AB: Brüche und Variablen (Basis)

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)

\( \frac{1}{2} + \frac{x}{2} = \frac{5}{2} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{2} + \frac{x}{2} = \frac{5}{2} \Rightarrow x = 4 \)

b)

\( \frac{1}{3} + \frac{x}{6} = \frac{2}{3} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{3} + \frac{x}{6} = \frac{2}{3} \Rightarrow x = 2 \)

c)

\( \frac{1}{4} + \frac{x}{5} = \frac{9}{20} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{4} + \frac{x}{5} = \frac{9}{20} \Rightarrow x = 1 \)

d)

\( \frac{1}{20} + \frac{x}{4} = \frac{6}{20} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{20} + \frac{x}{4} = \frac{6}{20} \Rightarrow x = 1 \)

e)

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

f)

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

g)

\( \frac{1}{7} + \frac{14}{x} = \frac{6}{14} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{7} + \frac{14}{x} = \frac{6}{14} \Rightarrow x = 49 \)

h)

\( \frac{3}{6} + \frac{6}{x} = \frac{15}{12} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{3}{6} + \frac{6}{x} = \frac{15}{12} \Rightarrow x = 8 \)

i)

\( \frac{x}{4} + \frac{1}{12} = \frac{48}{36} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{4} + \frac{1}{12} = \frac{48}{36} \Rightarrow x = 5 \)

j)

\( \frac{x}{7} + \frac{10}{14} = \frac{8}{7} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{7} + \frac{10}{14} = \frac{8}{7} \Rightarrow x = 3 \)

k)

\( \frac{x}{39} + \frac{5}{78} = \frac{18}{156} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{39} + \frac{5}{78} = \frac{18}{156} \Rightarrow x = 2 \)

l)

\( \frac{1}{20} + \frac{x}{80} = \frac{30}{160} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{1}{20} + \frac{x}{80} = \frac{30}{160} \Rightarrow x = 11 \)

2.

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

a)

\( \frac{7}{2} - \frac{x}{2} = \frac{3}{2} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{7}{2} - \frac{x}{2} = \frac{3}{2} \Rightarrow x = 4 \)

b)

\( \frac{5}{3} - \frac{x}{6} = \frac{4}{3} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{5}{3} - \frac{x}{6} = \frac{4}{3} \Rightarrow x = 2 \)

c)

\( \frac{30}{6} - \frac{x}{3} = \frac{10}{6} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{30}{6} - \frac{x}{3} = \frac{10}{6} \Rightarrow x = 10 \)

d)

\( \frac{5}{2} - \frac{x}{4} = \frac{1}{2} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{5}{2} - \frac{x}{4} = \frac{1}{2} \Rightarrow x = 8 \)

e)

\( \frac{7}{3} - \frac{2}{x} = \frac{19}{9} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{7}{3} - \frac{2}{x} = \frac{19}{9} \Rightarrow x = 9 \)

f)

\( \frac{15}{6} - \frac{5}{x} = \frac{15}{9} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{15}{6} - \frac{5}{x} = \frac{15}{9} \Rightarrow x = 6 \)

g)

\( \frac{6}{14} - \frac{14}{x} = \frac{1}{7} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{6}{14} - \frac{14}{x} = \frac{1}{7} \Rightarrow x = 49 \)

h)

\( \frac{18}{5} - \frac{37}{x} = \frac{7}{4} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{18}{5} - \frac{37}{x} = \frac{7}{4} \Rightarrow x = 20 \)

i)

\( \frac{x}{4} - \frac{1}{12} = \frac{6}{36} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{4} - \frac{1}{12} = \frac{6}{36} \Rightarrow x = 1 \)

j)

\( \frac{x}{50} - \frac{10}{125} = \frac{5}{250} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{50} - \frac{10}{125} = \frac{5}{250} \Rightarrow x = 5 \)

k)

\( \frac{x}{39} - \frac{5}{78} = \frac{18}{156} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{x}{39} - \frac{5}{78} = \frac{18}{156} \Rightarrow x = 7 \)

l)

\( \frac{30}{200} - \frac{x}{50} = \frac{1}{20} \Rightarrow x = \textcolor{#DDD}{\left\lbrack \quad \right\rbrack} \) \( \frac{30}{200} - \frac{x}{50} = \frac{1}{20} \Rightarrow x = 5 \)

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