1

a

b

$2 \times 7 = 14$ stukjes; $\frac{3}{7} + \frac{1}{2} = \frac{6}{14} + \frac{7}{14} = \frac{13}{14}$

2

a

$\frac{a}{3} + \frac{a}{9} = \frac{3 a}{9} + \frac{a}{9} = \frac{4 a}{9}$	$\frac{3}{a} + \frac{9}{a} = \frac{12}{a}$
$\frac{5 a}{6} + \frac{a}{9} = \frac{15 a}{18} + \frac{2 a}{18} = \frac{17 a}{18}$	$\frac{3}{a} + \frac{2}{5} = \frac{15}{5 a} + \frac{2 a}{5 a} = \frac{15 + 2 a}{5 a}$
$\frac{a}{3} - \frac{2 a}{9} = \frac{3 a}{9} - \frac{2 a}{9} = \frac{a}{9}$	$\frac{3 a}{a} - \frac{1}{2} = 3 - \frac{1}{2} = 2 \frac{1}{2}$
$\frac{4 a}{3} - \frac{5 a}{9} = \frac{12 a}{9} - \frac{5 a}{9} = \frac{7 a}{9}$	$\frac{2}{7} - \frac{1}{a} = \frac{2 a}{7 a} - \frac{7}{7 a} = \frac{2 a - 7}{7 a}$

b

$\frac{2}{3} + \frac{1}{12} = \frac{8}{12} + \frac{1}{12} = \frac{9}{12} = \frac{3}{4}$	$\frac{p}{5} + \frac{1}{2} = \frac{2 p}{10} + \frac{5}{10} = \frac{2 p + 5}{10}$
$\frac{3}{7} - \frac{1}{4} = \frac{12}{28} - \frac{7}{28} = \frac{5}{28}$	$\frac{1}{5} - \frac{1}{7} = \frac{7}{35} - \frac{5}{35} = \frac{2}{35}$
$\frac{1}{2} - \frac{1}{4} + \frac{1}{5} = \frac{10}{20} - \frac{5}{20} + \frac{4}{20} = \frac{9}{20}$	$\frac{4}{5} - \frac{3}{7} = \frac{28}{35} - \frac{15}{35} = \frac{13}{35}$
$\frac{2}{5} + \frac{1}{2} = \frac{4}{10} + \frac{5}{10} = \frac{9}{10}$	$\frac{3}{5} - \frac{q}{7} = \frac{21}{35} - \frac{5 q}{35} = \frac{21 - 5 q}{35}$
$\frac{4}{5} + \frac{1}{2} = \frac{8}{10} + \frac{5}{10} = \frac{13}{10} = 1 \frac{3}{10}$	$\frac{p}{2} + \frac{q}{3} = \frac{3 p}{6} + \frac{2 q}{6} = \frac{3 p + 2 q}{6}$

c

$\frac{3}{p} + \frac{1}{3} = \frac{9}{3 p} + \frac{p}{3 p} = \frac{p + 9}{3 p}$	$\frac{4}{5} - \frac{2}{p} = \frac{4 p}{5 p} - \frac{10}{5 p} = \frac{4 p - 10}{5 p}$
$\frac{p}{2} + \frac{2}{p} = \frac{p^{2}}{2 p} + \frac{4}{2 p} = \frac{p^{2} + 4}{2 p}$	$p + \frac{1}{p} = \frac{p}{1} + \frac{1}{p} = \frac{p^{2}}{p} + \frac{1}{p} = \frac{p^{2} + 1}{p}$
$\frac{1}{p} + \frac{1}{q} = \frac{q}{p q} + \frac{p}{p q} = \frac{p + q}{p q}$	$\frac{2}{p} - \frac{3}{q} = \frac{2 q}{p q} + \frac{3 p}{p q} = \frac{3 p + 2 q}{p q}$
$\frac{q}{p} + \frac{p}{q} = \frac{q^{2}}{p q} + \frac{p^{2}}{p q} = \frac{p^{2} + q^{2}}{p q}$	$q - \frac{1}{p} = \frac{q}{1} - \frac{1}{p} = \frac{p q}{p} - \frac{1}{p} = \frac{p q - 1}{p}$

3

a

b

$\frac{24}{180} = \frac{2}{15}$ deel door de IJssel

c

$\frac{36}{180} = \frac{1}{5}$ deel door de Nederrijn

d

$\frac{2}{3} + \frac{2}{15} + \frac{1}{5} = 1$ samen

4

$\frac{3}{10}$	$\frac{6}{5} = 1 \frac{1}{5}$
$\frac{2}{10} = \frac{1}{5}$	$\frac{8}{5} = 1 \frac{3}{5}$
$\frac{a}{10}$	$\frac{2 a}{5}$ (of $\frac{2}{5} a$ )
$\frac{10 a}{5 a} = 2$	$\frac{20 a}{4 a} = 5$

5

$\frac{3 \cdot 2 \cdot 1}{4 \cdot 5 \cdot 2} = \frac{6}{40} = \frac{3}{20}$	$\frac{a}{5} \cdot \frac{a}{3} \cdot \frac{3}{2} = \frac{a \cdot a \cdot 3}{5 \cdot 3 \cdot 2} = \frac{3 a^{2}}{30} = \frac{a^{2}}{10}$
$\frac{3}{7} \cdot \frac{a}{6} \cdot \frac{6}{10} = \frac{18 a}{420} = \frac{3 a}{70}$	$\frac{3}{2} \cdot \frac{8}{3} \cdot \frac{41}{10} = \frac{3 \cdot 8 \cdot 41}{2 \cdot 3 \cdot 10} = \frac{984}{60} = 16 \frac{2}{5}$
$\frac{a \cdot b \cdot c}{b \cdot c \cdot a} = \frac{a b c}{a b c} = 1$	$\frac{11}{4} \cdot \frac{2}{5} \cdot \frac{4}{3} = \frac{11 \cdot 2 \cdot 3}{4 \cdot 5 \cdot 3} = \frac{88}{60} = 1 \frac{7}{15}$
$\frac{1 \cdot 2 \cdot a \cdot 4 b}{2 \cdot b \cdot 4 \cdot 5} = = \frac{8 a b}{40 b} = \frac{a}{5}$	$\frac{15}{4} \cdot \frac{2 a}{5} \cdot \frac{6}{b} = \frac{15 \cdot 2 a \cdot 6}{4 \cdot 5 \cdot b} = \frac{180 a}{20 b} = \frac{9 a}{b}$

6

a

10 potten, want $10 \cdot \frac{1}{2} = 5$ liter.

b

aantal liters in kleine pot	$\frac{1}{2}$	$\frac{1}{3}$	$\frac{2}{3}$	$1 \frac{1}{4}$
aantal potten	10	15	$7 \frac{1}{2}$	4

c

9 ; 8 ; 2 ; $\frac{1}{4}$

7

a

4 potten, $3 \frac{1}{3} : \frac{5}{6} = 4$

b

$2 \frac{1}{10}$ ; $11 \frac{1}{5}$ ; 2 ; $1 \frac{7}{8}$

8

a

5 ; $\frac{1}{3}$ ; $\frac{2}{3}$ ; $\frac{3}{10}$ ; $\frac{2}{9}$

b

1

c

1 ; $\frac{1}{a}$

9

$16 \cdot \frac{3}{2} = \frac{48}{2} = 24$	$28 \cdot \frac{3}{1} = \frac{84}{1} = 84$
$16 \cdot \frac{7}{4} = \frac{112}{4} = 28$	$28 \cdot \frac{3}{4} = \frac{84}{4} = 21$
$\frac{2}{3} \cdot \frac{7}{4} = \frac{14}{12} = 1 \frac{1}{6}$	$28 \cdot \frac{3}{8} = \frac{84}{8} = 10 \frac{1}{2}$
$\frac{4}{7} \cdot \frac{3}{2} = \frac{12}{14} = \frac{6}{7}$	$28 \cdot \frac{10}{3} = \frac{280}{3} = 93 \frac{1}{3}$

10

$\frac{a}{3} + \frac{a}{4} = \frac{4 a}{12} + \frac{3 a}{12} = \frac{7 a}{12} (= \frac{7}{12} a)$	$\frac{a}{3} \cdot \frac{a}{4} = \frac{a^{2}}{12} (= \frac{1}{12} a^{2})$
$\frac{a}{3} : \frac{a}{4} = \frac{a}{3} \cdot \frac{4}{a} = \frac{4 a}{3 a} = \frac{4}{3} = 1 \frac{1}{3}$	$1 + \frac{4}{a} = \frac{a}{a} + \frac{4}{a} = \frac{a + 4}{a}$
$a - \frac{4}{a} = \frac{a^{2}}{a} - \frac{4}{a} = \frac{a^{2} - 4}{a}$	$a \cdot \frac{4}{a} = \frac{4 a}{a} = 4$
$a : \frac{4}{a} = a \cdot \frac{a}{4} = \frac{a}{1} \cdot \frac{a}{4} = \frac{a^{2}}{4} (= \frac{1}{4} a^{2})$	${(\frac{a}{3})}^{2} = \frac{a}{3} \cdot \frac{a}{3} = \frac{a^{2}}{9} (= \frac{1}{9} a^{2})$
$12 : (3 : a) = 12 : \frac{3}{a} = 12 \cdot \frac{a}{3} = \frac{12 a}{3} = 4 a$	$12 : (a : 3) = 12 : \frac{a}{3} = 12 \cdot \frac{3}{a} = \frac{36}{a}$
$\frac{6}{\frac{2}{a}} = 6 : \frac{2}{a} = 6 \cdot \frac{a}{2} = \frac{6 a}{2} = 3 a$	$1 : (a + \frac{1}{a}) = 1 : (\frac{a^{2} + 1}{a}) = 1 \cdot \frac{a}{a^{2} + 1} = \frac{a}{a^{2} + 1}$

11

a

Omdat de rechthoeken gelijkvormig zijn, is de verhoudingen korte zijde : lange zijde voor beide rechthoeken gelijk.
Dat geeft $2 : p = p : 8$ ofwel $\frac{2}{p} = \frac{p}{8}$ .

b

$\frac{q}{5}$	$=$	$\frac{4}{2 q}$	MAAL $5$
$q$	$=$	$\frac{20}{2 q} = \frac{10}{q}$	MAAL $q$
$q^{2}$	$=$	$10$	wortel nemen
$q = \sqrt{10} \approx 3,16$		( $q = ‐ \sqrt{10}$ voldoet niet)

12

$\frac{3}{x} = \frac{5}{x + 1} \to 3 \cdot (x + 1) = 5 \cdot x \to 3 x + 3 = 5 x \to 3 = 2 x \to x = 1 \frac{1}{2}$
$\frac{y}{y + 3} = \frac{3}{4} \to 3 \cdot (y + 3) = 4 \cdot y \to 3 y + 9 = 4 y \to y = 9$