfrac{x(2x-1)^2}{x^2+2x-3}divdfrac{x^2-4}{3x}
,$ is

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

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

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

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{3x^2}{(x-1)(x+2)}
text{, }xneleft{ -3,-2,dfrac{1}{2},1,2 right}
.$

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

Using $A=dfrac{bh}{2}
,$ or the formula for the area of a triangle, with a height of $dfrac{5x-35}{x+3}$ and a base of $dfrac{4x^2}{x^2-16x+63}
,$ then the area is

$$
begin{align*}
A&=dfrac{left(dfrac{4x^2}{x^2-16x+63}right)left(dfrac{5x-35}{x+3}right)}{2}
.end{align*}
$$

The factored form of the expression above is

$$
begin{align*}
A&=left(dfrac{4x^2}{(x-9)(x-7)}right)left(dfrac{5(x-7)}{x+3}right)div2
\\
A&=left(dfrac{4x^2}{(x-9)(x-7)}right)left(dfrac{5(x-7)}{x+3}right)left(dfrac{1}{2}right)
.end{align*}
$$

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

$$
begin{align*}
A&=left(dfrac{cancel{2}(2)x^2}{(x-9)(cancel{x-7})}right)left(dfrac{5(cancel{x-7})}{x+3}right)left(dfrac{1}{cancel{2}}right)
\\
A&=dfrac{10x^2}{(x-9)(x+3)}
.end{align*}
$$

Since the denominator of the original equation cannot be zero, then the area of the triangle, $A,$ with its restriction is $A=dfrac{10x^2}{(x-9)(x+3)}
text{, }xneleft{ -3,7,9 right}
.$

$$
A=dfrac{10x^2}{(x-9)(x+3)}
text{, }xneleft{ -3,7,9 right}
$$

Using $rho=dfrac{m}{v}
,$ or the formula for density, with a mass of $dfrac{p+1}{3p+1}$ and a density of $dfrac{p^2-1}{9p^2+6p+1}
,$ then the volume is

$$
begin{align*}
dfrac{p^2-1}{9p^2+6p+1}&=dfrac{dfrac{p+1}{3p+1}}{v}
\\
vqty(dfrac{p^2-1}{9p^2+6p+1})&=qty(dfrac{dfrac{p+1}{3p+1}}{v})v
\\
vqty(dfrac{p^2-1}{9p^2+6p+1})&=dfrac{p+1}{3p+1}
\\
v&=dfrac{frac{p+1}{3p+1}}{frac{p^2-1}{9p^2+6p+1}}
\\&=
dfrac{p+1}{3p+1}divdfrac{p^2-1}{9p^2+6p+1}
\\&=
dfrac{p+1}{3p+1}cdotdfrac{9p^2+6p+1}{p^2-1}
\\&=
dfrac{p+1}{3p+1}cdotdfrac{(3p+1)(3p+1)}{(p+1)(p-1)}
.end{align*}
$$

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

$$
begin{align*}
v&=dfrac{p+1}{1}cdotdfrac{(1)(3p+1)}{(p+1)(p-1)}
&qty(text{cancel $3p+1$})
\\&=
dfrac{1}{1}cdotdfrac{(1)(3p+1)}{(1)(p-1)}
&qty(text{cancel $p+1$})
\\&=
dfrac{3p+1}{p-1}
.end{align*}
$$

Since the denominators of both the original and simplified equations cannot be zero, then the volume, $v,$ with its restriction is $v=dfrac{3p+1}{p-1}
text{, }pneleft{ -1,-dfrac{1}{3} ,1right}
.$

$$
v=dfrac{3p+1}{p-1}
text{, }pneleft{ -1,-dfrac{1}{3} ,1right}
$$

The error is in the third step when the expression $x^2-y^2$ was divided by $(x-y).$ This is not allowed because the value of $(x-y)$ is zero which will cause division by zero, and hence, will make the expression undefined.

The given expression, $dfrac{dfrac{m^2-mn}{6m^2+11mn+3n^2}divdfrac{m^2-n^2}{2m^2-mn-6n^2}}{dfrac{4m^2-7mn-2n^2}{3m^2+7mn+2n^2}}
,$ is equivalent to

$$
begin{align*}
&
dfrac{m^2-mn}{6m^2+11mn+3n^2}divdfrac{m^2-n^2}{2m^2-mn-6n^2}divdfrac{4m^2-7mn-2n^2}{3m^2+7mn+2n^2}
\\&=
dfrac{m^2-mn}{6m^2+11mn+3n^2}cdotdfrac{2m^2-mn-6n^2}{m^2-n^2}divdfrac{4m^2-7mn-2n^2}{3m^2+7mn+2n^2}
\\&=
dfrac{m^2-mn}{6m^2+11mn+3n^2}cdotdfrac{2m^2-mn-6n^2}{m^2-n^2}cdotdfrac{3m^2+7mn+2n^2}{4m^2-7mn-2n^2}
.end{align*}
$$

The factored form of the expression above is

$$
begin{align*}
&
dfrac{m(m-n)}{(3m+n)(2m+3n)}cdotdfrac{(2m+3n)(m-2n)}{(m+n)(m-n)}cdotdfrac{(3m+n)(m+2n)}{(4m+n)(m-2n)}
.end{align*}
$$

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

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

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

$$
dfrac{m(m+2n)}{(m+n)(4m+n)}
text{, }mneleft{ -2n, -n,-dfrac{n}{4},-dfrac{n}{3},-dfrac{3n}{2},n,2n right}
$$

We are given the equation for Newton’s law of gravitation

$$
begin{equation*} F_g=Gdfrac{m_1m_2}{r^2} end{equation*}
$$

$$
begin{align*} text{ } m_1,;m_2 &= text{mass; of; two; objects}\
G &= text{gravitational; constant}\
r &= text{distance between } m_1 text{ and } m_2
end{align*}
$$

The mass of Mercury $m_M$ is 2.2 times greater than the mass of Pluto $m_P$ which implies

$$
begin{equation*} m_M=2.2m_Pend{equation*}
$$

Meanwhile, Pluto’s distance from the sun $r_P$ is 102.1 times as much as that of Mercury $r_M$ which implies

$$
begin{equation*} r_P=102.1 r_Mend{equation*}
$$

We would like to find the ratio of Sun’s gravitational forces of Mercury to that of Pluto, that is, $dfrac{F_{g,M}}{F_{g,P}}$

Let $m_S$ be the mass of the sun.

The gravitational force on Mercury is
$$
begin{equation*}F_{g,M}=dfrac{Gm_Sm_M}{r_M^2}end{equation*}
$$

Since $m_P=dfrac{m_M}{2.2}$ and $r_P=102.1r_M$, the gravitational force on Pluto is

$$
begin{equation*}F_{g,P}=dfrac{Gm_Sm_P}{r_P^2}=dfrac{Gm_S(m_M/2.2)}{(102.1r_M)^2}=dfrac{1/2.2}{102.1^2}dfrac{Gm_Sm_M}{r_M^2}end{equation*}
$$

Therefore

$$
begin{equation*}F_{g,P}=dfrac{1/2.2}{102.1^2}F_{g,M}implies dfrac{F_{g,M}}{F_{g,P}}=dfrac{102.1^2}{1/2.2}=22933.7end{equation*}
$$