 # Fraction Rules Solutions

Solutions to the fraction rules question are presented with steps and explanations.

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1. )
$\quad \dfrac{0}{3} + \dfrac{1}{3} = \dfrac{0+1}{3} = \dfrac{1}{3}$

2. )
$\quad \dfrac{2}{0} + 5$ undefined because the fraction $\dfrac{2}{0}$ has a denominator equal to zero and is therefore undefined.

3. )
given:   $\dfrac{3}{5} + 2$
Rewrite the integer $2$ as a fraction with denominator $5$
$\quad \dfrac{3}{5} + 2 = \dfrac{3}{5} + 2 \times \dfrac{5}{5}$
$\quad = \dfrac{3 + 2 \times 5}{5} = \dfrac{13}{5}$

4. )
given:   $\dfrac{3}{2} + 2.1$
Rewrite the decimal number $2.1$ as a fraction
$\quad \dfrac{3}{2} + 2.1 = \dfrac{3}{2} + \dfrac{2.1}{1}$
$\quad = \dfrac{3}{2} + \dfrac{2.1 \times 10}{1 \times 10}$
$\quad = \dfrac{3}{2} + \dfrac{21}{10}$
Rewrite the two fractions with a common denominator
$\quad = \dfrac{3}{2} \times \dfrac{10}{10} + \dfrac{21}{10} \times \dfrac{2}{2}$
Multiply fractions and simplify
$\quad = \dfrac{30}{20} + \dfrac{42}{20}$
$\quad = \dfrac{30 + 42}{20} = \dfrac{72}{20}$
Which may be reduced to
$\quad = \dfrac{18}{5}$

5. )
given:   $0.1 x + \dfrac{2x}{3}$
Rewrite the term $0.1 x$ as a fraction
$\quad 0.1 x + \dfrac{2x}{3} = \dfrac{1}{10} x + \dfrac{2x}{3} = \dfrac{x}{10} + \dfrac{2x}{3}$
Rewrite the fractions with common denominator
$\quad = \dfrac{x}{10} \times \dfrac{3}{3} + \dfrac{2x}{3} \times \dfrac{10}{10}$
Simplify
$\quad = \dfrac{3 x}{30} + \dfrac{20 x}{30}$
$\quad = \dfrac{3 x + 20 x}{30} = \dfrac{23 x}{30}$

6. )
given:   $3 x + \dfrac{x}{4}$
Rewrite the fractions with common denominator
$\quad 3 x + \dfrac{x}{4} = 3 x \times \dfrac{4}{4} + \dfrac{x}{4}$
Simplify
$\quad = \dfrac{12 x}{4} + \dfrac{x}{4}$
$\quad = \dfrac{12x + x}{4} = \dfrac{13 x }{4}$

7. )
given:   $3x - \dfrac{5 x}{4}$
Rewrite the fractions with common denominator
$\quad 3x - \dfrac{5 x}{4} = 3 x \times \dfrac{4}{4} - \dfrac{5 x}{4}$
Simplify
$\quad = \dfrac{12 x}{4} - \dfrac{5 x}{4}$
Subtract the fractions
$\quad = \dfrac{12x - 5 x}{4} = \dfrac{7 x }{4}$

8. )
given:   $\dfrac{3}{5} \times \dfrac{4}{9}$
Apply multiplication rule of fractions
$\quad \dfrac{3}{5} \times \dfrac{4}{9} = \dfrac{3 \times 4}{5 \times 9} = \dfrac{3 \times 4}{5 \times 3 \times 3}$
The numerator and denominator have a common factor $3$ and therefore the fraction may be reduced by dividing numerator and denominator by $3$
$\quad = \dfrac{ (3 \times 4) \div 3 }{ (5 \times 9) \div 3}$
Simplify
$\quad = \dfrac{ 4 }{ 15 }$

9. )
given:   $6 \times \dfrac{3}{7}$
Rewrite $6$ as a fraction
$\quad 6 \times \dfrac{3}{7} = \dfrac{6}{1} \times \dfrac{3}{7}$
Apply multiplication rule of fractions
$\quad = \dfrac{6 \times 3}{ 1 \times 7 }$
Simplify
$\quad = \dfrac{18}{7}$

10. )
given:   $\dfrac{2x}{5} \times \dfrac{1}{2}$
Apply rule of multiplication of fractions
$\quad \dfrac{2x}{5} \times \dfrac{1}{2} = \dfrac{2x \times 1}{5 \times 2}$
The numerator and denominator have a common factor $2$ and the fraction may then be reduced by dividing numerator and denominator by $2$
$\quad = \dfrac{(2x \times 1) \div 2}{(5 \times 2)\div 2}$
Simplify
$\quad = \dfrac{x}{5}$

11. )
given:   $\dfrac{6}{7} \div 3$
Rewrite $3$ as a fraction
$\quad \dfrac{6}{7} \div 3 = \dfrac{6}{7} \div \dfrac{3}{1}$
Apply division rule of fraction (change into a multiplication of the first fraction by the reciprocal of the second fraction)
$\quad = \dfrac{6}{7} \times \dfrac{1}{3}$
Apply the multiplication rule of fractions
$\quad = \dfrac{6 \times 1}{7 \times 3}$
Factor the numerator
$\quad = \dfrac{3 \times 2 \times 1}{7 \times 3}$
Divide numerator and denominator by the common factor $3$ to reduce the fraction
$\quad = \dfrac{ (3 \times 2 \times 1) \div 3}{ (7 \times 3) \div 3}$
Simplify
$\quad = \dfrac{2}{7}$

12. )
given:   $x \div \dfrac{1}{9}$
Use division rule of fractions
$\quad x \div \dfrac{1}{9} = x \times \dfrac{9}{1}$
Rewrite $x$ as a fraction
$\quad = \dfrac{x}{1} \times \dfrac{9}{1}$
Apply the multiplication rule of fractions
$\quad = \dfrac{x \times 9}{1 \times 1}$
Simplify
$\quad = \dfrac{9 x}{1}$
Simplify fraction with denominator equal to 1
$\quad = 9x$

13. )
given:   $\dfrac{2x}{5} \div \dfrac{1}{9}$
Use division rule of fractions
$\quad \dfrac{2x}{5} \div \dfrac{1}{9} = \dfrac{2x}{5} \times \dfrac{9}{1}$
Apply the multiplication rule of fractions
$\quad = \dfrac{2x \times 9}{5 \times 1}$
Simplify
$\quad = \dfrac{18 x}{5}$

14. )
given:   $- \dfrac{-3}{5} + \dfrac{-3}{5}$
Apply rule of signs to rewrite the fraction $\dfrac{-3}{5}$ included in the given expression as $- \dfrac{3}{5}$
$\quad - \dfrac{-3}{5} + \dfrac{-3}{5} = - ( - \dfrac{3}{5}) + \dfrac{-3}{5}$
Simplify
$\quad = \dfrac{3}{5} + \dfrac{-3}{5}$
$\quad = \dfrac{3 +(- 3)}{5} = \dfrac{0}{5} = 0$

15. )
given:   $\dfrac{2}{-9} + \dfrac{7}{9}$
Apply rule of signs to rewrite the fraction $\dfrac{2}{-9}$ included in the given expression as $\dfrac{-2}{9}$
$\quad \dfrac{2}{-9} + \dfrac{7}{9} = \dfrac{- 2}{9} + \dfrac{7}{9}$
$\quad = \dfrac{-2 + 7}{9} = \dfrac {5}{9}$

16. )
given:   $\dfrac{-5}{-2} - \dfrac{7}{2}$
Apply rule of signs to rewrite the fraction $\dfrac{-5}{-2}$ included in the given expression as $\dfrac{5}{2}$
$\quad \dfrac{-5}{-2} - \dfrac{7}{2} = \dfrac{5}{2} - \dfrac{7}{2}$
Subtract the fractions
$\quad = \dfrac{5 - 7}{2}$
Apply rule of signs to simplify
$\quad = \dfrac{-2}{2} = - \dfrac{2}{2} = -1$

17. )
given:   $\dfrac{-2x}{3} - \dfrac{ - 5x}{- 3}$
Apply rule of signs to rewrite the fraction $\dfrac{ - 5x}{- 3}$ included in the given expression as $\dfrac{ 5x}{3}$
$\quad \dfrac{-2x}{3} - \dfrac{ - 5x}{- 3} = \dfrac{-2x}{3} - \dfrac{ 5x}{3}$
Subtract
$\quad = \dfrac{-2x - 5x}{3}$
Simplify
$\quad = \dfrac{-7x}{3}$
Which may also be written as (using rule of signs)
$\quad = - \dfrac{7x}{3}$

18. )
given:   $2 - \dfrac{ 4 + \dfrac{1}{3}}{1+\dfrac{1}{2}}$

Rewrite $4$ in the numerator $4 + \dfrac{1}{3}$ as a fraction with denominator equal to $3$ and $1$ in the denominator $1+\dfrac{1}{2}$ as a fraction with denominator equal to $2$

$\quad 2 - \dfrac{ 4 + \dfrac{1}{3}}{1+\dfrac{1}{2}} = 2 - \dfrac{ 4 \times \dfrac{3}{3} + \dfrac{1}{3}}{1 \times \dfrac{2}{2} +\dfrac{1}{2}}$
Simplify
$\quad = 2 - \dfrac{ \dfrac{12}{3} + \dfrac{1}{3}} {\dfrac{2}{2} +\dfrac{1}{2}}$
$\quad = 2 - \dfrac{ \dfrac{13}{3}} {\dfrac{3}{2}}$
Use rule of division of fractions
$\quad = 2 - \dfrac{13}{3} \times {\dfrac{2}{3}}$
Multiply fractions
$\quad = 2 - \dfrac{26}{9}$
Rewrite $2$ as a fraction with denominator equal to $9$
$\quad = 2 \times \dfrac{9}{9} - \dfrac{26}{9}$
Simplify
$\quad = \dfrac{18}{9} - \dfrac{26}{9}$
Subtract and simplify
$\quad = \dfrac{18-26}{9} = \dfrac{-8}{9} = - \dfrac{8}{9}$

19. )
given:   $x - \dfrac{ 2x + \dfrac{x}{2}}{x - \dfrac{2x}{3}}$

Rewrite $2x$ in $2x + \dfrac{x}{2}$ as a fraction with denominator equal to $2$ and $x$ in $x - \dfrac{2x}{3}$ as a fraction with denominator equal to $3$

$\quad x - \dfrac{ 2x + \dfrac{x}{2}}{x - \dfrac{2x}{3}} = x - \dfrac{ 2x \dfrac {2}{2} + \dfrac{x}{2}}{x \dfrac{3}{3} - \dfrac{2x}{3}}$
Simplify
$\quad = x - \dfrac{ \dfrac {4x}{2} + \dfrac{x}{2}}{\dfrac{3 x}{3} - \dfrac{2x}{3}}$
Add and subtract fractions with common denominator

$\quad = x - \dfrac{ \dfrac {4x + x}{2} }{\dfrac{3 x - 2x}{3} }$
Simplify
$\quad = x - \dfrac{ \dfrac {5x}{2} }{\dfrac{x}{3} }$
Use rule of division of fractions
$\quad = x - \dfrac {5x}{2} \times \dfrac{3}{x}$
Simplify
$\quad = x - \dfrac {15 x}{2 x}$
Reduce the fraction $\dfrac {15 x}{2 x}$ dividing its numerator and denominator by $x$
$\quad = x - \dfrac {15 x \div x}{2 x \div x}$
Simplify
$\quad = x - \dfrac {15}{2}$
Rewrite $x$ as a fraction with denominator equal to $2$
$\quad = x \times \dfrac{2}{2} - \dfrac {15}{2}$
Simplify
$\quad = \dfrac{2x }{2} - \dfrac {15}{2}$
Subtract fractions
$\quad = \dfrac{2x - 15}{2}$

solutions to the above questions included.