AB: Brüche und Variablen (Erweitert)

Das Rechnen mit Variablen hast du hoffentlich schon einmal gesehen. Man schreibt an die Stelle, wo eine Zahl eingesetzt werden soll, einen Buchstaben. Meist verwendet man den Buchstaben x.

1.

Löse die folgenden Additions-Aufgaben, indem du berechnest, was für x eingesetzt werden soll:

a)

\( \frac{1}{2} + \frac{x}{2} = \frac{15}{2} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{2} + \frac{x}{2} = \frac{15}{2} \Rightarrow x = 14 \)

b)

\( \frac{1}{3} + \frac{x}{6} = \frac{4}{3} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{3} + \frac{x}{6} = \frac{4}{3} \Rightarrow x = 6 \)

c)

\( \frac{1}{4} + \frac{x}{5} = \frac{18}{40} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{4} + \frac{x}{5} = \frac{18}{40} \Rightarrow x = 1 \)

d)

\( \frac{1}{20} + \frac{x}{4} = \frac{22}{40} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{20} + \frac{x}{4} = \frac{22}{40} \Rightarrow x = 2 \)

e)

\( \frac{1}{5} + \frac{3}{x} = \frac{4}{10} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{5} + \frac{3}{x} = \frac{4}{10} \Rightarrow x = 15 \)

f)

\( \frac{1}{15} + \frac{3}{x} = \frac{2}{5} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{15} + \frac{3}{x} = \frac{2}{5} \Rightarrow x = 9 \)

g)

\( \frac{1}{17} + \frac{299}{x} = \frac{36}{14} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{17} + \frac{299}{x} = \frac{36}{14} \Rightarrow x = 119 \)

h)

\( \frac{2}{33} + \frac{33}{x} = \frac{1}{11} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{2}{33} + \frac{33}{x} = \frac{1}{11} \Rightarrow x = 1089 \)

i)

\( \frac{x}{8} + \frac{1}{24} = \frac{60}{36} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{8} + \frac{1}{24} = \frac{60}{36} \Rightarrow x = 13 \)

j)

\( \frac{x}{7} + \frac{100}{14} = \frac{80}{7} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{7} + \frac{100}{14} = \frac{80}{7} \Rightarrow x = 30 \)

k)

\( \frac{x}{168} + \frac{5}{84} = \frac{18}{56} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{168} + \frac{5}{84} = \frac{18}{56} \Rightarrow x = 44 \)

l)

\( \frac{1}{40} + \frac{x}{80} = \frac{30}{120} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{40} + \frac{x}{80} = \frac{30}{120} \Rightarrow x = 18 \)

2.

Löse die folgenden Subtraktions-Aufgaben, indem du berechnest, was für x eingesetzt werden soll:

a)

\( \frac{70}{2} - \frac{x}{2} = \frac{43}{2} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{70}{2} - \frac{x}{2} = \frac{43}{2} \Rightarrow x = 27 \)

b)

\( \frac{50}{3} - \frac{x}{6} = \frac{42}{3} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{50}{3} - \frac{x}{6} = \frac{42}{3} \Rightarrow x = 16 \)

c)

\( \frac{100}{6} - \frac{x}{3} = \frac{90}{6} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{100}{6} - \frac{x}{3} = \frac{90}{6} \Rightarrow x = 5 \)

d)

\( \frac{15}{2} - \frac{x}{4} = \frac{10}{2} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{15}{2} - \frac{x}{4} = \frac{10}{2} \Rightarrow x = 10 \)

e)

\( \frac{477}{3} - \frac{118}{x} = \frac{369}{9} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{477}{3} - \frac{118}{x} = \frac{369}{9} \Rightarrow x = 1 \)

f)

\( \frac{30}{6} - \frac{10}{x} = \frac{30}{9} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{30}{6} - \frac{10}{x} = \frac{30}{9} \Rightarrow x = 6 \)

g)

\( \frac{25}{8} - \frac{168}{x} = \frac{6}{12} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{25}{8} - \frac{168}{x} = \frac{6}{12} \Rightarrow x = 64 \)

h)

\( \frac{18}{50} - \frac{37}{x} = \frac{7}{40} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{18}{50} - \frac{37}{x} = \frac{7}{40} \Rightarrow x = 200 \)

i)

\( \frac{x}{3360} - \frac{1}{120} = \frac{6}{320} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{3360} - \frac{1}{120} = \frac{6}{320} \Rightarrow x = 91 \)

j)

\( \frac{x}{500} - \frac{100}{125} = \frac{50}{250} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{500} - \frac{100}{125} = \frac{50}{250} \Rightarrow x = 500 \)

k)

\( \frac{x}{156} - \frac{5}{39} = \frac{18}{312} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{156} - \frac{5}{39} = \frac{18}{312} \Rightarrow x = 29 \)

l)

\( \frac{30}{400} - \frac{x}{300} = \frac{1}{40} \Rightarrow x = \color{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{30}{400} - \frac{x}{300} = \frac{1}{40} \Rightarrow x = 15 \)

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